Really advanced techniques of integration (definite or indefinite) Okay, so everyone knows the usual methods of solving integrals, namely u-substitution, integration by parts, partial fractions, trig substitutions, and reduction formulas. But what else is there? Every time I search for "Advanced Techniques of Symbolic Integration" or "Super Advanced Integration Techniques", I get the same results which end up only talking about the methods mentioned above. Are there any super obscure and interesting techniques for solving integrals?

As an example of something that might be obscure, the formula for "general integration by parts " for $n$ functions $f_j, \ j = 1,\cdots,n$ is given by
$$
\int{f_1'(x)\prod_{j=2}^n{f_j(x)}dx} = \prod_{i=1}^n{f_i(x)} - \sum_{i=2}^n{\int{f_i'(x)\prod_{\substack{j=1 \\ j \neq i}}^n{f_j(x)}dx}}
$$
which is not necessarily useful nor difficult to derive, but is interesting nonetheless.

So out of curiosity, are there any crazy unknown symbolic integration techniques?

Here are a few. The first one is included because it's not very well known and is not general, though the ones that follow are very general and very useful.

A great but not very well known way to find the primitive of $f^{-1}$ in terms of  the primitive of $f$, $F$, is (very easy to prove: just differentiate both sides and use the chain rule):
$$
\int f^{-1}(x)\, dx = x \cdot f^{-1}(x)-(F \circ f^{-1})(x)+C.
$$

Examples:

  $$
\begin{aligned}
\displaystyle \int \arcsin(x)\, dx
&= x \cdot \arcsin(x)- (-\cos\circ \arcsin)(x)+C \\
&=x \cdot \arcsin(x)+\sqrt{1-x^2}+C.
\end{aligned}
$$

  $$
\begin{aligned}
\int \log(x)\, dx
&= x \cdot \log(x)-(\exp \circ \log)(x) + C \\
&= x \cdot \left( \log(x)-1 \right) + C.
\end{aligned}
$$

This one is more well known, and extremely powerful, it's called differentiating under the integral sign. It requires ingenuity most of the time to know when to apply, and how to apply it, but that only makes it more interesting. The technique uses the simple fact that
$$
\frac{\mathrm d}{\mathrm d x} \int_a^b f \left({x, y}\right) \mathrm d y = \int_a^b \frac{\partial f}{\partial x} \left({x, y}\right) \mathrm d y.
$$

Example:

  We want to calculate the integral $\int_{0}^{\infty} \frac{\sin(x)}{x} dx$. To do that, we unintuitively consider the more complicated integral $\int_{0}^{\infty} e^{-tx} \frac{\sin(x)}{x} dx$ instead.

  Let $$ I(t)=\int_{0}^{\infty} e^{-tx} \frac{\sin(x)}{x} dx,$$ then $$ I'(t)=-\int_{0}^{\infty} e^{-tx} \sin(x) dx=\frac{e^{-t x} (t \sin (x)+\cos (x))}{t^2+1}\bigg|_0^{\infty}=\frac{-1}{1+t^2}.$$

  Since both $I(t)$ and $-\arctan(t)$ are primitives of $\frac{-1}{1+t^2}$, they must differ only by a constant, so that $I(t)+\arctan(t)=C$. Let $t\to \infty$, then $I(t) \to 0$ and $-\arctan(t) \to -\pi/2$, and hence $C=\pi/2$, and $I(t)=\frac{\pi}{2}-\arctan(t)$.

  Finally,
  $$
\int_{0}^{\infty} \frac{\sin(x)}{x} dx = I(0) = \frac{\pi}{2}-\arctan(0) = \boxed{\frac{\pi}{2}}.
$$

This one is probably the most commonly used "advanced integration technique", and for good reasons. It's referred to as the "residue theorem" and it states that if $\gamma$ is a counterclockwise simple closed curve, then $\displaystyle \int_\gamma f(z) dz = 2\pi i \sum_{k=1}^n \operatorname{Res} ( f, a_k )$ . It will be difficult for you to understand this one without knowledge in complex analysis, but you can get the gist of it with the wiki article.
Example:

  We want to compute $\int_{-\infty}^{\infty} \frac{x^2}{1+x^4} dx$.  The poles of our function $f(z)=\frac{x^2}{1+x^4}$ in the upper half plane are  $a_1=e^{i \frac{\pi}{4}}$ and $a_2=e^{i \frac{3\pi}{4}}$. The residues of our function at those points are
  $$\operatorname{Res}(f,a_1)=\lim_{z\to a_1} (z-a_1)f(z)=\frac{e^{i \frac{-\pi}{4}}}{4},$$
  and
  $$\operatorname{Res}(f,a_2)=\lim_{z\to a_2} (z-a_2)f(z)=\frac{e^{i \frac{-3\pi}{4}}}{4}.$$
  Let $\gamma$ be the closed path around the boundary of the semicircle of radius $R>1$ on the upper half plane, traversed in the counter-clockwise direction. Then the residue theorem gives us ${1 \over 2\pi i} \int_\gamma f(z)\,dz=\operatorname{Res}(f,a_1)+\operatorname{Res}(f,a_2)={1 \over 4}\left({1-i \over \sqrt{2}}+{-1-i \over \sqrt{2}}\right)={-i \over 2 \sqrt{2}}$ and $ \int_\gamma f(z)\,dz= {\pi \over \sqrt{2}}$.
  Now, by the definition of $\gamma$, we have:
  $$\int_\gamma f(z)\,dz = \int_{-R}^R \frac{x^2}{1+x^4} dx + \int_0^\pi {i (R e^{it})^3 \over 1+(R e^{it})^4} dz = {\pi \over \sqrt{2}}.$$
  For the integral on the semicircle
  $$
\int_0^\pi {i (R e^{it})^3 \over 1+(R e^{it})^4} dz,
$$
  we have
  $$
\begin{aligned}
\left| \int_0^\pi {i (R e^{it})^3 \over 1+(R e^{it})^4} dz \right|
&\leq \int_0^\pi \left| {i (R e^{it})^3 \over 1+(R e^{it})^4} \right| dz \\
&\leq \int_0^\pi {R^3 \over R^4-1} dz={\pi R^3 \over R^4-1}.
\end{aligned}
$$
  Hence, as $R\to \infty$, we have ${\pi R^3 \over R^4-1} \to 0$, and hence $\int_0^\pi {i (R e^{it})^3 \over 1+(R e^{it})^4} dz \to 0$.
  Finally,
  $$
\begin{aligned}
\int_{-\infty}^\infty \frac{x^2}{1+x^4} dx
&= \lim_{R\to \infty} \int_{-R}^R \frac{x^2}{1+x^4} dx  \\
&= \lim_{R\to \infty} {\pi \over \sqrt{2}}-\int_0^\pi {i (R e^{it})^3 \over 1+(R e^{it})^4} dz =\boxed{{\pi \over \sqrt{2}}}.
\end{aligned}
$$

My final "technique" is the use of the mean value property for complex analytic functions, or Cauchy's integral formula in other words:
$$
\begin{aligned}
f(a)
&= \frac{1}{2\pi i} \int_\gamma \frac{f(z)}{z-a}\, dz \\
&= \frac{1}{2\pi} \int_{0}^{2\pi} f\left(a+e^{ix}\right) dx.
\end{aligned}
$$

Example:

  We want to compute the very messy looking integral $\int_0^{2\pi} \cos (\cos (x)+1) \cosh (\sin (x)) dx$. We first notice that
  $$
\begin{aligned}
&\hphantom{=} \cos [\cos (x)+1] \cosh [\sin (x)] \\
&=\Re\left\{
\cos [\cos (x)+1] \cosh [\sin (x)]
-i\sin [\cos (x)+1] \sinh [\sin (x)]
\right\} \\
&= \Re \left[ \cos \left( 1+e^{i x} \right) \right].
\end{aligned}
$$
  Then, we have
  $$
\begin{aligned}
\int_0^{2\pi} \cos [\cos (x)+1] \cosh [\sin (x)] dx
&= \int_0^{2\pi} \Re \left[ \cos \left( 1+e^{i x} \right) \right] dx \\
&= \Re \left[ \int_0^{2\pi} \cos \left( 1+e^{i x} \right) dx \right] \\
&= \Re \left( \cos(1) \cdot 2 \pi \right)= \boxed{2 \pi \cos(1)}.
\end{aligned}
$$

If we have $$u(x)=\prod_{i=1}^nf_i(x)$$
Then, through the product rule:
$$u'(x)=\sum_{i=1}^{n}\prod_{j=1}^{n}f_j^{(\delta_{ij})}(x)$$
Where $\delta_{ij}$ is the Kronecker delta, $f^{(0)}(x)=f(x)$, and $f^{(1)}(x)=f'(x)$.

Thus,
$$\int\sum_{i=1}^{n}\prod_{j=1}^{n}f_j^{(\delta_{ij})}(x)dx=\prod_{i=1}^nf_i(x)+C$$
It's really specific, and I haven't used it yet, but it's still a pretty neat formula (especially with that product notation).

Here's another. Assume $I=\{1,2,\dots,n\}$ for any $n\in\Bbb N$, and $\forall i,j\in I, i\neq j\iff \alpha_i\neq \alpha_j$
$$G=\int\prod_{i\in I}\frac{1}{x-\alpha_i}dx$$
Using partial fraction decomposition, we get $$G=\sum_{i\in I}\beta_{ij}\int\frac{dx}{x-\alpha_i}$$
Where $$\beta_{ij}=\prod_{i\neq j\in I}\frac{1}{\alpha_i-\alpha_j}$$
Then of course, $$G=\sum_{i\in I}\beta_{ij}\ln|x-\alpha_i|$$
Which is also pretty fun to look at (but is also little more useful).


If $\{x\}$ is an open set in $X$, for all $x\in X$, then all subsets of $X$ is open in $X$ Today for example the teacher ask us to nejdeme next example, but none of us knew, so we left for example homework, but try again but I can not solve the example, so please someone help me, the example is next

Show that if $\{x\}$ is an open set in $X$, for all $x\in X$, then all subsets of $X$ is open in $X$.

Plase help me. Thanky very much. Thanks for your anwers

HINT: Recall that the definition of an open set in a metric space is as follows.

  $U$ is open if for every $x\in U$ there exists some $\varepsilon>0$ such that $B(x,\varepsilon)\subseteq U$.

Now use the assumption that every singleton is open, to conclude that every set is open.

In a metric space $(X,d)$ every singleton set $\{x\}$ is a closed set. To verify it take any point $y \in X, y \neq x$. Take the distance $d(x,y)$ and see if you are getting an open ball about $y$ to prove $X - \{x\}$ is open.

You can write any subset of $X$ as an union of singleton set. Singleton set is open in our matric space.

We have a result that arbitrary union of open set is open. To prove it, take an arbitrary union of open set and see if you can construct open ball about the point which is contained in the arbitrary union.

So any set in the space will be open.


Is it bad form to write mysterious proofs without explaining what one intends to do? Often when doing assignments, I find myself deliberately writing in a "mysterious" way. By this I mean that the reader usually will not understand what exactly is going on and what for, until the very end where all the things come together.

A simple example is if I wish to prove that $S$ is true by showing that it is equivalent to $s$ being true, and then proving that $s$ is true. Often I will find myself doing this by writing something that reads like ...

  "Consider s, this seemingly random object that I present to you out of
  nowhere. Let us work on this for the next 1-2 pages, don't ask me why
  ..... [1-2 pages later] .... and that proves that $s$ is true. Now by noticing this and that, we see that this implies $S$ is true.  Surprise! QED"

Is this bad form? It also seems a bit pretentious to be, because the reason I think I sometimes do this is because these are the proofs I have mostly met in textbooks. Rarely is the proof sketched before it's given, very often new, foreign, confusing objects are introduced without introductions and motivations, and it's usually near the end of the proof that I would get my "aha" moment. The problem with this is of course that if one does not know the why behind some of the steps during the first reading, then one will have a harder time remembering how the pieces fit together throughout the proof.

But of course my instructors are not students reading (undergaduate) textbooks, and therefore they can perhaps deal with these sorts of mysterious proofs? Maybe they even prefer them, rather than having to waste time on reading me informally writing a few sentences prior to the proof outlining my ideas, giving a proof sketch, etc? I also do not wish to run the risk of sounding patronising or arrogant: "look at me and my geniusly complicated proof that I will now explain to you step by step".

The purpose of proofs is communication. If your proof is obscure, then you have failed to communicate.

Strive to be as clear as possible, including motivation for complicated arguments, if necessary.

Since all answers agree that this is a bad idea, I'm tempted to add a rare case for the opposite.

Say you have been working for 7 years on a famous conjecture that has tortured mathematicians for 300+ years without anyone else having even come close to proving it. But you have managed to finally tackle it, in a 150 pages paper.

In that case, it seems like a good idea to have a "mysterious" proof - and provide it in, say 3 days lectures - that escalate into proving the unexpected and exciting result.

At Last, Shout of 'Eureka!'


Determine the value of $a$ for which the function is continuous at $x=0$ Determine the value of $a$ for which the function is continuous at $x=0$
  $$
f(x) =
\begin{cases}
a \sin\frac{\pi}{2}(x+1) & \text{if} \ x\le0, \\ \\
\frac{\tan x - \sin x}{x^3} &\text{if} \ x\gt0. \
\end{cases}
$$

Can you help me with this problem?

HINT:

$$\lim_{x\to0^+}\frac{\tan x - \sin x}{x^3}$$

$$=\lim_{x\to0^+}\frac{\sin x}x\cdot\lim_{x\to0^+} \frac{1-\cos x}{x^2}\cdot\lim_{x\to0^+}\frac1{\cos x}$$

$$=\lim_{x\to0^+} \frac{1-\cos x}{x^2}  \text{ as }\lim_{x\to0}\frac{\sin x}x=1=\lim_{x\to0}\cos x$$

$$=\frac12\left(\lim_{x\to0^+} \frac{\sin \frac x2}{\frac x2}\right)^2\text{ as }\cos2y=1-2\sin^2y$$

$$=\frac12$$

$$\lim_{x\to0^-}\sin\left(\frac\pi2(x+1)\right)=1$$

Now use this.


How many ways can 8 people be seated in a row? I am stuck with the following question,

  How many ways can 8 people be seated in a row?
  if there are 4 men and 4 women and no 2 men or women may sit next to each other.

I did it as follows,
As 4 men and 4 women must sit next to each other so we consider each of them as a single unit.
Now we have we 4 people(1 men group, 1 women group, 2 men or women) they can be seated in 4! ways.Now each of the group of men and women can swap places within themselves so we should multiply the answer with 4!*4!
This makes the total  4!*4!*4! =13824 .
Please help me out with the answer. Are the steps clear and is the answer and the method right?

Thanks

If there is a man on the first seat, there has to be a woman on the second, a man on the third so forth. Alternatively, we could start with a woman, then put a man, then a woman and so forth. In any case, if we decide which gender to put on the first seat, the genders for the others seats are forced upon us. So there are only two ways in which we can divide the eight aligned seats into "man-seats" and "woman-seats" without violating the rule that no two men and no two women may sit next to each other.

Once you have chosen where to seat the men and where the women you can permute the two groups arbitrarily, giving $4!$ possibilities each. So in total the number of possible constellations is
$$
2\cdot 4!\cdot 4!=1152.
$$

Is it that no two men and no two women should be seated together.
Yep? Then do this..
Permutation of men in 4 places be 4!. This leaves us with 5 gaps(3 in between and 2 on either side) in which we got to arrange 4 women.. P(5,4) i.e. 5! ways.
Finally   5!*4! is the answer!.


Question about smooth functions and their signs with given initial conditions Problem Statement
Suppose $f:\mathbb{R}\to\mathbb{R}$ is a smooth function (infinitely differentiable) where $f(x)\geq 0$ for all $x\in\mathbb{R}$, $f(0)=0$, and $f(1)=1$. Show that there is some $n\in\mathbb{N}$ and an $x_0\in\mathbb{R}$ such that $f^{(n)}(x_0)<0$.
Proof Attempt
We can consider two cases:
Case (1): There exists a $z<0$ such that $f(z)>0$. By the MVT, there exists a point $z_0\in(z,0)$ such that $$f'(z_0)=\frac{f(z)-f(0)}{z-0}=\frac{f(z)}{z}<0 $$ so we are done.
Case (2): For all $x<0$, $f(x)=0$. In this case, $f'(x)=0$ for all $x<0$, which implies that, by left sided derivatives and the fact that $f'$ is smooth, $f'(0)=0$. It then follows that,  since $f'(x)=0$ on $(-\infty,0]$, we have $f^{(n)}(x)=0$ for all $x\in(-\infty,0]$.
For $n=1$, define $z_1$ as the real number in $(0,1)$ such that $$f'(z_1)=\frac{f(1)-f(0)}{1-0}=1$$For $n\in\mathbb{N}$, define $z_{n+1}$ by the real number in $(0,z_n)$ such that $$f^{(n+1)}(z_{n+1})=\frac{f^{(n)}(z_n)-f^{(n)}(0)}{z_n-0}=\frac{f^{(n)}(z_{n})}{z_n}>0$$ whose existence is given by the mean value theorem. This defines the sequence $(z_n)$, which converges to $0$ and is strictly decreasing.

Edit 2: Continuing, we see that by definition, we have $$f^{(n+1)}(z_{n+1})=\frac{f^{(n)}(z_n)}{z_n}>f^{(n)}(z_n)$$ for any $n\in\mathbb{N}$ with $n\geq 2$, since $0<z_n<1$, thus the sequence $f^{(n)}(z_n)$ is strictly increasing and always greater than or equal to $1$. Now, $f^{(n)}(0)=0$, but since $(z_n)$ can get arbitrarily close to $0$ and $f^{(n)}(z_n)\geq 1$ for all $n\in\mathbb{N}$, we have $f^{(n)}(0)=0<1\leq f^{(n)}(z_n)$ for large enough $n$.
This is has almost shown that $f$ is not differentiable for some value of $n$, but I am not quite sure that it has exactly shown that. If it has, then we are done, as Case (2) cannot happen. I would like help either formalizing this or fixing this if it is not true.

This is where my argument, to me, seems to break down. Intuitively, the slope of the secant lines between $f^{(n)}(z_n)$ and $f^{(n)}(0)$ is increasing without bound as $n$ increases. My idea was to create a sequence of points which converged to $0$, but broke the differentiability at $0$. I would greatly appreciate any tips in proceeding, critiques of my proof outline, or alternative proof methods. Thank you in advance.
Edit 1: In Case (2), I am not explicitly making any assumptions that would lead to a contradiction, but why is this proof going in that direction? Either Case (2) cannot happen, which I find intuitively true, but my intuition often fails me, or one of my assumptions in the beginning of Case (2) is false.
Edit: Just to close this off, one of my faulty assumptions was that I assumed the sequence I constructed converged to $0$, which it doesn't have to.

This is actually a very hard question. So hard, that it appeared in the 2018 Putnam Exam.
There are lots of solutions to this and I do not know which one suits your knowledge best. So just choose.


Is a number meeting these conditions divisible by forty-nine? I am not a mathematician, I'm a linguistics PhD student. As part of my research I need to put various convoluted sentences through various syntactic transformations and see then check whether people think they are true or not. Mathematical statements (well, some of them) suit my purposes very well, because they are less context dependent and can be straightforwardly assigned a truth value (i.e. be deemed true or false). The problem is that I'm not a mathematician. When these sentences get a bit convoluted, I have a bit of a problem knowing whether they are true or false myself (before they undergo various syntactic transformations).

I have a particular sentence which states that if a given number is:

an integer
divisible by 7 (meaning it will yield an integer if divided by 7)
a square number

then it is divisible by 49. I intuitively believe this to be correct (although I can't explain why). Is this actually true? I don't want to waste everybody's time by starting with an untrue untransformed sentence.

You can derive this fairly directly from Euclid's Lemma, which says that if a product $a \cdot b$ is divisible by prime $p$, then either $a$ or $b$ (or both) is divisible by $p$.

So, $n^2$ divisble by 7 means $n \cdot n$ divisible by 7 which (by Euclid's Lemma) means $n$ divisible by 7 and thus $n^2$ divisible by 49.

Of course it took even Euclid a little work to prove the Lemma

suppose this number is x.

Now we know:

x is an integer (Assumption: positive integer)
x is divisible by 7
x is a square number

From Point 1 we know its a positive integer. From that we can say x is 1 or greater than 1. (x=1 or x>1)
From Point 2 we know x is divisible by 7. That brings us to conclude that x is at least 7. Or multiple of 7 since its divisible by 7. We can say (x=7 or x=7*y) Here y can be any positive integer (y=1 or greater)
Now from Point 3 we know that its a square. that means it has some form of (x= 7^2) or (x=7^(2) .y) or (x=7^2 . z^2). where y = z^2.

What I am trying to say here is

7 is a prime number.
x is a square and divisible by 7.

This forces the number x to be a product of 7^2. Which makes it divisible by 49. because 49 is 7^2.


Average of an Unbounded sequence of postive numbers. Could someone check my proof.  More of a sketch really.

Can it happen that $s_n>0$ for all $n$ and that $\limsup s_n=\infty$, although $\lim\sigma_n=0$, where $\sigma_n={\sum_{k=1}^ns_n\over n+1}$.

Let $M>0$, $\{s_{n_k}\}$the subsequence such that $s_{n_k}<M$ for all $k$, $\{s_{m_k}\}$ the subsequence such that $s_{m_k}\ge M$ for all $k$.
Finally for each $n$, let $a(n)=\max\{k:s_{n_k}<n\}$ and $b(n)=\max\{k:s_{m_k}<n\}$. For $n$ large enough,

$$\begin{align}
\sigma_n & ={\sum_{k=0}^{a(n)}s_{n_k}\over n+1}+{\sum_{k=0}^{b(n)}s_{m_k}\over n+1}\\
& >{\sum_{k=0}^{b(n)}s_{m_k}\over n+1}\\
&\ge{(b(n)+1)M\over b(n)+1}=M.
\end{align}$$

Therefore, $$\lim_{n\to\infty}\sigma_n\ne 0.$$

It appears $\sigma_n$ is supposed to be the average of the first $n+1$ terms-you should state that. On the right you are using $n$ for the value used for separating the subsequences, so the first line is not correct.  Your definitions in the sentence starting "Finally" do not work.  It is possible that there are infinitely many $s_n \lt M$, so $a(n)$ is not defined.  We know there are infinitely many $s_n \gt M$.


What is the matrix of the monster Lie algebra? In Richard Borcherds' proof of monstrous moonshine, he constructs a "monster Lie algebra", which is a $\mathbb Z^2$-graded, infinite-dimensional Lie algebra with a contravariant bilinear form acted on by the monster group. The monster Lie algebra is a generalized Kac-Moody algebra, so is associated to a infinite symmetric "matrix" $A = (a_{ij})_{ij}$ satisfying certain conditions.

Is a recurrence relation (or an explicit formula) for the $a_{ij}$ known?

I guess your difficulty might be that in the papers, the root spaces are illustrated without telling you the GKM matrix coefficients. Actually these coefficients can be obtained directly from the basic properties of the monster Lie algebra, and it depends on how much you know about the construction. Indeed the whole matrix can be explicitly described using the coeffficients of the (normalized) $j$-function, which is well known (at least to experts). If you feel it hard to work it out yourself, the direct pictorial answer can be seen in Roger Carter's book, or p.22--23 of this article:

http://jurisiche.people.cofc.edu/jur696.pdf

for free. Besides the original papers by Borcherds, the articles by Elizabeth Jurisich and the book by Urmie Ray will tell more details.


Where are the axioms? It is said that our current basis for mathematics are the ZFC-axioms.

Question: Where are these axioms in our mathematics? When do we use them? I have now studied math for a year, and have yet to run into a single one of these ZFC axioms. How can this be if they are supposed to be the basis for everything I have done so far?

I think this is a very good question. I don't have the time right now to write a complete answer, but let me make a few quick points (and I'll add more when I get the chance):

Most mathematics is not done in ZFC. Most mathematics, in fact, isn't done axiomatically at all: rather, we simply use propositions which seem "intuitively obvious" without comment. This is true even when it looks like we're being rigorous: for example, when we formally define the real numbers in a real analysis class (by Cauchy sequences, Dedekind cuts, or however), we (usually) don't set forth a list of axioms of set theory which we're using to do this. The reason is, that the facts about sets which we need seem to be utterly tame: for example, that the intersection of two sets is again a set.
ZFC arose in response to a historical need. The history of modern logic is fascinating, and I don't want to do it injustice; let me just say (wildly oversimplifying) that you only really need to axiomatize mathematics if there's real danger of different people using different axioms implicitly, without realizing it. One standard example here is the axiom of choice, which very reasonable people alternately find perfectly intuitive and clearly false. So ZFC, very roughly speaking, won the job of being the "default" axiom system for mathematics: you're perfectly free to prove theorems using (say) NF instead, but it's considered gauche if you don't explicitly say that's what you're doing. Are there reasons to prefer some other system to ZFC? I'm a very pro-ZFC person, but even I'd have to say yes. The point isn't that ZFC is perfect, though; it's that it's strong enough to address the vast majority of our mathematical needs, while also being reasonable enough that it doesn't cause huge problems most of the time. This strength, by the way, is crucial: we don't want to have to keep updating our axiomatic framework to allow us to do basic mathematics, so overshooting in terms of strength is (I would argue) preferable (the counterargument is that overshooting runs a greater risk of inconsistency; but that's an issue for a different question, or at least a bit later when I have more time to write).
Even in the ZFC context, ZFC is usually overkill. OK, let's say you buy the ZFC sales pitch (I certainly did - and I just love the complimentary toaster!). Then you have some class of theorems you want to prove, and - after expressing them in the language of ZFC (which is frankly a tedious process, and one of the practical objections to ZFC emerges from this) - proceed to prove them from the ZFC axioms. But then you notice that you didn't use most of the ZFC axioms at all! This is in fact the norm - replacement especially is overkill in most situations. This isn't a problem, though: ZFC doesn't claim to be minimal, in any sense. And in fact the study of what axioms are needed to prove a given theorem is quite rich: see e.g. reverse mathematics.

Tl;dr: I would say that the sentence "ZFC is the foundation for modern mathematics," while broadly correct, is hiding a lot of context. In particular:

Most of the time you're not going to actually be using axioms at all.
ZFC's claim to prominence is primarily sociological/historical; we could equally well have gone with NF, or something completely different.
The ZFC axioms are wildly overpowered for most of mathematics; in particular, you probably won't really need the whole of ZFC for almost anything you do.
Most of all: ZFC doesn't come first. Mathematics comes first; ZFC is a mathematical theory that, among other things, "absorbs" the vast majority of mathematics in a certain way. But you can do math without ZFC. (It's just that you run the risk of accidentally invoking an "obvious" set-theoretic principle which isn't so obvious, and so conflicting with other results which invoke the "obvious" negation of your "obvious" axiom. ZFC provides a general language for us to do math in, so we don't have to worry about things like this. But in practice, this almost never occurs.)

Note that you could also ask this question with regard to formal logic - specifically, classical first-order logic - in general; and there has been a lot written about this (I'll add some citations when I have more time). But that's going very far afield.

Really tl;dr (and, I should add, in conflict with a number of people - this is my opinion): foundations do not enable, but rather serve, mathematics.

ZFC offers a conceptual framework with the sanity of FOL where you can talk about objects small and large, and even consider classes as syntactic sugar. There are many statements that are nevertheless not

decided in ZFC, like the Continuum Hypothesis, V = L, Large Cardinal Axioms, etc.. So when you do mathematics, you might add further foundational axioms, besides working in the conceptual framework of ZFC.

See also:

Logical Dreams - Past and Future of Set Theory
Saharon Shelah - 2002
https://arxiv.org/abs/math/0211398


Show that all groups of order $3$ are isomorphic with each other. I wanted to find an isomorphism between two arbitrary groups of order $3$ and then conclude but , I don’t know how I can concretely define this application.

Let $G$ and $H$ be two groups of order $3$.
Let $g\in G$ such that $g\neq 1$.
Method 1: By using Lagrange's Theorem, $|\langle g\rangle|$ divides $|G|=3$. Since $\langle g\rangle$ is nontrivial, it must have size $3$ and therefore $G=\langle g\rangle$.
Method 2: Let $G=\{1,g,f\}$. Since the Cayley table of a group is a Latin square, by filling the Cayley table of $G$, we can obtain that $f=g^2$ and $g^3=1$.
So by either method, we get that $G=\{1,g,g^2\}$. Similarly, we get $H=\{1,h,h^2\}$.
Thus the obvious guess for the isomorphism from $G$ to $H$ will be $\phi:G\rightarrow H$ where $$1\mapsto 1,\ g\mapsto h,\ g^2\mapsto h^2.$$
The map is clearly a bijection. To complete the proof, we need to show that it is a homomorphism. If you know modular arithmetic, you can show by the method below: Let $g^i,g^j\in G$,
$$\phi(g^ig^j)=\phi(g^{(i+j)\bmod n})=h^{(i+j)\bmod n}=h^ih^j.$$
Otherwise, you can prove by considering all the possibilities. Since $G$ is abelian, it is sufficient to prove for the following cases:
$$\phi(1\cdot 1),\ \phi(1\cdot g),\ \phi(1\cdot g^2),\ \phi(g\cdot g),\ \phi(g\cdot g^2),\ \phi(g^2\cdot g^2).$$

Let the elements be $\{e,a,b\}$ with $e$ the identity.  Show there is only one way to fill in the addition table that satisfies the group axioms.


Is $f:\mathbb{Z}_{\geq 0}\times \mathbb{Z} \rightarrow \mathbb{Z}$ defined as $f(m,n)=2^m(2n-1)$ bijective? Let $f:\mathbb{Z}_{\geq 0}\times \mathbb{Z} \rightarrow \mathbb{Z}$ defined as $f(m,n)=2^m(2n-1)$. Is $f$ bijective?

So every integer can be written in the form $2^i \xi$ where $\xi$ is an odd integer. And such a representation is unique. So clearly this is a bijection. Am I right?

The function is  not onto. $f(m,n)=0$ does not have a solution.

The function is not bijective, consider $f(k,0)=2^{k}(2(0)-1)$. Note injectivity is violated.

Recall the definition of injectivity is $f(x)=f(y)$ iff $x=y$. But clearly we have $f(k,0)=f(p,0)=0$ since both $2^{p}(2(0)-1)=0$ and $2^{k}(2(0)-1)=0$.

Surjectivty holds, because we can choose $f(0,n)=(2n-1)=r$ has a solution for every $n$


What is the math behind the game Spot It? I just purchased the game Spot It. As per this site, the structure of the game is as follows:

  Game has 55 round playing cards. Each card has eight randomly placed symbols. There are a total of 50 different symbols through the deck. The most fascinating feature of this game is any two cards selected will always have ONE (and only one) matching symbol to be found on both cards.

Is there a formula you can use to create a derivative of this game with different numbers of symbols displayed on each card.

Assuming the following variables:

S = total number of symbols
C = total number of cards
N = number of symbols per card

Can you mathematically demonstrate the minimum number of cards (C) and symbols (S) you need based on the number of symbols per card (N)?

The celebrated Ray-Chaudhuri–Wilson theorem states that $C \leq S$, contradicting your numbers.

An almost matching construction is as follows. Pick some prime number $n$. Our universe, of size $n^2+n+1$, consists of pairs of numbers in $\{0,\ldots,n-1\}$ plus $n+1$ singletons $\{0,1,\ldots,n-1,\infty\}$ ("points at infinity"). For each $0 \leq a \leq n-1$ and $0 \leq b \leq n-1$ we will have a card of size $n+1$ containing the pairs $\{(x,ax+b \mod{n})\}$ and the singleton $a$. There are also $n$ special cards, for each $0 \leq c \leq n-1$, containing the pairs $\{(c,x)\}$ and the singleton $\infty$. One super special card contains all $n+1$ singletons.

Clearly two cards with the same $a$ intersect only at the singleton. Two cards with different $a$s intersect at the unique solution to $a_1x+b_1 = a_2x+b_2 \pmod{n}$. Two special cards intersect only at the singleton, and a normal and a special card intersect at $(c,ac+b)$. Finally, the super special card intersects the rest at a singleton.

In total, we have $n^2+n+1$ cards and symbols, each card containing $n+1$ symbols, and two cards intersecting at exactly one symbol. In your case $n=7$ and so the number of cards and symbols should be $7^2+7+1 = 57$.

N squared minus N + 1 is the correct formula for calculating both the number of images and the number of cards.  Here is the specification for generating the cards themselves:
Below are the specifications for generating each card of Spot-it.  Although the commercial Spot-it game contains 8 symbols per card, these specifications work for any desired number of symbols/card.  The rules are that every card contains one and only one symbol matching every other card.  Also, all symbols have an equal probability of being the matching symbol.

Let N = number of symbols per card
Let C = the total number of symbols as well as the total number of cards to be generated.
Then C = N2 – N + 1

Generate a matrix with N columns and C rows
Column 1:  Symbol 1 for N rows
            Symbol 2 for N-1 rows
                   .
                             .
                   .
            Symbol N for N-1 rows
Column 2:  Row 1 = Symbol (Column #)
            Row 2 = Symbol N+1
            Row 3 = Symbol 2N
                      Row 4 = Symbol 3N-1
                      Row 5 = Symbol 4N-2
                      Etc. through row 2N-1
             Repeat above rows N-2 times
Column 3:  Row 1:  Symbol (Column #)
             Row 2:  Previous column Row 2 plus 1
                      Row 3:  Previous column Row 3 plus 1
             Etc. through row 2N-1
             Repeat above rows N-2 times
Column 4:  Same as Column 3
       .
       .
       .
Column N

The Spot-it Cards are the Rows


Endomorphisms of the maximal ideal of a local ring Let $R$ be a commutative local  ring with maximal ideal $\mathfrak{m}$. Is it true in general that $\text{Hom}_R(\mathfrak{m},\mathfrak{m})\cong \text{Hom}_R(\mathfrak{m}, R)$? What if the Krull dimension of $R$ is equal to one?

Let $R = \mathbb{Z}_2$, $\mathfrak{m} = 2\mathbb{Z}_2$.  Then $x\mapsto x/2$ is an $R$-module homomorphism $\mathfrak{m}\to R$ whose image isn't contained in $\mathfrak{m}$.


How could it be possible for the Goldbach conjecture to be undecidable? The answer to the question "Could it be that Goldbach conjecture is undecidable?" claims that it is possible for something such as the Goldbach conjecture to be undecidable, meaning that assuming that it is true and assuming that it is false would both lead to no contradiction.

But if it is undecidable, then, if we assume that it is false, it would mean that exists an even number that cannot be written as the sum of two primes. If a natural number exists, then it can be written down using a finite number of digits (any natural number is definable). This means that that number exists whether or not the conjecture is true, so if we assumed that it was true, there would be a contradiction, so it therefore can't be undecidable.

What is the flaw in what I just said?

What is the flaw in what I just said?

Your argument correctly shows that the Goldbach conjecture cannot be false and undecidable. However, it can be true and undecidable.

This is equivalent to saying that it can be true in the standard model of, say, Peano arithmetic, but false in some nonstandard model: in a nonstandard model there may be a nonstandard counterexample to Goldbach's conjecture which is not an ordinary integer, and so which cannot be written down in the usual sense.

The reason it can be true and yet assuming it's false doesn't lead to a contradiction is that if it's undecidable, Peano arithmetic doesn't know it's true, so Peano arithmetic can't use the fact that it's true in a proof.

if we assume that it is false, it would mean that exists an even number that cannot be written as the sum of two primes. If a natural number exists, then it can be written down using a finite number of digits

Ah not quite.  If the proof the negation of Goldbach's conjecture is constructive, and if all the assumptions of the proof are constructive, then what you say is true.  Converting a nonconstructive proof into a constructive proof is easy, just convert every nonconstructive inference step into an axiom.  But converting a nonconstructive set of axioms into a constructive set of axioms is not trivial.  Here are some examples of what would have to be true about every single assumption in the proof for the result to be constructive:

For every assumption of the form $A \lor B$, you must be able to actually compute which of $\{A,~B\}$ is true.  As a corollary:
For every predicate in the proof $P$, where the assumption $P(x) \lor \lnot P(x)$ is assumed, you must be able to actually compute which of $\{P(x),~\lnot P(x)\}$ is true
Every assumption of the form $A(x) \implies B(x)$, whenever $A(x)$ is computable, $B(x)$ is also computable
Every assumption of the form $A(x)$, $A(x)$ must be computable

And there's more.  Even 1 assumption could make the axiom set non-constructive.  Let $G(m)$ be the predicate that holds when $m$ is less than 3, odd, or the sum of 2 primes.  What if the (dis)proof of Goldbach's conjecture defines a predicate:

$$Q(n) = \forall m > n ~:~ G(m)$$

and made the assumption $Q(x) \lor \lnot Q(x)$.  Can you write a computer program that, inputting $x$, outputs true if $Q(x)$ and false otherwise?  If not, then the assumptions are not constructive (wrt your computing ability, that's a separate consideration though), and then you can't necessarily infer that the digits of the Goldbach counterexample are computable.


Is 10 closer to infinity than 1? This may be considered a philosophical question but is the number "10" closer to infinity than the number "1"?

There is more than one notion of "closer than."

Metric: We are not able to extend the metric on $\Bbb R$ to the extended line $[-\infty,\infty]$ in a meaningful way (i.e. where bigger elements of $\Bbb R$ are closer to $+\infty$ than smaller elements). Even if we extend the concept of "metric" to include infinite distances, this doesn't work.

Order-theoretic: A partial order $\le$ on a set is a relation satisfying certain properties. It is a way of ordering the elements of the set, put simply. A linear ordering is one in which every two elements are comparable: for any $x$ and $y$, one of $x\le y$ or $y\le x$ is true at least. The extended real line gets its linear ordering from $\Bbb R$, with $+\infty$ made maximal and $-\infty$ made minimal. If $x,y\le z$ then we can say that $y$ is closer than $x$ to $z$ if $x\le y\le z$ and $x\ne y$. On this view, $10$ is indeed closer to $+\infty$ than $1$ (and we can reverse directions to say $1$ is closer to $-\infty$ than $10$).

One observation to be made here is that we essentially decreed larger numbers closer to $+\infty$ purely by fiat. Rather than making this whole discussion trivial, though, it actually illustrates the power of modern algebra in converting imagination into (mathematical) reality. It makes intuitive sense for bigger numbers to be closer to $+\infty$, and there is a consistent way to encode this mathematically, so we simply make it so and therefore have what we want.

Topological: A metric on a space induces a topology on that space. In particular, a metrizable topology. There are more general topologies though, so we can do and allow more things. However not everything can be compared in terms of "closer than." One possible interpretation is that $x$ is closer than $y$ to $z$ if neighborhoods of $z$ containing $y$ also contain $x$ but not vice-versa. It is generally too optimistic to expect elements to be comparable in this way, but the topology on $\Bbb R$ and the extended line is nice enough to make this an almost useful concept.

The problem with this approach is that neigborhoods can be "disconnected," so we can simply put a neighborhood around $+\infty$ containing both $1$ and $10$ and then delete a patch around $10$, thus telling us that neither $1$ nor $10$ is closer than the other. But if we strengthen our notion to only connected neighborhoods, then we get what we want: every connected neighborhood of $+\infty$ containing $1$ must also contain $10$, but not vice-versa. [This originally slipped my mind, but Mike helpfully pointed it out.]

It is worth noting that the topology on the extended line is induced by the ordering on it (topologies are induced by orderings by considering "intervals" to be a topological base). So this actually isn't fundamentally any newer information than the previous context. Furthermore, different metrics can induce the same topology, and it is possible to make $[-\infty,+\infty]$ a metric space in which $10$ is closer than $1$ to $+\infty$. For example, $d(a,b):=|\tan^{-1}(a)-\tan^{-1}(b)|$ does this, as Rahul points out

My thought is that it is closer to both, dependent on your origin of view. To better define this question we would need a third variable, say PoV for Point of View.

Take an infinite length (aka: a Line) in a straight line. We must therefore assume there is no starting point as this fundamentally destroys our idea of 'infinite'. So where on this infinite length do we place the 1 and the 10? They must be relative to each other given our laws of mathematics i.e. 10 comes after 1 and both are equal distance either side of 5 and so on.

But either side of 10 and 1 is infinite and so infinite has two ends (in the case of a line) and therefore is the same distance to both -infinity (meters or whatever).

Now this raises another question. if we are measuring in a finite length on an infinite line then how do we decide where 1 and 10 must be placed. It could be placed at any length and at any distance apart and the line either side must always be infinite. So then, could the distance between the two points be infinite? Yes.

Personally I think the word 'infinite' is a cause of unnecessary philosophical questions :D

I also think The word infinite should be labelled as a vector or mathematical expression, rather than a number which most people treat it so. 1 and 10 are plotted points on a virtual graph whereas infinity is a mathematically defined expression (whether it be a curve or a straight line is irrelevant).

My 2 cents. Take it with a grain of salt as I am not a mathematician; I'm just a software developer!


How to find the unique sums in the values 1,2,4,8,16, 32 I apologize but I'm not sure what you would even call this problem. I have some data that provide a numeric code for race as follows:

hispanic(1) + american_indian_or_alaska_native(2) + asian(4) + black_or_african_american(8) + native_hawaiian_or_other_pacific_islander(16) + white(32)

So for example, a 5 represents a person who identifies as Asian and Hispanic (4 + 1), and a 25 represents a person who identifies as Pacific Islander, Black, and Hispanic (16 + 8 + 1).

I am trying to write a program that will retrieve what races are present from the given number. I figure there must be an equation that can determine the combination without detailing every unique combination, but I might be wrong! I tried to think about using modulo but I didn't get very far.

Thanks, and if you have suggestions for tags, please comment as I'm not sure where this fits into mathematics.

*edit Thanks everyone! This really helped me to think about the problem and generate an efficient solution. Answering my question didn't depend on using SAS but here is the SAS code I ended up using, which I think shows intuitively how to solve the problem:

data want;
    set have;

/* convert decimal to 6-place binary */
    eth_bin = put(ethnicity, binary6.);

/* if 1st digit is 1 then race is present, and so on*/
    if substr(eth_bin, 1, 1) = 1 then white = "Yes";
    if substr(eth_bin, 2, 1) = 1 then pacific_islander = "Yes";
    if substr(eth_bin, 3, 1) = 1 then black = "Yes";
    if substr(eth_bin, 4, 1) = 1 then asian = "Yes";
    if substr(eth_bin, 5, 1) = 1 then american_indian = "Yes";
    if substr(eth_bin, 6, 1) = 1 then hispanic = "Yes";
run;

Notice that the numbers $1, 2, 4, 8, 16, 32$ are all perfect powers of $2$. This means that from any sum of them, the individual components can always be retrieved because there is a unique way to express a number as a binary number. One way to retrieve the racial info is keep dividing by $2$ and getting the remainder. Every remainder of $1$ indicates a certain race (depending on the stage of division). If the remainder is $0$, that race is not present. To elaborate, roughly you follow this algorithm:

Let $n$ be your number with the racial info.
While $n$ is non-zero, re-assign $n$ to $\lfloor \frac{n}{2} \rfloor$ and make a note of the remainder. If the remainder is $1$, then that race is present, where the race starts at Hispanic and so on.
When $n=0$ you are finished, there are no more races.

Update: Henning Makholm's suggestions to use bitwise operators is in fact a more elegant solution, programmatically, but may be harder to grasp intuitively.

hispanic(1)
american_indian_or_alaska_native(2)
asian(4)
black_or_african_american(8)   native_hawaiian_or_other_pacific_islander(16)
white(32)
An example (not an entire program, just a routine) for QB64:
[
IF (yourcode%% = 0) THEN
-->PRINT "no preference"
ELSE
-->FOR i%% = 0 TO 5
----->j%% = 2 ^ i%%
----->IF (j%% AND yourcode%%) THEN
-------->SELECT CASE j%%
----------->CASE 1: v\$ = "hisp"
----------->CASE 2: v\$ = "amer"
----------->CASE 4: v\$ = "asian"
----------->CASE 8: v\$ = "black"
----------->CASE 16: v\$ = "pacific"
----------->CASE 32: v\$ = "white"
-------->END SELECT
-------->PRINT v\$
----->END IF
-->NEXT 'i%%
END IF
]
You can abbreviate your categories or spell them out in full. All pertinent categories will be listed for a given code.


Splitting of primes in an extension with "unknown" ring of integers Consider the Galois extension $\mathbb{Q}(\zeta_{7})/\mathbb{Q}$.

I am looking for the decomposition of the prime ideal $5\mathbb{Z}$ in the integral closure of $\mathbb{Z}$ in $\mathbb{Q}(\zeta_{7})$.

Here is what I thought:

First, since the extension is Galois, the ramification index and inertia degree of all the primes $\mathfrak{q_{i}}$ above $5\mathbb{Z}$ are equal and hence we have the formula
\begin{equation}
efg=6
\end{equation}

Now, suppose I don't know the "description" of the integral closure of $\mathbb{Z}$ in $\mathbb{Q}(\zeta_{7})$. Is it still possible to know these numbers $e$, $f$ and $g$?

How can we use the decomposition and inertia groups to find these numbers? Doesn't it require that we already know the "form" of the prime ideal factors, so that we can tell if it gets fixed or not by the action of the Galois group?

If we can't, then what about passing to a completion? Still can't? If not, I feel like these groups aren't really helping since they require that we already know the decomposition of the ideal, which is what I'm looking for.

Thank you.

The (original) argument of Sam for irreducibility of $(x^7-1)/(x-1)$ over $\Bbb F_5$ is not correct. The statement is right, however. Here’s a different approach.

The degree of the irreducible factor(s) of $x^6+x^5+x^4+x^3+x^2+x+1$ is the field extension degree when you adjoin a seventh root of unity to $\Bbb F_5$. The first power of $5$ for which $7|(5^n-1)$ is $n=6$. (You’ll notice that this is the multiplicative order of $5$ in $\Bbb F_7^*$.) Thus $\Bbb F_{5^6}$ has a seventh root of unity, but no smaller field does. So the polynomial in question is irreducible.

As for Shoutre’s original question, it’s by no means necessary to find an integral basis for the ring of integers of $\Bbb Q(\zeta_7)$. The question of the splitting of $5$ is purely local, so you can pass immediately to the same question over $\Bbb Q_5$.

I'm not sure it's possible to know the numbers $e$, $f$, and $g$ without a "description" of the ring of integers of $\mathbb{Q}(\zeta_7)/\mathbb{Q}=K$ (say). To find the factorization of $\langle 5 \rangle$ in $K$, we can use the Dedekind-Kummer theorem which essentially says that the factorization of $\langle 5 \rangle$ in $K$ is the same as the factorization of the minimal polynomial of $\zeta_7$ mod $5$. This calculation gives us:

$$
\begin{split} &&\text{ } x^6+x^5+x^4+x^3+x^2+x+1 \text{ (mod }5\text{)} \\ =&&\text{ } x^2+x+1+x^3+x^2+x+1 \text{ (mod }5\text{)} \\ =&&\text{ } x^3+2x^2+2x+2 \text{ (mod }5\text{)} \end{split}
$$

Checking each of $x=0,1,2,3,4$ shows that the above polynomial is irreducible mod $5$ and this implies that $\langle 5 \rangle$ does not factor in $K$.

Although this looks as though it does not rely on $\mathbb{O}_K$ (the ring of integers), the theorem above does not hold for a finite number of primes (those primes which divide the conductor of $\mathbb{O}_K$). This means that we need to check that $5$ is not one of these exceptions and to do this we need to know what $\mathbb{O}_K$ is. In our case we are fine because $\mathbb{O}_K=\mathbb{Z}(\zeta_7)$. In general, I don't think it's that difficult to work out what $\mathbb{O}_K$ is, so there is no problem using it here to work out the factorization (i.e. you should always be able to obtain this information).


Can parametric equations graph all kinds of lines? I saw this question which had a similar viewpoint, but was limited to straight lines and polynomials. Now we know that we can graph some pretty crazy stuff with parametric equations. For example:

and even

So naturally one would think that, if we put pen to paper, and drew something, anything, there would be a set of parametric equations to describe that shape or figure (of course, I'm not saying that finding the equations will be easy, but they should exist). So, is this true?

As far as I know, yes. Just about any line you could draw could be approximated by a parametric curve consisting of polynomials. For the Cartesian equivalent, look up "infinite series polynomials". A higher degree of accuracy can be achieved by adding more polynomial terms. Cusps (corners) could be achieved by including such functions as $\sqrt{x^2}$. Jump discontinuities could be achieved by including such functions as $\sqrt{x^2}/x$. A single parametric curve could then look as if it was multiple lines.

Disclaimer: All equations produced this way would be approximations of the originally drawn line, but could be made arbitrarily accurate. Coming up with these functions could be extremely tedious to do by hand, because any change made to any portion of the function would require adjustments to the rest of the curve.


Which one is less than others? Which one is less than others?

$\frac{3}{5} , \frac{2}{3} , \frac{6}{13} , \frac{23}{38}$

Yes the answer is $\frac{6}{13}$ but the real question is this:

I've a 12 years old brother and he just asked me this question so he is kid and I need to explain this with the most simple way.

First I thought using common Denominator but it's hard and time consuming then I find out that if we multiply all numbers by 2, all numbers would be greater than their Denominator except for $\frac{6}{13}$ so this most be less than others.

Is there another way to explain this?

The easiest way to see whether $\frac ab<\frac cd$ (where I assume that $b,d>0$) is to test whether $ad<bc$. This is really what one does when bringing fractions to the common denominator $bd$, but one need not actually change any fraction when going further to compare it to other fractions.

There may be a common denominator smaller than $bd$, but in general it is not worth while to go looking for it if you just want to compare two fractions.

Concretely to look for the minimal element one could determine that $\frac35<\frac23$ since $3\times3<5\times2$, then that $\frac35>\frac6{13}$ since $3\times13>5\times 6$, and finally that $\frac6{13}<\frac{23}{38}$ since $6\times38<13\times23$, showing that $\frac6{13}$ is the smallest of the four fractions. The last inequality is actually fairly gross, and the computation can be somewhat simplified (at least for mental calculation) by using an intermediate fraction: $\frac6{13}<\frac12<\frac{23}{38}$ since both $6\times2<13\times1$ and $1\times38<2\times23$ (using $\frac23$ would work as well). But of course searching a simple intermediate fraction is not guaranteed to be successful.

Note that you don't have to find the least common denominator, just any common denominator.

So you can just multiply each numerator by all of the other denominators, and compare the results. (These would be the numerators corresponding to the denominator that would be the product of all of the separate denominators.)

In your case, you would compare the numbers
$$3\cdot(3\cdot 13\cdot 38)=4446$$
$$2\cdot(5\cdot 13\cdot 38)=4940$$
$$6\cdot (5\cdot 3\cdot 38)=3420$$
$$23\cdot (5\cdot 3\cdot 13)=4485$$

The smallest is the third, whic corresponds to the original fraction $\frac{6}{13}$.


Every nonzero vector in $\mathbb{R}^n$ can be scaled to be arbitrarily close to $\mathbb{Z}^n-\{0\}$. Problem. Show that for all nonzero $v\in\mathbb{R}^n$ and all $\epsilon>0$, there exists $r>0$ such that $\operatorname{dist}(rv,\mathbb{Z}^n-
\{0\})<\epsilon.$

In other words, we must show that for all $\epsilon>0$ there exists $w\in\mathbb{Z}^n-\{0\}\subseteq\mathbb{R}^n$ such that $|v-w|<\epsilon$.

  Bonus question. Can we take $w$ to be arbitrarily large?

Attempts. The case $n=1$ is trivial: we can take $r=1/v$ so that $rv=1$. But that approach obviously doesn't generalize to $n>1$. Any idea?

More generally, if $v$ is rational, say $v_i=a_i/b_i$, then we can take $r=b_1\cdots b_n$ so that $rv\in\mathbb{Z}^n-\{0\}$.

What you refer to as the 'dist()' function is more commonly referred to as the vector norm, a function which takes a vector and produces the 'length' of a vector.  When you scale a vector by a constant, the norm of the vector scales by that constant (at least for the usual norm in this case).  It follows that the argument given for n=1 can be used nearly verbatim, all you need to do is translate from scaling the vector to scaling the norm.


Combinatorics in a $n \times n$ grid Natural number $n>2018$ is given. Numbers $1,2,\ldots,n^2$ are written (in an arbitrary order) into the fields of the $n\times n$ grid. Prove that it is possible to choose $n$ fields so that there's one field in each row and column and that there aren't any four consecutive terms of an arithmetic sequence in the chosen fields.

I have no idea how to start this. My attempts to find any correlations between the amount of four term sequences and the amount of fields chosen were all unsuccessful and I'm stuck.

Sketch (probabilistic method): Let $S$ be the set of arithmetic sequences of length 4 among the numbers $1,2,\cdots, n^2$. Clearly, for large $n$, $|S| \sim \frac{n^2}{3} \cdot \frac{n^2}{2} = \frac16 n^4$, since there are approximately $\frac{n^2}{3}$ choices for the common difference and on average $\frac{n^2}{2}$ choices for the linear shift. Now choose the $n$ fields uniformly at random (over the total space of $n!$). For a fixed $s \in S$, the chosen $n$ fields contain the 4 terms of $s$ with probability at most $\frac{1}{n(n-1)(n-2)(n-3)} \sim n^{-4}$. A simple union-bound gives us a probability of at least $$1 - \sum_{s \in S}\frac{1}{n(n-1)(n-2)(n-3)} \sim 1 - \sum_{s \in S} n^{-4} \sim 1 - (\frac{1}{6}n^4)(n^{-4}) = \frac{5}{6} > 0$$ that the chosen $n$ fields avoid all elements of $S$, i.e. all the arithmetic sequences of length 4.

I think that the problem boils down to proving that the number of choices of n fields containing four consecutive terms of an arithmetic sequence is smaller than n!


Proving $\lim_{n\rightarrow\infty}a_n=l\Rightarrow \lim_{n\rightarrow\infty}a^2_n=l^2$ $$a_n\ge0, l\ge0$$If $$\lim_{n\rightarrow\infty}a_n=l$$
Then $$\lim_{n\rightarrow\infty}a^2_n=l^2$$

Here is a trick, which will help you solve this problem.

$$a_n^2 - l^2 = (a_n - l)^2 + 2l(a_n - l)$$
using this you should be able to show that $a_n^2 - l^2 \rightarrow 0$.

EDIT: To elaborate further, by the triangle inequality we now know that
$$\lvert a_n^2 - l^2\rvert \leq \lvert (a_n - l)^2 \rvert + \lvert 2l(a_n-l)\rvert \hspace{10mm} (1)$$
So we just need to show that each of the two terms on the right get small as $n$ gets large. You should be able to make both of them small by using the fact that $a_n \rightarrow l$.

FINAL EDIT: To see a complete proof. Fix $\epsilon > 0$. Choose $N_0$ so that for $n\geq N_0$, $$\lvert a_n - l\rvert \leq \sqrt{\epsilon};$$ and $N_1$ large so that  for $n\geq N_1$, $$\lvert a_n - l \rvert \leq \frac{1}{\lvert 2 l \rvert} \epsilon.$$Then if $n\geq\max(N_0,N_1)$, the inequality (1) above gives us $$\lvert a_n^2 - l^2 \rvert \leq \epsilon + \epsilon = 2 \epsilon$$
and since $\epsilon$ was arbitrary we conclude that $\lvert a_n^2 - l^2\rvert \rightarrow 0$.

Suppose: $$\lim_{n\rightarrow\infty}a_n^2\not\rightarrow l^2$$
Exists an $\epsilon>0$ such that: $$|n^2-l^2|\ge\epsilon \Leftrightarrow l^2-\epsilon\ge n^2 \ge l^2 + \epsilon$$

Left side of the equation is false, because if it were true, $\epsilon<0$. Which is false.

Right side is false because:$$n^2-l^2\ge\epsilon \Leftrightarrow -l^2 \ge \epsilon-n^2 \ge 0$$

Which is again false, because $n,l\ge0$

EDIT: If this is wrong, please let me know why.


A space $X$ that contains a copy of $\ell_1$, does not contain a complemented copy of $\ell_1$, and whose dual is not weakly sequentially complete I want to find an example of a Banach space $X$ which contains a copy of $\ell_1$, does not contain a complemented copy of $\ell_1$, and so that $X^*$ is not weakly sequentially complete.

Let $Y$ be a quasi-reflexive space which is not reflexive (for example, the James space). Then neither $Y$ nor $Y^*$ is weakly sequentially complete. Since $Y$ is quasi-reflexive, it does not contain copies of $\ell_1$. Now, set $X=C[0,1]\oplus Y$. Then $X^*$ is not weakly sequentially complete because $Y^*$ is not. Further, $X$ does not contain complemented copies of $\ell_1$.

Such artificial construction was necessary here because $C[0,1]^*$ is weakly sequentially complete.

A Banach space with an  unconditional basis containing a copy of $\ell_1$ contains a complemented copy of $ell_1.$  See JMAA342(2008) pp.83-88


Existence of Differentiable Function Q: If $f$ is Riemann integrable over $[0,1]$, then there exists a differentiable function $F$ on $[a, b]$ such that $F'=f$ and $\int_a^b f(x)dx = F(b)-F(a)$.

Here's what I'm thinking so far:

If $\alpha = x$, then you can say that $f$ is also Riemann-Stieltjes integrable on $[0,1]$.
Then, since $f$ must be a bounded real-valued function due to being Riemann integrable, we can use the theorem that states "For $0\le x \le 1$, put $F(x) = \int_0^x f(t) dt$. Then $f$ is continuous on $[0,1]$
In order for $F$ to be differentiable, $f$ must be continuous on $[0,1]$, so we add that as an assumption
Then, by Fundamental Theorem of Calculus, we can say  $\int_0^1 f(x) \; dx = F(1) - F(0)$.

I realize there are lots of issues here, but I'm hoping I'm at least on the right track.  Any direction or hints are appreciated.

Your question has a negative answer. For instance the function defined by $f(x)=\cases{-1,& $-1\le x<0$ \cr 1,&$ 0\le x\le 1$}$ is Riemann integrable over $[-1,1]$, but is not the derivative of any function (since derivatives have the intermediate value property as a result of Darboux's Theorem).

Note that, A bounded function $f$ of $[a,b]$ is Riemann integrable on $[a,b]$ iff the function is continuous almost everywhere. This is known as Lebesgue's criteria.


Finding the real value of x I replaced the numbers with variables so i let $y=2$ and $z=12$ and i got the following after simplifying the left side:

$$2^{144}-\frac{1}{2^{144}}=8^x-8^{-x}$$

but i'm not seeing how this can help me find $x$.


Local Weierstrass preparation I could not find a reference of the local version of Weierstrass preparation theorem It is used for example in Denef & van den Dries "$p$-adic and real subanalytic sets".
Can you help, thank you

See '$p$-adic numbers, $p$-adic analysis, and zeta-functions' by Koblitz. Chapter IV sections 3 and 4 have a good exposition on newton polygons and the Weierstrass preparation theorem.

A good elementary proof of the Weierstrass preparation theorem can be found in the Jun-ichi Igusa's An introduction to the theory of Local Zeta Functions (Google books link).

See Chapter 2, section 2.3.


Two defective balls out of ten There are 2 defective balls in a batch of 10. To inspect, someone will randomly check 2 balls to see if they're defective and will reject if at least 1 ball they check is defective.
What's the probability of finding at least 1 defective ball when randomly selecting 2 from the batch of 10?

The solution should deal with permutations, but I'm not sure how to initially approach this.

Hint:  the number of ways you could get a defective ball is
\begin{align}
\binom{2}{1}\binom{8}{1} + \binom{2}{2}\binom{8}{0}.
\end{align}

hey actually it is a quite easy question u dont need to worry
wat we have to do is select two balls out of 10 that is 10c2 which comes out to be 45 and then non defective once that is 8c2 which comes out to be 28
after doing this the probablity of non defctive comes out to be 28/45 and for non defctive just subtract from 1*here u r done*


The connection between the Riemann hypothesis and the harmonic series I have heard most people say that the harmonic series, given by
$\displaystyle\sum_{n=1}^{\infty}{n^{-1}}$, is central or somehow related to a proof of the Riemann hypothesis. I have been trying to understand the sense in which such an assertion is held but to no avail. If anybody could enlighten me about this connection I would be grateful.

The asymptotic of $\sum_{n \le x} \frac1n$ as $x \to \infty$ is easily found with some $x^{-10}$ error term.
The PNT and the Riemann hypothesis (more generally the analytic continuation of $\frac{1}{\zeta(s)},\log \zeta(s),\frac{\zeta'(s)}{\zeta(s)}$) is about the asymptotic of $\prod_{p \le x} \frac{1}{1-p^{-1}} = \sum_{n, \text{Lpf}(n) \le x} \frac1n$ where $\text{Lpf}$ is the largest prime factor.

Well the generalized harmonic series, given by
$$S_{n,2j}=\sum_{k=1}^n k^{2j}; j\in\mathbb{N}\tag{1},$$
is central to a (to be verified) proof of the Riemann hypothesis. See  (Ref.1:a preprint at arXiv).

The method used in Ref.1 is based on the approach of Polya and Hurwitz. It is summarized in a MSE question.

For Generalized Riemann Hypothesis, the Dirichlet character $\chi$ weighted generalized harmonic series appeared.
$$S_{n,2j}(\chi)=\sum_{k=1}^n \chi(k)k^{2j+b}; b=0,1;j\in\mathbb{N}_0\tag{2}.$$


What are some examples of a mathematical result being counterintuitive? As I procrastinate studying for my Maths Exams, I want to know what are some cool examples of where math counters intuition.

My first and favorite experience of this is Gabriel's Horn that you see in intro Calc course, where the figure has finite volume but infinite surface area (I later learned of Koch's snowflake which is a 1d analog). I just remember doing out the integrals for it and thinking that it was unreal. I later heard the remark that you can fill it with paint, but you can't paint it, which blew my mind.

Also, philosophically/psychologically speaking, why does this happen? It seems that our intuition often guides us and is often correct for "finite" things, but when things become "infinite" our intuition flat-out fails.

Here's a counterintuitive example from The Cauchy Schwarz Master Class, about what happens to cubes and spheres in high dimensions:

Consider a n-dimensional cube with side length 4, $B=[-2,2]^n$, with radius 1 spheres placed inside it at every corner of the smaller cube $[-1,1]^n$. Ie, the set of spheres centered at coordinates $(\pm 1,\pm 1, \dots, \pm 1)$ that all just barely touch their neighbor and the wall of the enclosing box. Place another sphere $S$ at the center of the box at 0, large enough so that it just barely touches all of the other spheres in each corner.

Below is a diagram for dimensions n=2 and n=3.

Does the box always contain the central sphere? (Ie, $S \subset B$?)

Surprisingly, No! The radius of the blue sphere $S$ actually diverges as the dimension increases, as shown by the simple calculation in the following image,

The crossover point is dimension n=9, where the central sphere just barely touches the faces of the red box, as well as each of the 512(!) spheres in the corners. In fact, in high dimensions nearly all of the central sphere's volume is outside the box.

Independence in Statistics.
If I have a box with red & black socks, and I take one, and then another one the probability of both socks being red is sometimes different from the case when I, instead, pick a pair. (EX.If all socks are connected to every other with a string, like a complete graph, and then I pick some string, automatically choosing the socks at either end as the pair selected).


Why is a finite integral domain always field? This is how I'm approaching it: let $R$ be a finite integral domain and I'm trying to show every element in $R$ has an inverse:

let $R-\{0\}=\{x_1,x_2,\ldots,x_k\}$,
then as $R$ is closed under multiplication $\prod_{n=1}^k\ x_i=x_j$,
therefore by canceling $x_j$ we get $x_1x_2\cdots x_{j-1}x_{j+1}\cdots x_k=1 $,
by commuting any of these elements to the front we find an inverse for first term, e.g. for $x_m$ we have $x_m(x_1\cdots x_{m-1}x_{m+1}\cdots x_{j-1}x_{j+1}\cdots x_k)=1$, where $(x_m)^{-1}=x_1\cdots x_{m-1}x_{m+1}\cdots x_{j-1}x_{j+1}\cdots x_k$.

As far as I can see this is correct, so we have found inverses for all $x_i\in R$ apart from $x_j$ if I am right so far. How would we find $(x_{j})^{-1}$?

Remember that cancellation holds in domains. That is, if $c \neq 0$, then $ac = bc$ implies $a=b$. So, given $x$, consider $x, x^2, x^3,......$. Out of finiteness there would be a repetition sometime: $x^n  = x^m$ for some $n >m$. Then, by cancellation, $x^{n-m} =1$, and $x$ has an inverse.

Why to find the inverse of $xj$
Can't it be solved this way
$(x1.x2...xk) - xj = 0$
Then $xj(x1..xj-1.xj+1..xk  -1)=0$
As $xj$ is not a $0$ and $R$ is integral
We get $x1..xj-1.xj+1..xk = 1$
Choosing an arbitrary $xm$
$Xm(x1..xm-1.xm+1..xj-1.xj+1..xk)=1$
So $xm$ has an inverse


query about the cosine of an irrational multiple of an angle? de Moivre's identity
$$
(\cos \theta + i \sin \theta)^n = \cos n\theta + i \sin n\theta
$$
only applies as written when $n \in \mathbb{Z}$. if the exponent is a fraction $\frac{m}{n}$ then there will be $n$ values of $(\cos \theta + i \sin \theta)^{\frac{m}{n}}$, whilst for irrational $\alpha$ the set $\{e^{i\alpha (\theta +2\pi k)} \}$ is a dense subset of the unit circle.

nevertheless $e^{i \alpha \theta}$
is distinguished amongst the values of $(e^{i\theta})^\alpha$ even when $\alpha$ is not an integer. so can we proceed as follows?

when $0 \lt \theta \lt \frac{\pi}4$
$$
(\cos \theta + i \sin \theta)^\alpha = \cos^{\alpha}\theta(1+i \tan \theta)^{\alpha} = \cos \alpha \theta + i \sin \alpha \theta \tag{?}
$$
using the binomial theorem, which converges for the specified range of values for $\theta$, we have:

$$
\cos \alpha \theta = \cos^{\alpha}\theta \sum_{m=0}^\infty (-1)^m \binom{\alpha}{2m} \tan^{2m}\theta \tag{1}
$$
if (1) is true, then if $0 \lt \alpha, \beta \lt \frac{\pi}4$ we would have
$$
\cos^{\alpha}\beta \sum_{m=0}^\infty (-1)^m \binom{\alpha}{2m} \tan^{2m}\beta = \cos^{\beta}\alpha \sum_{m=0}^\infty (-1)^m \binom{\beta}{2m} \tan^{2m}\alpha
$$

Since your question is something along "how to prove step (?)" I will solve that part.  I can't do anything beyond that.

$$\cos^{\alpha}\theta(1+i\tan\theta)^{\alpha}=[\cos\theta(1+i\tan\theta)]^{\alpha}=(\cos\theta+i\sin\theta)^{\alpha}$$$$\cos*\tan=\cos*\frac{\sin}{\cos}=\sin$$

Simple exponent rule of "factoring" out the exponent and distributing the terms.


Can there be a month with 6 Mondays? (or 6 of any other days?) This year, this month (November) we've 5 Mondays so I get 5 pay checks on 2nd, 9th, 15th, 22nd, 29th (which feels very good - I get paid one extra check :-).

I wonder if it's possible some years to have 6 Mondays? I haven't seen it but I wonder if there will be such year?

If Monday is not possible, then is any other week day possible to happen 6 times in one month?

(I'm mathematics enthusiast, no deep knoweldge - thanks!)

To fit the maximum number of Mondays in a month, you'd want the first day of the month to be a Monday. Then it is easy to see:

7 days after that, you have 2 Mondays in your month, total days: 8
7 days after that, you have 3 Mondays in your month, total days: 15
7 days after that, you have 4 Mondays in your month, total days: 22
7 days after that, you have 5 Mondays in your month, total days: 29

The next step would lead to total days: 36, so: No, you can at most have 5 Mondays in a month.

In the above, replace Monday with your favorite day of the week and you'll see Mondays are not alone in this, the same goes for Tuesdays, Wednesdays,Thursdays, Fridays, Saturdays and Sundays.

Mathematically, no it can't.

Practically, however, yes it can.

The calendar is defined by law, and there is precedent for the calendar being changed. In September 1752, Great Britain and the British Empire switched from the Julian to the Gregorian calendar, thereby skipping 3-13 September. This month only had 2 Mondays.

If dates can be removed by a law, they can also be added by law. If a country decided to switch the other way, right now, having already had 4 Mondays this month (today is 27 November 2015, Gregorian), we'd jump back to the 14th, and have two more Mondays (Julian and Gregorian weekdays are the same - Today is Friday the 14th Julian, Friday 27 Gregorian).

The Julian calendar is rarely used now, it drifts relative to the tropical year (that is, the day that the sun crosses the equator mid-autum(fall)/mid-spring keeps moving earlier and earlier).


If G has a composition series so does H and G/H Let $G$ be a group having a composition series and $H$ a normal subgroup of $G$. Show that there exists a composition series in which one term is $H$.

My idea is to use the Schreier refinement theorem.

Let $G=G_0 \triangleright G_1 \triangleright G_2 \triangleright \dots \triangleright G_m = \{ e \}$ be a composition series for $G$. Now consider $G\triangleright H \triangleright \{ e\}$, if this is a composition series we are done.

If not, by the Schreier refinement theorem there exists equivalent refinements of these two series. Call these refinements $R_1$ and $R_2$, where $R_1$ is the refinement of the composition series. Now because these two refinements are equivalent the factors of $R_2$ are isomorphic to factors of $R_1$. Now because $R_1$ is a refinement of a composition series this means all of its factors are simple groups. Thus all the factors of $R_2$ are simple. Thus if we remove groups from $R_2$ in a way such that we eliminate trivial factors this should give us a composition series for $G$ that contains $H$.

Does this solution seem correct?

Thanks for the feedback!

I don't see why you want to remove factors of $R_{2}$. Anyway, you have $R_{2}$ which is a subnormal serie that contains H, where all the factors are simple as you said.
There is a theorem which states that a subnormal serie is a composition serie iff all its factors are simple. So we have immediately that $R_2$ is a composition serie.

The proof of the the theorem used is easy:
Let be $G \trianglerighteq G_{1} \trianglerighteq ... \trianglerighteq 1$. Its factors are $G_{i}/G_{i+1}$.

$G_{i}/G_{i+1}$ simple $\Leftrightarrow$ $\nexists H/G_{i} \trianglelefteq G_i/G_{i+1}$ $\Leftrightarrow$  $\nexists G_{i+1} \vartriangleleft H \vartriangleleft G_i $ which means that I can't make a proper refinement. For the second iff the third isomorphism theorem it's used.

Any normal series can be refined to a composition series, so this gives that any normal subgroup fits into a composition series of $G$. Taking the tail including $H$ gives a composition series for $H$, and taking quotients above $H$ gives one for $G/H$.


Solve the given integral. Find:
$$
\lim_{n \to \infty} n^{2} \int_{0}^{1} \frac{dx}{(1+x^{2})^{n}}
$$

This is indeterminate form $\infty \times 0$.

How can I initiate the problem.
Just need an hint.

The same limit can also be written as

$$
\lim_{n \to \infty} \frac{\int_{0}^{1} \frac{dx}{(1+x^{2})^{n}}}{\frac{1}{n^2}}
$$

And, then this becomes a $0/0$ form of limit so now we can apply L-Hospitals Rule, but just take care about the fact that differentiation has to be carried out with respect to $n$ because the limit is applied to $n$ and correspondingly to the entire expression. So, the differentiation with respect to $n$ can be taken inside the integral and then can be differentiated.
Hope this helps you !


$n$ balls are thrown randomly into $k$ bins - how many are empty? A large number of variants of this question were already asked here, including these - one, two, which are close, but none seem to answer my question.
Assume that $n$ balls are thrown randomly and independently into $k$ bins.

What is the probability of finding $x$ empty bins?
What is the expectation of the number of empty bins?

We do the expectation, without finding the distribution. Let $X_i=1$ if Bin $i$ is empty, and let $X_i=0$ otherwise. Then the number of empty bins is $X_1+\cdots+X_k$, and the expected number is $E(X_1)+\cdots+E(X_k)$.

The probability Bin $i$ is empty is $\left(\frac{k-1}{k}\right)^n$. Thus $E(X_i)=\left(\frac{k-1}{k}\right)^n$. Multiply  by $k$ for the expected number of empty bins.

The OP is more concerned with the first question, what is the probability of $x$ empty bins.

Let me try. Start with $x=1$. Thus, all remaining $k-1$ bins are NOT empty. The total number of put $n$ balls into $k-1$ bins is standard and many solutions online, the number is $C_{n+k-2}^{k-2}$. This includes the situations that some bins are EMPTY. So if every bin is NOT empty, thus the total number of ways is to put $n-(k-1)$ balls into $k-1$ bins (assume every bin has already 1 ball). It has $C_{n-1}^{k-2}$ ways. Thus, $$Prob(x=1) = n*\frac{C_{n-1}^{k-2}}{C_{n+k-2}^{k-2}}$$
The reason we multiply by $n$ is because of $C_n^1$ ways to choose which box is empty.

I believe other situations for $x$ can be derived in a similar way.


zeros of derivative and increasing function Let $f: \mathbb{R} \to \mathbb{R} $ be a differentiable function. Is it true that $f$ is strictly increasing on $\mathbb R$ if and only if $f'(x) \geq 0$ on $\mathbb R$ and the set $X = \{x \in \mathbb{R} \,| \, f'(x) = 0\}$ is countable?

Suppose $f$ is differentiable and strictly increasing. Then yes, we must have $f'\ge 0$ everywhere. However, we can have $f'=0$ on an uncountable set, in fact a set of positive measure.

Example: Let $U$ be an open set containing the rationals, with $m(U)<1.$ Let $E=\mathbb R\setminus U.$ Then $E$ is closed. Define

$$f(x)=\int_0^x d(t,E)\,dt.$$

Since $d(t,E)$ is continuous, $f'(x) = d(x,E)$ everywhere. We thus have $f'=0$ on $E$ and $f'>0$ on $U.$

Now $m(E)=\infty$ and hence is uncountable. The fact that $f'>0$ on the open dense set $U$ implies $f$ is strictly increasing, and we have our example.

Edit: the op has edited the question to strictly increasing from simply increasing thus my answer is no longer valid. I think this isn't true and here is why, Take $f(x) = -x^2$ when $x<0$, $f(x) = 0$ when $x \in [0,1]$ and $f(x) = x^2-1$ when $x>1$, this function is clearly continuous, differentiable and increasing (not strictly). Now for the set of points between [0,1] f'(x) = 0, but x is an element of the reals so any closed interval such as [0,1] is uncountable.


What will be set builder notation for this set? Following set is neither of even nor of odd so how can I express this by using set builder notation?

{0,3,6,9,12}

What you have there already is a form of set builder notation. If you’re asking how you can express it in set builder notation using some property or properties of the numbers in the set, notice that they are the multiples of $3$ between $0$ and $12$, inclusive:

$$\{0,3,6,9,12\}=\{n\in\Bbb N:n\le 12\text{ and }3\mid n\}\;,$$

where $3\mid n$ means that $n$ is a multiple of $3$, in case you’re not familiar with that notation. Another possibility is

$$\big\{3k:k\in\{0,1,2,3,4\}\big\}\;;$$

yet another is

$$\{n\in\Bbb N:\exists k\in\Bbb N(n=3k)\text{ and }n\le 12\}\;.$$

You can also write it in this way: $\{\; 3x \;|\; x = 0, 1, 2, 3, 4 \;\}$.


Is the limit $\lim_{x \to 1}{(2-x)}^{\tan(\pi/2)x}$ undefined? I'm preparing for a calculus exam and came across this limit:

$$\lim_{x \to 1}{(2-x)}^{\tan(\pi/2)x}$$

Of course $\tan(\pi/2)$ is undefined, but the excercise was a multiple choice with options $\infty$, 0, e, $e^\pi$ and $e^{2/\pi}$

So WolframAlpha says $\tan(\pi/2)$ is $\infty$, I figured this is a $1^\infty$ indeterminate and first tried to rewrite it like this:

$$
\lim_{x \to 1}{(1+(1-x))}^{(1-x)^{-1}(1-x)\tan(\pi/2)x}
$$

And I have $$\lim_{x \to 1}{(1+(1-x))}^{(1-x)^{-1}}=e$$

But then I got stuck trying to calculate $$\lim_{x \to 1}(1-x)\tan(\pi/2)x$$

I've also tried other methods but everything I try I end up stuck in a similar situation. Also, all the ways I can think of to calculate this limit seem too far-fetched (including the one above).

So, is the excercise wrong? If it's undefined, is it because of $\tan(\pi/2)$ or is there some other argument?

When it says $\tan(\pi/2)x$, it's meant to be interpreted as $\tan((\pi/2)x)$, not $(\tan(\pi/2))x$ (yes, it is terrible notation).  So the expression $(x-1)\tan((\pi/2)x)$ is a $0\cdot\infty$ indeterminate form as $x\to 1$ and you can try using L'Hopital's rule.  If it were $(\tan(\pi/2))x$, the expression would of course be undefined for all values of $x$, and so it would be meaningless to talk about the limit.

(If this confusion is not what you are actually asking about, apologies and I will delete this answer!)

You can do without taking logarithms too! Do it by writing the limit as follows
$$
\lim_{x\to 1}(1-x)\tan\left(\frac{\pi x}{2}\right) =\lim_{x\to 1} \frac{1-x}{\cot\left(\frac{\pi x}{2}\right)}
$$
and then apply L'Hopital's rule to get the required answer. It works: I wanted to upload a scan of the solution but the size of loadable images is limited to 2Mb.


Prove that R is a commutative ring. Suppose that $R$ is a ring. Prove that if $a\in R$, $a^2 = a$, then R is a commutative ring.

So, I know that this means that the ring multiplication is commutative. So... is this saying that for ANY $a\in R$, $a^2 = a$? Which means that every element of R is its own multiplicative inverse... But inverses, if they exist, are unique, so then the only element of R is unity. And if all you have is unity, then of course multiplication is commutative?

I'm probably way off here.

EDIT:

So, following the advice in the comments, I produced something that looks like this:

Since R is a ring, if $a,b\in R$, then $ab\in R$. Thus

$(ab)^2 = ab$

$\to (ab)(ab)=ab $

$\to (ab)(ab)a=aba $

$\to a(ba)(ba)=a(ba)$

$\to a(ba)^2=a(ba)$

So, AM I allowed to just say "by cancellation" $(ba)^2=ba$ therefore R is a commutative ring?

You need to prove that $ab=ba$ for all $a, b \in R$. I'm not sure I follow the approach that people are attempting to outline in the comments, and certainly the phrase "by cancellation" is very fishy. Here's the approach I know works:

Prove that $R$ has characteristic $2$ (or $1$, but that's pretty boring). This follows from $2^2=2$. Edit: Or, as pointed out in the comments, if you don't assume $R$ has unity, you can still show that $a+a=(a+a)^2 = a^2+a^2+a^2+a^2 = a+a+a+a$, witnessing $a+a=0$ for every $a \in R$.
Compare $(a+b)$ with $(a+b)^2$, which must be equal. Since you don't yet know that multiplication is commutative, you have $(a+b)^2 = a^2+ab+ba+b^2 = a+ab+ba+b$. Now subtract $a$ and $b$ to get $ab+ba=0$. In characteristic $2$, this is equivalent to $ab=ba$.

@Indigo i hope this is usefull to you,
for $a,b\in R$ , $a^{2}=a$ and $b^{2}=b$ , since $R$ is ring, $ab\in R$.
$ab=(a^{2})(b^{2})$  and  $ab=(ab)^{2}$
We get
$(ab)^{2}=(a^{2})(b^{2})$
$abab=aabb$
By cancelation
$ba=ab$


How find this value $\frac{a^2+b^2-c^2}{2ab}+\frac{a^2+c^2-b^2}{2ac}+\frac{b^2+c^2-a^2}{2bc}$ let $a,b,c$ such that
$$\left(\dfrac{a^2+b^2-c^2}{2ab}\right)^2+\left(\dfrac{b^2+c^2-a^2}{2bc}\right)^2+\left(\dfrac{a^2+c^2-b^2}{2ac}\right)^2=3,$$

find the value
$$\dfrac{a^2+b^2-c^2}{2ab}+\dfrac{a^2+c^2-b^2}{2ac}+\dfrac{b^2+c^2-a^2}{2bc}$$

is true?

Yes, I tink this problem can prove
$$(a+b+c)(a+b-c)(a+c-b)(b+c-a)=0$$

so $$\dfrac{a^2+b^2-c^2}{2ab}+\dfrac{a^2+c^2-b^2}{2ac}+\dfrac{b^2+c^2-a^2}{2bc}=1or -3$$
How many nice methods prove $$(a+b+c)(a+b-c)(a+c-b)(b+c-a)=0$$ ?

and  I  see this easy problem
http://zhidao.baidu.com/question/260913315.html

First, we simplify the initial equation by multiplying out by the denominator. Let

$$f(a,b,c) = c^2(a^2+b^2-c^2)^2 + a^2(b^2+c^2-a^2)^2 + b^2(c^2+a^2-b^2)^2 - 12a^2b^2c^2$$

In Ron Gordon's deleted post, he realized that $f(a,b,c) = 0 $ if $a+b=c$, $b+c=a$, $c+a=b$. This strongly suggests that $a+b-c, b+c-a, c+a-b$ are factors (sort of Remainder Factor Theorem). And indeed they are. We have

$$f(a,b,c) = -(a+b-c)(a-b+c)(-a+b+c)(a+b+c)(a^2+b^2+c^2)$$

which you can check in Wolfram.

Since the denominators are non-zero, we have $(a^2+b^2+c^2)>0$. Thus $$f(a,b,c) = 0 \Leftrightarrow (a+b-c)(a-b+c)(-a+b+c)(a+b+c) = 0 $$

We now split into cases.

Case 1. $(a+b-c)(a-b+c)(-a+b+c) = 0$

(Once again, multiply by denominators, and using Ron's observation.) Defining

$$g(a,b,c) = c(a^2+b^2-c^2) + a(b^2+c^2-a^2) + b(c^2+a^2-b^2) - 2abc $$

gives us

$$g(a,b,c) = - (a+b-c)(a-b+c)(-a+b+c) = 0 $$

Hence, the answer is 1.

Case 2. $a+b+c = 0$.

Then each term can be simplified in the form $\frac{ (a+b)^2 - c^2 - 2ab}{2ab} = -1$, hence the answer is -3.

I think you might be talking about a triangle, with $a$, $b$, $c$ as it's sides of a triangle. If the angles opposite to side a, b, c are A, B,C, then $$\cos A=\frac{b^2+c^2-a^2}{2cb}$$ and similarly, $$\cos B=\frac{a^2+c^2-b^2}{2ac}$$ and $$\cos C=\frac{b^2+a^2-c^2}{2ab}$$ If it is true then $\cos A = \cos B = \cos C =1$ from the first term (since minimum and maximum of cos x with any x is plus minus 1). So the answer is either -3, -2,-1 ,1,2 or 3.

Form here, we are getting a variety of answers, but since here only one of the angle can be negative(obtuse), the answer could only be 1 or 3. (since the sum could only be 1+1-1 or 1+1+1).

Now since all three angles of the triangle cannot be 0, so they must be one angle 180 and the rest 0. So 1 is the only answer.

NOTE: The above is only true when a,b and c are angles of a triangle. And the above is the called the cosines rule of a triangle. Although the triangle discussed above is in fact a straight line, but it can be noticed that the sum of the two sees containing the 180 angle is equal to the third side.


Intuitively, why are the two limit definitions of $e^x$ equivalent? Thanks for reading!

Intuitively, why does...

$$\lim_{n \rightarrow \infty} \left(1+\frac{1}{n}\right)^{xn}=\lim_{n \rightarrow \infty} \left(1+\frac{x}{n}\right)^{n}=e^x$$

Note, I'm not asking why $e^x$ is one of the two limits. I understand the first limit, or at least I think I do.

In terms of continuous growth however (I don't just want a mathematical proof - the more intuitive the answer is the better), I'd like to understand why the two limits are equivalent!

Why is letting some principal amount grow by $\frac{1}{n}$ times its current value $xn$ times equal to letting that principal grow by $\frac{x}{n}$ times its current value $n$ times if we allow $n \rightarrow \infty$?

Thanks!

Forget that $n$ is an integer. For $x > 0$ we have

$$\lim_{y\to\infty} \left(1+\frac{x}{y}\right)^y = \lim_{y\to\infty} \left(1+\frac{1}{\frac{y}{x}}\right)^{x\cdot \frac{y}{x}}$$

If $y \to \infty$ then $z:= \frac{y}{x} \to \infty$ as well so change of variables gives that this is equal to $$\lim_{z\to\infty} \left(1+\frac1z\right)^{xz}$$
which is the desired expression.

I will say my first thought.

If fix $x$,  one can take $\exp (x)$ for "constant" intuitively. And
 it is known that for a real number $x'$ there are infinite sequences convergent to it, I mean there are many many sequences we do not construct (or notice) with the same limit $x'$,
so let $a_n = (1+\frac{x}{n})^n$, $b_n = (1+\frac{1}{n})^{xn}$, sequences $\{ a_n \} ,  \{ b_n \}$  are just two cases( we are familiar with ) among them.

Maybe there is nothing special comparing to the rest.


compute the sum of a series $$
\sum_{n=0}^\infty \frac{x^n}{n(n+1)}
$$
what is the sum of this?
I suspect this has something to do with $e^x = \sum x^n/n! $ but I don't know how to go from there

$$
\sum\limits_{n=0}^{\infty}x^n=\frac1{1-x}
$$
Integrate once to obtain
$$\sum_{n=0}^{\infty}\frac{x^{n+1}}{n+1}=-\log\left({1-x}\right)$$
Integrate again to obtain
$$\sum_{n=0}^{\infty}\frac{x^{n+2}}{(n+1)(n+2)}=\left( 1-x \right)\log  \left( 1-x \right)  +x$$
$$\sum_{n=1}^{\infty}\frac{x^{n+1}}{n(n+1)}=\left( 1-x \right)\log  \left( 1-x \right)  +x$$
$$\sum_{n=1}^{\infty}\frac{x^{n}}{n(n+1)}=\frac{\left( 1-x \right)\log  \left( 1-x \right)}x  +1$$

(Sum should be from $n=1$.)

$$\frac1{n(n+1)} = \frac1{n}-\frac1{n+1} $$

so that

$$\begin{align}\sum_{n=1}^{\infty} \frac{x^n}{n (n+1)} &= \sum_{n=1}^{\infty} \frac{x^n}{n} - \sum_{n=1}^{\infty} \frac{x^n}{n+1} \\ &= -\log{(1-x)} - \frac1{x} \sum_{n=1}^{\infty} \frac{x^{n+1}}{n+1}\\ &=-\log{(1-x)} - \frac1{x} \left [-\log{(1-x)}-x \right ]\\ &= 1+\left ( \frac1{x}-1\right )\log{(1-x)} \end{align}$$

Note that the radius of convergence is $1$, i.e., the sum only converges when $|x| \le 1$.


Existence of integers $a$ and $b$ such that $p = a^2 +ab+b^2$ for $p = 3 $ or $p\equiv 1 \mod 3$ **Eisenstein primes, existence of integers **

I am working in subring $$R = \{a + b\zeta : a,b \in \mathbb{Z}\}$$ of $\mathbb{C}$ where $\zeta = \frac{1 + \sqrt{-3}}{2} \in \mathbb{C}$. I want to show that there exist integers $a$ and $b$ with $p = a^2 + ab+ b^2$ where $a,b \in \mathbb{Z}$ and that $p =3$ or $p \cong 1 \mod 3$. In this problem, we have a norm defined by $$N(a+ b\zeta) = a^2 +ab +b^2.
$$ I have been stuck in this problem for a while now. I was wondering if anyone could give me a hint on how to find the existence of such integers. I have been thinking of using the norm somehow, like considering some $x \in R$ such that $N(x) = p$ and from there try to show something about $a$ and $b$, but I haven't really made any progress. Any help with this will be ggreatly appreciated!

If $p=3$ then $a=b=1$.

If $p\equiv 1\pmod 3$ then
$$(-3/p)=(3/p)(-1)^{(p-1)/2}=(p/3)(-1)^{(p-1)/2}(-1)^{(p-1)/2}=(1/3)=1$$

(Here, $(a/b)$ means the Legendre symbol).

Then, $-3$ is a square mod $p$. Assume that $m^2\equiv -3\pmod p$. Let $x=(m-1)(p+1)/2$. We see that
$$x^2+x+1\equiv 0\pmod p$$
or
$$(x-\zeta)(x+\zeta)=kp$$
Note that $\Bbb Z[\zeta]$ is an UFD. If $p$ were prime in $\Bbb Z[\zeta]$ then $p$ would divide $x+\zeta$ or $x-\zeta$. This is clearly impossible, so $p$ is not prime in $\Bbb Z[\zeta]$.

Then
$$p=(a+b\zeta)(c+d\zeta)$$

Taking norms
$$p^2=(a^2+ab+b^2)(c^2+cd+d^2)$$
and both factors in RHS must be $p$.

Example: Take $p=97$. We see that $26^2+3=679=97\cdot 7$. And $25/2\equiv 61\pmod {97}$ so $61^2+61+1=97\cdot 39$. Then $97$ divide $(61+\zeta)(61-\zeta)$. Since
$$\frac{61\pm\zeta}{97}\notin\Bbb Z[\zeta]$$
then $97$ can be factored in $\Bbb Z[\zeta]$. This method does not give the factorization, but we can try by brute force.
$$97=8^2+3\cdot 8+3^2$$

$a^2 + ab+ b^2 \equiv \{0_{a\equiv 0}|1_{a\equiv\pm 1}\} + \{-1_{\text{if }a\equiv -b\equiv \pm 1}|0_{a\equiv 0\lor b\equiv 0}|1_{b\equiv \pm 1}\} + \{0_{b\equiv 0}|1_{b\equiv\pm 1}\} \mod 3$.

If one of $a$ or $b$ is equiv to $0$ and the other isnt then the result is $0+0 + 1$ (or $1+0+1$)$\equiv 1 \mod 3$.

If $a \equiv b$ then the result is $\{0|1\} + \{0|1\} + \{0|1\}\equiv \{0|3\}\equiv 0 \mod 3$.

And if $a \equiv -b \equiv \pm 1$ the result is $1 -1 +1\equiv 1 \mod 3$.

So the result follows precisely when $a\not \equiv b \mod 3$.

What am I missing about the question?

If $p = 3$.  possibilities are $(\pm 1,\pm 1),(\pm 2, \mp 1)$.


Simple limit problem: $\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})$ While trying to help my sister with her homework she gave me the next limit: $$\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})$$

I know the conventional way of solving it would be (That's what i showed her):

$$\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})=\lim_{x\to2}\left(\frac{x+2-4}{x^2-4}\right)=\lim_{x\to2}\left(\frac{x-2}{(x+2)(x-2)}\right)=\lim_{x\to2}\left(\frac{1}{x+2}\right)=\frac14$$

But she gave me the next answer:
$$\begin{align}
\lim_{x\to2}(\frac{1}{x-2}-\frac{4}{x^2-4})&=\lim_{x\to2}\frac{1}{x-2}-4\lim\frac{1}{x^2-4}\\
&=\lim_{x\to2}\frac{1}{x-2}-4\lim_{x\to2}\frac{1}{x+2}\lim_{x\to2}\frac{1}{x-2}\\
&=\lim_{x\to2}\frac{1}{x-2}-4\frac14\lim_{x\to2}\frac{1}{x-2}\\
&=\lim_{x\to2}\frac{1}{x-2}-\lim_{x\to2}\frac{1}{x-2}\\
&=0
\end{align}$$

I actually couldn't explain her why is she wrong. Cause technically it looks fine. What am i missing?

The error was in the very first step. We can say that $$\lim_{x\to a}\bigl(f(x)+g(x)\bigr)=\lim_{x\to a}f(x)+\lim_{x\to a}g(x)$$ provided that both of the limits on the right-hand side exist. In this case--$f(x)=\frac1{x-2}$ and $g(x)=-\frac4{x^2-4}$ with $a=2$--neither of these limits exist.


Where does the Pythagorean theorem "fit" within modern mathematics? I am interested in how today's professional mathematicians view the Pythagorean theorem, in terms of how the theorem fits within the axiomatic framework of mathematics.  I often come across textbooks that define length by the Pythagorean theorem, so that the theorem is in essence a definition or axiom.  In more modern mathematics such as linear algebra, is the Pythagorean theorem generally just used as the definition of length?  Is it more conventional today to treat the Pythagorean theorem as a definition (or axiom) rather than a theorem?  Are there any modern proofs of the Pythagorean theorem that don't rely on Euclidean geometry (like a proof that utilizes linear algebra/the dot product, etc.)?

Length, in Euclidean geometry, is a relation, not a number. We say that two segments are of the same length if they are congruent to each other, and congruence is one of the undefined notions of Euclidean geometry. The Pythagorean theorem comes about as soon as you decide to map lengths to positive real numbers, and it does so because of how length and angles are related by the congruence and similarity axioms (e.g., how the Pythagorean theorem is proved).

Lengths, in vector spaces, can be anything that satisfies the axioms for a norm. But to have a Pythagorean theorem, you need a notion of perpendicularity of vectors, that is, an inner product, which exists only if the norm satisfies the parallelogram law. For example, the taxicab norm, which is given by $|(a,b)|=|a|+|b|$ does not come from an inner product and there is no Pythagorean theorem for that notion of length.

The natural question is why does the Pythagorean theorem in Euclidean geometry correspond to the Pythagorean theorem in a finite dimensional vector space over the reals with an inner product? (there is a theorem that states that in a finite-dimensional vector space there is only one inner product up to isomorphism, so we really can talk about THE Pythagorean theorem).

The answer is that the finite dimensional vector space over the reals with an inner product encodes the congruence and similarity axioms of Euclidean geometry. The basic exposition of (most of) this can be found in the first half of Chapter 2 of Emil Artin's wonderful book Geometric Algebra.

Take a geometry of points and lines for which the following statements are true:

Any two points determine a line.
Parallel postulate: for any point $p$ not on line $\ell_1$ there exists a unique line $\ell_2$ that passes through $p$ but doesn't intersect $\ell_1$.
Non-triviality: there exist $3$ non-collinear points.
Desargues' Theorem (the wiki article is horrible)

The first three statements allow you to define translations (transformations of the plane that send lines to parallel lines and have no fixed points), while the fourth allows you to a construct a field over which the space of translations is a $2$-dimensional vector space, and such that if you associate translations $OP$ and $OQ$ with $(0,1)$ and $(1,0)$, then any point $R$ has a translation $OR$ which can be written as $(a,b)$ with $a$ and $b$ in the field.

In this way, any affine Desarguesian plane (thing that satisfies statements 1, 2, 3 and 4) can be identified with a $2$-dimensional vector space over some field (and conversely any $2$-dimensional vector space over some field corresponds to an affine Desarguesian plane).

Alright. Now, Euclidean geometry satisfies Desargues' Theorem, so which field does it correspond to? Well, it turns out that the fact that the geometry is ordered (i.e., we have a notion of an ordering of points on a line) means that the field has to be ordered and thus is a subfield of the real numbers (this is where we need something like the continuity or Hilbert's completeness axiom which roughly state that things that should intersect do intersect, and imply that the base field is all of the real numbers).

Then the whole inner product shebang turns out to just be a codification of the various congruence axioms. The norm is given by choosing some vector to make a unit vector (segment) and then considering to which vector in the same direction other vectors (segments) are congruent to. The inner product is a matter of encoding the notion of angle between vectors as an inner product, i.e., as a function linear in both vectors, which essentially takes the notions of similarity, congruences and unit circles.

It turns out that you can do all of this in the reverse direction, and so the Euclidean plane really is a 2-d vector space over the reals with an inner product.

Addendum: a comment on symmetry groups that is too long to leave as a comment.

Every norm on a vector space has its own unit blob, that is, the set of vectors with norm less than one. Geometrically, blobs that correspond to a norm satisfy the following properties: they are convex, they are absorbing (every vector is a multiple of a vector in the blob), and they don't contain any lines through the origin. It is a theorem that any such blob corresponds to a norm.

As we know, the transformations of a vector space are linear transformations, which means that the symmetry group of an object in your vector space is going to consist of the linear transformations that bijectively send an object to itself. In other words, the group of symmetries of an object is the set of invertible linear transformations that fix that object.

Now, the blob of the Euclidean norm is the unit disk, and the unit circle is the boundary of the disk. The unit circle is the set of vectors with Euclidean norm $1$, and in general, the boundary of the unit blob for a norm will be the set of vectors with norm $1$. The group of symmetries of the boundary of the blob then will be the set of invertible linear transformations that send vectors with norm $1$ to vectors with norm $1$.

A slight caveat is that you don't want to consider just any old invertible linear transformations because when you add a norm to a vector space, you are in effect defining a topology, that is, a way to define continuity. As a result you want to work with continuous linear transformations. BUT! In a finite dimensional vector space, no matter what the norm, we have that every linear transformation is continuous.

What this means is that for any norm on, say the $2$-d vector space over the reals, the symmetry groups of the boundary of the unit blob are all invertible linear transformations of the same space. Hence, we can directly compare these groups to each other since they live in the same space (of invertible linear transformations of $\mathbb R^2$).

Example: the boundary of the unit blob for the Euclidean norm is the unit circle, i.e., all points such that $x^2 + y^2=1$. The boundary of the unit blob of the Taxicab norm is the unit diamond, i.e., all points such that $|x|+|y|=1$.

Now if you think about it, any linear transformation that sends the unit diamond to itself will also send the unit circle to itself. Hence, the symmetry group for the Euclidean norm contains the symmetry group for the Taxicab norm, and in fact it contains it properly since, for example, rotation by $45$ degrees fixes the circle, but doesn't fix the diamond.

It all has to do with non-euclidean geometry.  The basic idea is that the distance function, or to say the same thing as a professional mathematician, the "metric", is super important in geometry.  You can have any kind of distance function you like, as long as it has three key properties (look them up).  This is a pretty big subject, and it all goes back to the independence of Eulcid's 5th axiom on parallel lines.

The distance between two points depends on the curvature of the underlying space, that's the key idea.

It's not possible to prove that the distance between two points is given by the Pythagorean theorem; because the actual distance between two actual points is the domain of physics, not axiomatic mathematics.  The best we can do in mathematics is define what we mean by distance (with a distance function), and since it's a definition, there's nothing to prove.

The story goes that Gauss realized Euclid's 5th postulate was independent because there are many geometries where it fails, e.g. on a circle or hyperboloid.  But Gauss didn't want to just tell the world he had solved the oldest problem in math, so what he did was when Riemann came to him looking for a job as a professor, he told Riemann to list 3 possible topics to give a talk on as a sort of interview for the job.  So Riemann wrote two topics down, but he couldn't think of a good third topic, and he put down "foundations of geometry" even though he hadn't thought very hard about it, just because he had some nice idea's which he had not fully formulated.  And since Gauss already been thinking about that, he picked the foundations of geometry; and the rest is math history.  Riemann went home and came up with a theory significantly beyond what Gauss had already known.


Derivative of the matrix exponential with respect to its matrix argument I was trying to find the Frechet derivative of $f = \exp(X)$, where $X \in \mathbb{R}^{n\times n}$ is positive definite. I thought it ought to be $\exp(X)$.

I see results where the derivative is with respect to a scalar argument, but this question has not been asked before.

I tried to see if I could find $Df_X$ starting with
$Df_X[H] = \exp(X+H) - \exp(X)$.  If I can show that the right hand side evaluates to $I + XH + X^2H/2 + \ldots = \exp(X)H$, I am done.

After I use the series definition, however,  I am lost  because I see no reason to  assume that $A$ and $H$ commute.

Please help.

EDIT
Following the suggestion in the comment, I try to compute the Gateaux derivative as $\exp(X + tH)$ by writing down the first few terms.

$\exp(X+tH) = I + (X + tH) + (X^2 + tXH + tHX + t^2H^2)/2 + \cdots$

$\dfrac{d}{dt}\exp(X+tH)\Big|_{t=0} = H + (XH+HX)/2 + \cdots$

And now am stuck again. It seems the expression on the right cannot be rearranged to give what I want.
I think it  is the derivative of the trace of the exponential, not the exponential itself that yields $\exp(X)$

Let $f : \mathbb R^{n \times n} \to \mathbb R^{n \times n}$ be defined by $f (X) = \exp(X)$. Hence,

$\begin{array}{rl} f (X + h V) &= \exp(X + h V)\\ &= I_n + (X + h V) + \frac{1}{2!} (X + h V)^2 + \frac{1}{3!} (X + h V)^3 + \cdots\\ &= I_n + X + h V + \frac{1}{2!} X^2 + \frac{h}{2!} (X V + V X) + \frac{h^2}{2!} V^2 + \frac{1}{3!} X^3 + \\ &\,\,\,\,\,+ \frac{h}{3!} (X^2 V + X V X + V X^2) + \frac{h^2}{3!} (X V^2 + V X V + V^2 X) + \frac{h^3}{3!} V^3 + \cdots\\ &= f (X) + h \left( V + \frac{1}{2!} (X V + V X) + \frac{1}{3!} (X^2 V + X V X + V X^2) + \cdots\right) + \cdots\end{array}$

Thus, the directional derivative of $f$ in the direction of $V$ at $X$ is given by

$$\begin{array}{rl} D_V f (X) &= \displaystyle\lim_{h \to 0} \frac{1}{h} \left( f (X + h V) - f (X) \right)\\ &= V + \frac{1}{2!} (X V + V X) + \frac{1}{3!} (X^2 V + X V X + V X^2) + \cdots\end{array}$$

We then write

$$D_V f (X) = M_0 + \frac{1}{2!} M_1 + \frac{1}{3!} M_2 + \frac{1}{4!} M_3 + \cdots$$

where

$$\begin{array}{rl} M_0 &= V\\ M_1 &= X V + V X =: \{X,V\}\\ M_2 &= X^2 V + X V X + V X^2\\ &\vdots\\ M_k &= \displaystyle\sum_{i=0}^k X^{k-i} V X^i\end{array}$$