What's the probability of "at least" and "exactly" one event occurring? If I know the probability of event $A$ occurring and I also know the probability of $B$ occurring, how can I calculate the probability of "at least one of them" occurring?

I was thinking that this is $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and }B)$.

Is this correct?

If it is, then how can I solve the following problem taken from DeGroot's Probability and Statistics:

  If $50$ percent of families in a certain city subscribe to the morning newspaper, $65$ percent of the families subscribe to the afternoon newspaper, and $85$ percent of the families subscribe to at least one of the two newspapers, what proportion of the families subscribe to both newspapers?

In a more mathematical language, we are given $P(\text{morning})=.5$, $P(\text{afternoon})=.65$, $P(\text{morning or afternoon}) = .5 + .65 - P(\text{morning and afternoon}) = .85$, which implies that $P(\text{morning and afternoon}) = .3$, which should be the answer to the question.

Is my reasoning correct?

If it is correct, how can I calculate the following?

  If the probability that student $A$ will fail a certain statistics examination is $0.5$, the probability that student $B$ will fail the examination is $0.2$, and the probability that both student $A$ and student $B$ will fail the examination is $0.1$, what is the probability that exactly one of the two students will fail the examination?

These problems and questions highlight the difference between "at least one of them" and "exactly one of them". Provided that "at least one of them" is equivalent to $P(A \text{ or } B)$, but how can I work out the probability of "exactly one of them"?

You are correct.

To expand a little: if $A$ and $B$ are any two events then

$$P(A\textrm{ or }B) = P(A) + P(B) - P(A\textrm{ and }B)$$

or, written in more set-theoretical language,

$$P(A\cup B) = P(A) + P(B) - P(A\cap B)$$

In the example you've given you have $A=$ "subscribes to a morning paper" and $B=$ "subscribes to an afternoon paper." You are given $P(A)$, $P(B)$ and $P(A\cup B)$ and you need to work out $P(A\cap B)$ which you can do by rearranging the formula above, to find that $P(A\cap B) = 0.3$, as you have already worked out.

probability of only one event occuring is as follows:
  if A and B are 2 events then probability of only A occuring can be given as
P(A and B complement)= P(A) - P(A AND B )


$\lim_{x \to 0}\sin(\frac{1}{x})$=? What is the limit:

$\lim_{x \to 0}\sin(\frac{1}{x})$?

I plotted this function on https://www.desmos.com/

And this was the result:

I searched the web for this function and it was straight forward stated as "The limit doesn't exist" without any theoretical proof. Could anyone provide a theoretical proof or something a bit more concrete?

Recall that when a limit

$$\lim_{x\to x_0} f(x)=L$$

exists it is unique and it is the same for all the subsequences, that is

$$\forall x_n \to x_0 \implies f_n=f(x_n) \to L$$

Therefore to prove that a limit doesn't exist it suffices to show that at least two subsequences exist with different limit.

In this case let consider

$$x_n=\frac2{\pi n}\to 0^+$$

then

$$\sin\left(\frac{1}{x_n}\right)=\sin\left(n\frac \pi 2\right)$$

What can we conclude form here? (try for example with $n=4k$ and $n=4k+1$)

Let’s look at half the limit: approaching from the right, say. Note that $1/x\to\infty$ as $x\to 0^+$, so we may rewrite the limit as

$$\lim_{x\to 0^+} \sin(1/x) = \lim_{x\to\infty} \sin(x).$$

Can you see why $\sin(x)$ doesn’t have a limit in this case?


Another interesting task connected with De Bruijn sequence Rule "zero is better than one":

Consider a sequence which is made up only of elements from the set {0, 1} and constructed regarding to the following rules:

1) It starts with n ones.

2) We put one at the end of the sequence if and only if by putting zero we get a subsequence which length is n and it has been already written.

3) If there is no turn that produces a new subsequence which length is n we stop.

Prove that after doing this algorithm we will get De Brujin sequence with parameters n and k = 2.

Note: Subsequence is a continuous part of the sequence.

So when I first started playing with De Bruijn sequences, before I even knew that name or that anyone else had played with them, that was the first way I found that produced them. My grandfather (who also had not heard of them but likes these kinds of puzzles) and I had tried to write up a proof that it worked. By the time we had finally gotten one I had gone onto making a more general solution to the problem, and I can't seem to find where the proof we made went, but I will try to piece it back together again.

By the way I am sure this proof is super redundant and inefficient, but whatever. It made more sense when trying to expand to other algorithms that produced any De Bruijn sequence, which is what we did later.

The first major thing we wanted to prove was that this method never produced a repeat, which is kind of redundant here since you just went out and required that you add in a manner that skips repeats, whereas our first method required counting the times you saw a certain subsequence of length n-1.

The second thing we wanted to prove was that it ended at some point, which it must since there are only finite possible subsequences that one can produce without repeats within the finite n and k given.

The third thing, which here is the only non-trivial thing to prove, was that it created every possible subsequence of length n in base k. We did this by first establishing that the last n-1 digits of the sequence must all be ones because the sequence cannot repeat, and the last n-1 digits must have occurred twice before to ensure that no digit could be placed at the end. Then because the last n-1 digits must have occurred twice before they must be the same as the first n-1 digits. this is because if they were not the first n-1 digits, then on two other occasions they must have had a digit preceding them and then again at the end, this would mean 3 different digits must have proceeded them, and since only 2 are present in this system, it is impossible for any n-1 digits other than the first ones to be present at the end of the sequence.

Once it is established that the last n-1 digits are all ones it is possible to prove the number of n length sequences, and based on the fact there are no repeats, that the sequence is a complete De Bruijn sequence. each n-1 length subsequence ending in a one, other than the first, must be preceded by a digit. since two n-1 sequences composed of all ones other than the first have been shown to exist (I did not prove this here but it is trivial at this point) one can know that two different digits precede n-1 length strings of ones. that gives us 2 n length strings, let us focus on the one that is not the first n length string, so the one beginning with a 0. if we look at it's first n-1 digits one can know that the same subsequence occur somewhere else, since we already know that for a n-1 subsequence to be followed by a one, it must have been followed by a zero earlier (again not proven here but trivial by initial conditions) that way you have two new subsequences to perform the process on. You repeat the process of looking at the first n-1 digits of each subsequence, finding it's duplicate, seeing that each must have a different number preceding them, counting those n length subsequences and duplicating. Once a sequence ends in a 0, all the others you find in that step must also end in 0's (because each step decreases the number of 1's at the end by one) since you started with n-1 ones and 2 n length subsequences, and each time you perform these steps you double the number of n length subsequences you have to double n times (because each time you have subsequences that you find the first part of, find the copy, there of, getting an additional sequence for each sequence if you consider what follows, and then double again by looking at the part preceeding it. It does not actually double twice each time because the first and second doublings overlap between steps, but in the final step they don't getting and extra double). This gives you 2^n n length sequences, which is how many their should be in a De Bruijn Sequence for base 2.

Sorry this is a little roundabout and complex at the end, I need to clean up my language but this is mostly from memory and not looked over for good and consistent wording.


Find the function given the equation $(g \circ f)(x) = x + 5$ Let $g(x) = 4x + 1$.

Find $f(x)$, given that the equation $(g \circ f)(x) = x + 5$ is true for all values of $x$.

From my text, it looks like this would be expressed as $g[f(x)]$, but I don't see how you can determine the definition of $f(x)$. Do I need to assume a value of x and work backwards through the problem?

$(g\circ f)(x) = g(f(x))$.

So you know that $g(a) = 4a+1$, no matter what $a$ is. I particular, that means that
$$(g\circ f)(x) = g\bigl(f(x)\bigr) = 4(f(x)) + 1.$$
Now, you want $(g\circ f)(x) = x+5$, so that means that whatever $f(x)$ is, you want
$$4(f(x)) + 1 = x+ 5.$$
Can you figure out what $f(x)$ should be for this to be true?


Trig Substitution Integral Question My class is going over trig substitution, but I can't figure this one out, mostly because it's not in the correct form. Could someone help explain how to set up this problem?

$$ \int \frac {1}{x^2-25} dx $$

You don't need trig here: use partial fractions.

The integral
$$ \int \frac {1}{x^2-25} dx $$
isn't in the correct form yet, but you can manipulate it easily
$$ \int \frac {1}{x^2-25} dx =  \int \frac {1}{25((\frac{x}{5})^2-1)} dx = \frac{1}{25}\int \frac {1}{(\frac{x}{5})^2-1} dx$$
now, just substitute $u=\frac{x}{5}, du=\frac{dx}{5}$
$$ \frac{1}{5}\int \frac {1}{u^2-1} du$$
continuing to manipulate...
$$ \frac{1}{5}\int \frac {1}{u^2-1} du = \frac{1}{5}\int \frac {1}{\sec^2(\sec^{-1}u)-1} du=\frac{1}{5}\int \frac {1}{\tan^2(\sec^{-1}u)} du$$
From here on, it's just an inverse trigonometric substitution (albeit a messy one because it involves $\mid u \mid$). It would be far better to use inverse hyperbolic trigonometric substitution in this case, or as Robert Israel has already pointed out, partial fractions.


how to generate tessellation cells using the Poincare disk model? I'm a computer programmer, and while I like math, this is an area where my understanding of math falls short of what I need in order to apply it successfully.

I've been looking at M.C. Escher's "Circle Limit" drawings, which use a Poincare disk model to illustrate tilings of the hyperbolic plane. What I want to do is generate the coordinates (in the Cartesian plane, for a graphics display) of vertices in such a tiling. But I'm not even sure where to start. It seems like I need two things:

a way to generate the coordinates in the hyperbolic plane, for each vertex of several cells (polygons) in such a tiling; and
the formula to convert those coordinates to the Cartesian plane, using the Poincare Disk model.

For example, in Circle Limit I, each fish seems equivalent to a quadrilateral (four other fish touch its edges), and each vertex is surrounded by either 4 or 6 fish. I don't need to generate fish, just quadrilaterals. But I don't know how to do it.

Do you know of software to do this? Or can someone help me with an algorithm?

I want to generate a finite area of the plane (of course?). If an algorithm could help me generate $k$ rings of cells around the origin, that would be most convenient.

Thanks for any suggestions.

P.S. I'd also be interested in hyperbolic tilings besides Escher's, e.g. http://en.wikipedia.org/wiki/Heptagonal_tiling

P.P.S. I just found some source code for drawing hyperbolic tilings (see bottom of the page), which includes generating locations of vertices. I can extract these vertex coordinates by putting in print statements. But they're grouped by line segment rather than by cell, and converting from the former to the latter does not seem easy.

The most helpful solution I've found for my need -- cranking out the locations of vertices of each polygon -- is David Joyce's Hyperbolic Tessellations applet and its source code. This applet draws regular and quasiregular tessellations organized by polygons, so it's easy to put print statements into the update() method and output the locations of vertices of each polygon as the polygon is drawn. It's much less efficient than Hatch's code, but efficiency is not one of my requirements at this point. Clarity is much more important.

I found the explanations in Ajit Datar's master's thesis the most helpful for learning how the process of generating the tessellations works. There is also code that goes with it, which I have not yet looked at. Datar's program offers more flexibility than Joyce's applet: e.g. Datar's can build tessellations either with a polygon centered at the origin, or with a vertex at the origin; and can also tessellate "motifs" (polygons or polylines such as Escher's fish).

Hop onto Wikipedia, and look up "Möbius Transformations".

If you tile a flat plane with squares, a particular group translations and rotations leaves the pattern unaltered, right? That group is the symmetry group. If you tile the Poincaré circle with whatever, the symmetry group is made up of Möbius transformations.

Or:

On the Poincaré disk, a straight line is a circle that intersects the edge of the disk at right angles. Also, angles are preserved - if two "lines" are at 60 degrees, then they stay at 60 degrees even if you move the focus around. So given a line, a point on the line, and an angle, there's only one circular arc that's a "line".

Or:

Fool with the half-plane model first.


Proof that $2\sqrt2 + \sqrt7$ is an irrational number Prove that $2\sqrt2 + \sqrt7$ is an irrational number. I am trying to use contradiction to show that this is irrational. Also I am using the fact that $ 2\sqrt2 + \sqrt7 = \frac{1}{2\sqrt2 - \sqrt7}$.

Assume  $2\sqrt2 + \sqrt7 $ is rational. Then

$2\sqrt2 + \sqrt7 = \frac{m}{n}$

and

$2\sqrt2 - \sqrt7 = \frac{n}{m}$

if I add them together I get

$ 4\sqrt2 = \frac{m^2+n^2}{mn} $

then we can divide the 4 on both sides and label the new numerator and denominator as $ x , y$ then to prove $\sqrt2$ is irrational is trivial. I am not sure if this is the correct way of proving this. Like why does it work to add it's reciprocal? Is it because we are assuming it is rational and if it is rational so is its reciprocal? And rational numbers are closed under addition?

The proof works because if your number is a rational $\frac mn$, then its reciprocal, $\frac nm$ is also rational: it is the quotient of two integers. (The only way that it wouldn't be rational is in the case $m=0$, but that would imply that $2\sqrt2 = -\sqrt7$, which is clearly false.)

Similarly, when you add $\frac mn$ to $\frac nm$, you get another rational number, $\frac{m^2+n^2}{mn}$.

From

$$2\sqrt2 + \sqrt7 = \frac{n}{m}$$ and $$2\sqrt2 - \sqrt7 = \frac{m}{n},$$ by addition, you get

$$4\sqrt{2}=\frac{m}{n}+\frac{n}{m}.$$ That is, $\sqrt{2}$ should be rational.

${}{}{}{}$


Mapping the Upper Half plane to the Unit Ball Consider the sets $\mathbb{H} = \{ \ (x_{1}, ..., x_{n}) \in \mathbb{R}^{n} \ | \ x_{n} \geq 0\ \}$ and $B_{1}(\mathbf{0}) = \{ \ \mathbf{x} \in \mathbb{R}^{n}\ | \ ||\mathbf{x}|| \leq 1 \ \}$.

Does there exist a smooth homeomorphism $\alpha : \mathbb{H} \to B_{1}(\mathbf{0})$?

I don't think that there is.

Yes, there is. Think inverse stereographic projection. (this is for the open ball, and open half-space; misread the question).


Semisimple modules and short exact sequences Though I suspect that this is a folklore result, I've been unable to find a proof by digital book searching. I also mention (my apologies) that I am new to noncommutative algebras theory and I am not familiar with the literature.

  Let $0 \to M \to N \to P \to 0$ be a short exact sequence of left $A$-modules ($A$ some arbitrary ring). If $M$ and $P$ are semisimple, prove that $N$ is semisimple.

Closest I could find is the (well-known) fact that submodules/quotients of semisimple modules are semisimple.

I would be grateful for any indication or relevant reference.

The sequence $$0\to R/(I \cap J) \to R/I \oplus R/J \to R/(I+J) \to 0    $$
is exact. Since $R/I$ and $R/J$ are semisimple, $R/I \oplus R/J $ is semisimple, and then $R/(I+J)$ is semisimple.


What is the difference between a formula and a proof? As I understand it, a formula is a method for solving a mathematical problem expressed using alpha numeric characters like the quadratic formula is a method for solving quadratic equations when factoring will not work. I understand a proof to be a logical argument that may or may not produce a formula, but will produce a statement that something is true or false mathematically.

Take for example:
$$x^2+13x+22=0$$
This equation will not factor, so we would use:
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
to solve for x.

However, does the fact that this formula always allows you to solve for x constitute a proof or "scientific proof" that this formula speaks to mathematical and thus scientific truth? Or, must a formula have a proof in order to really be considered valid scientifically?

The only way that I know a formula works is via a proof. I might be able to verify by hand that it works for a number of cases, but that doesn't mean it always works; see here and here for some examples of this. This is not to say that experimental evidence is worthless - quite the contrary. But the special role of a proof is something which cannot be ignored.

Now, there are subtleties here. In order to prove something, I need to begin with axioms. What axioms are "acceptable?" The standard axiomatic foundation of mathematics is ZFC (but see here), but there are some "concrete" problems which can't be proved using these axioms alone (see e.g. here). The existence of such problems - and Goedel's theorem more generally - shows that ultimately, the notion of "proof" is more nuanced than we might think at first. For example, there could be a formula that always "works" for a given concrete problem, yet can't be proved to always work inside ZFC.

However, this situation tends to be the exception rather than the norm. And the answer to your question is no - formulas and proofs are quite different!

Man, here is a good example of formulas and proofs being different. Imagine trying to find a formula for a 5th degree polynomial, you keep dialing away at your calculator making up tables of values. You punch and punch away until your fingers a sore looking for patterns, just to find out that someone along time ago by the last name Galois used logic and proof to prove that there is no formula for a 5th degree polynomial!


Find the number of terms in this AP The AP is defined as follows:

$$
7 + 9 + ... + (2n + 1)
$$

If I remember correctly the answer given was $n-2$.

But I thought that since $n$ was defined as the number of terms in the progression, the answer should always be $n$.

How can the answer be $n - 2$ (or in fact anything other than $n$).

Thanks in advance for the help

Find out the common difference denoted by $d_1$.

Let the first term be represented by $A$ and let the last term be represented by $L$.

Calculate number of terms $T$ using the formula :
$T = \frac{L-A}{d_1} + 1$. We thus get $T = n-2$ using $L = 2n+1$ and $A= 7$.

As to why the answer is $n-2$, one can explain it in terms of the direct consequence of the formula. It is not always necessary that the number of terms in a sequence should be $n$. Your sequence is that of the odd numbers starting from $7$ and not from $3$. The latter case would have given you $n$ terms. Hope it helps.


Lindeberg CLT application Let $X_1,X_2,\dots$ be a sequence of independent Random Variables with
$$\mathbb{P}(X_k = -1) = \mathbb{P}(X_k = 1) = \frac{1 - 2^k}{2}$$
$$\mathbb{P}(X_k = 2^k) = \mathbb{P}(X_k = -2^k) = \frac{1}{2^{k+1}} $$

How can I show the CLT holds? Tried to verify Lindeberg's Condition without success.

First, use Borel-Cantelli Lemma to show that $\mathbb{P}[|X_k|>1,~\text{for some}~k\geq N]\leq 1/2^{N-1}$. Then, use the distributions of $\frac{S_n}{\sqrt{n}}$ and $\frac{\tilde{S_n}}{\sqrt{n}}$, and show the difference between them. Here $\tilde{S_n}=\sum_{i=1}^n\tilde{X_i}$ and $\tilde{X_k}$ satisfies $\mathbb{P}[\tilde{X_k}=1]=\mathbb{P}[\tilde{X_k}=-1]=1/2$.

There are three steps to show that.

Step 1, for any given $N$, there exists a $M>N$, such that $\frac{|S_n-\bar{S_n}|}{\sqrt{n}}<1/2^{N-1}$ for any $n>M$ while $\omega\in\Omega\backslash\mathcal{A}_N$. Here $\mathcal{A}_N\triangleq\{|X_k|>1 ~\text{for some}~k\geq N\}$.

Step 2, notice that $\mathbb{P}[\mathcal{A}_N]\leq 1/2^{N-1}$. Then you get $\mathbb{P}[\frac{S_n}{\sqrt{n}}\leq x]\leq \mathbb{P}[\frac{\tilde{S_n}}{\sqrt{n}}\leq x+1/2^{N-1}]+1/2^{N-1}$, as well as $\mathbb{P}[\frac{S_n}{\sqrt{n}}\leq x]\geq \mathbb{P}[\frac{\tilde{S_n}}{\sqrt{n}}\leq x-1/2^{N-1}]-1/2^{N-1}$.

Step 3, let $n$ tend $\infty$, and notice that $N$ can be arbitrary natural number. Then you can obtain the conclusion.


Is it true that $f(z)=u(z,0)+iv(z,0)$ for complex $z$ and why? As in the title, i came across this equivalence
$$f(z)=f(x+iy)=u(x,y)+iv(x,y)=u(z,0)+iv(z,0)$$ while reading my notes on complex analysis, and tried to see why is it true, but i couldn't figure it out. Maybe it could be true only in the case that f is holomorphic.

As pointed out by many, as written this doesn't make much sense (unless $z$ is real, in which case it's trivial). However, I think what is trying to be communicated is a certain calculational trick.

Let's start with the example $f(z) = e^z$. Then $u(x,y) = e^x \cos (y)$ and $v(x,y) = e^x \sin (y)$, so if we now redefine $u,v$ using these same formulae as functions of two complex variables, we see

$$ u(z,0) + i v(z,0) = e^z \cos(0) + i e^z \sin(0) = f(z).$$

In general, for any analytic function

$$f(z) = \sum_{n=0}^\infty (a_n + i b_n) z^n,$$

with $a_n,b_n$ real, for real $z$ we have

$$ u(z,0) = \sum_n a_n z^n \; \text{ and } \; v(z,0) = \sum_n b_n z^n.$$

Reinterpreting these now as complex analytic functions (i.e. defining $\tilde u : \mathbb C \to \mathbb C : z \mapsto \sum_n a_n z^n$ and likewise $\tilde v$) we have $$f(z) = \tilde u (z) + i \tilde v(z).$$

Thus we get the trick: if you can write down $u(x,0)$ and $v(x,0)$ as some formulae in terms of elementary analytic functions, replacing $x$ with $z$ in $u(x,0) + i v(x,0)$ will give you the correct formula for $f(z)$ on the whole plane.

It suffices to check when $f(z) = z^n$. Observe
\begin{align}
(x+iy)^n =& \sum^n_{k=1} \binom{n}{k}x^{n-k}(iy)^k = \sum_{k \text{ even}}\binom{n}{k}x^{n-k}(iy)^k + \sum_{k\text{ odd}}\binom{n}{k}x^{n-k}(iy)^k \\
=&\ \sum \binom{n}{2\ell}(-1)^\ell x^{n-2\ell}y^{2\ell} + i\sum \binom{n}{2\ell+1}(-1)^\ell x^{n-2\ell-1}y^{2\ell+1} \\
=&\ u(x, y) + iv(x, y)
\end{align}
and observe
\begin{align}
u(z, 0) + iv(z, 0) = z^n.
\end{align}
Thus, when $f(z) = z^n$, we have $f(z) = u(z, 0) + iv(z, 0)$.

In the case when $f$ is not analytic the claim is false. For instance,
\begin{align}
f(z) = \bar z  = x-iy  \ \ \text{ but } \ \ u(z, 0)+iv(z, 0) = z \neq \bar z.
\end{align}


A parent has 2 children. Given that one is a boy, what is the probability that the other is a girl? (Look for specifics below) So, I've been browsing through the net about this (including this site) and I've stumbling upon different answer. In this site, most of the answer were 2/3 but the arguments here are that BG and GB are different since the order of birth are of consequence.

However, what if one would not consider the order of birth but only the existence of the said children? What is the probability that there is one boy and one girl in a family with a children of 2 given that one is a boy and having a boy as a child is as likely as having a girl as a child?

Same case as another post here, I argue that it is 1/2.

It is not as if I'm omitting the GB and/or BG, but rather, I'm combining them since order is not necessary/important in this scenario. BG = GB.  Making the original sample space (with their likeliness); { BB (25%), GB/BG (50%), GG(25%) }. And given the condition (one is a boy), the sample space then is reduced to {BG (25%), GB (25%)} as having a boy child is as likely as having a girl.

Is it an invalid assumption?

The sample space is not reduced to $\{BG (25\%), GB (25\%)\}$, the sample space is reduced to $\{BB (25\%), GB/BG (50\%)\}$, or equivalently to $\{BB (25\%), BG (25\%), GB (25\%)\}$. Reduced to this case, the probability of a girl as the other child is $2/3$.

How does the sex of the given child [boy] affect the probability of the 2nd child.  It doesn't.  It doesn't matter whether the first child is a boy  or a girl.  There are still only two possible outcomes for the sex of the unknown child:  B or G.  The probability is simply 1/2.  It's similar to this question:  A coin is flipped two times.  Given that the first flip is heads, what is the probability that the second flip is heads.  Thinking that what already happened will affect what happens next in events of chance is called the Gambler's Fallacy.  They could have 15 heads in a row, the probability of the 16th coin being tails is still 1/2.


How do we find the derivative of a multivariate expression wrt another multivariate expression? I have $f(\theta,b) = 1 - \theta x - b$.  I want to find $$\frac{df(\theta,b)}{d[\frac{1-b}{\theta}]}$$
i.e. the change in $f$ wrt $\frac{1-b}{\theta}$.  In the language of limits,
$$
\lim_{h \to 0, a \to 0} \frac{f(\theta + h, b + a) - f(\theta, b)}{\frac{1 - b - a}{\theta + h} - \frac{1 - b}{\theta}}
$$
I saw this How to find the derivative of one expression with respect to another expression
which made me think I could try
$$\frac{df(\theta,b)}{d[\frac{1-b}{\theta}]} = \frac{\frac{df(\theta,b)}{d\theta}}{\frac{d[\frac{1-b}{\theta}]}{d\theta}} = \frac{\theta^2 x}{1 - b}$$
But if I do the same strategy wrt the other variable, I get
$$\frac{df(\theta,b)}{d[\frac{1-b}{\theta}]} = \frac{\frac{df(\theta,b)}{db}}{\frac{d[\frac{1-b}{\theta}]}{db}} = \theta$$ which is a contradiction.

The thing about decomposing $\frac{df(\theta,b)}{dg(\theta,b)}$ into two separate expressions, and then calculate the product is a fundamentally flawed way to calculate multivariate derivatives. It does often work, but not all the time. This article goes into more detail about it.
To respond to your question, with
$$g(\theta,b)=\frac{1-b}{\theta}$$
we know that
$$\theta=\frac{1-b}{g(\theta,b)}$$
Therefore, we plug it into $f$ and get
$$f(\theta,b)=1-\frac{1-b}{g(\theta,b)} x-b$$
And so,
$$\frac{df(\theta,b)}{dg(\theta,b)} = \frac{(1-b)x}{g(\theta,b)^2}$$
which when plugging in for $g(\theta,b)$, gets us the answer you found first,
$$\frac{\theta^2 x}{1-b}$$


How to efficiently find all elements in $\langle a\rangle$ for all $a$ in $\Bbb Z / 48 \Bbb Z$ We were asked in my class to find all elements of  $\langle a\rangle$ for all $a$ in $\Bbb Z / 48 \Bbb Z$.

The answer can be found here, and I understand how the cyclic groups are formed. Rather than tediously writing out all 48 cyclic groups, I am wondering what the pattern is, if there is one. I see that the group $\Bbb Z/48\Bbb Z$ is generated by $\langle1\rangle,\langle5\rangle,\langle7\rangle,\langle11\rangle,\langle13\rangle,\langle17\rangle,\langle19\rangle,\langle23\rangle,\langle25\rangle$, etc.

I know that $\langle a\rangle=\langle a^{-1}\rangle$, so $\langle 1\rangle = \langle 47\rangle, \langle 2\rangle = \langle 46\rangle...$

I feel like a pattern is glaring right at me, but I would love to hear from others! Thanks.

It all depends on whether the number is coprime to $48$.  Any number coprime to $48$ will generate the whole group.  Any number not coprime to $48$ will generate only multiples of the $\gcd$ of the number and $48$.


Is it possible to integrate this function without using spherical coordinates? I am looking to solve the integral

${\int\int\int{\rm e}^{-\sqrt {{x}^{2}+{y}^{2}+{z}^{2}}}} dxdydz$

My boundary conditions are the boundaries of a cube (not nicely aligned with the origin), so I'd like to avoid muddying the waters with spherical coordinates.

Any advice?

If you're interested, I'm essentially trying to solve the following problem.

I'm holding a flashlight underwater in a cubic swimming pool, what's the total amount of light being absorbed by the pool, assuming that some light escapes the pool? Treat the flashlight as an isotropic point source of light.

Attenuation of a beam in 1D is given by a simple exponential decay.

$\int \exp{(- \sqrt{x^2 + y^2 + z^2} ) } dx$

$ u = x^2 + y^2 + z^2$

$ du = 2x dx$

$ dx = du/2x$

$\int (\exp{(- \sqrt{u} ) }/u)  du =2 Ei(-\sqrt{u}) + C$ with $Ei(x)$ being the Exponential Integral


Roots of a polynomial having arbitrary constant term Let $f_r(x)=x(x-1)(x-2)(x-3)+r$, where $r$ is a real number. Then

$(A)$ $f_r(x)$ has a real root only for finitely many values of $r$;
$(B)$ $f_r(x)$ has a real root for every value of $r$;
$(C)$ $f_r(x)$ can never have a repeated root;
$(D)$ $f_r(x)$ can have a repeated root only for finitely many values of $r$.

My attempt:
Considering a function $\phi(x)=f_r(x)-r$, then the all the roots of $\phi(x)$ are real and distinct and so $\phi'(x)$ is also so. Also $\phi'(x)={f_r'}(x)$. So the roots of  ${f_r'}(x)$ are all real and distinct, but we don't say that  roots of $f_r(x)$ are all real and distinct (since converse may not not true). I am stuck here.
Another way is by choosing different value of $r$ solve $f_r(x)$ and discard one by one option, but this is too broad.
The problem appears in ISI B-Math 2006 admission test. Thank's for your valuable time.

Hint: consider the shape when $r=0$: it's a quartic with four roots, at $0, 1, 2, 3$, with positive leading coefficient. What, then, does that quartic look like?

Now translate that shape vertically. How does the number of roots change as we do this?

Consider the equation without the constant. It has a plot that looks like this.

What happens when you add the constant $r$. It moves the graph up, i.e values of $f(x)$ at all $x$ are increased by $r$. Now how much can you increase or decrease  the $r$. It should be obvious from looking at the graph that you can move the graph down as much as you want and you will always have two roots. But of you move the graph up by more than $1$ then it all the roots start to disappear. So $r$ can't be larger than $1$.

If you reduce the value of $r$ the function comes down, it is possible to chose a value for $r$ such that the maxima point just touches the x axis. This means that $x=1.5$ can be a repeated root of $f(x)$ for one value of $r$. Hence the option $D$ is the correct one.


Approximating the area below average of a concave function Given a non-decreasing concave function $f:[0,1]\rightarrow \mathbb{R}^+$. Define

\begin{align*}
F(n)=\displaystyle\sum_{i=1}^n \min\left\{\frac{1}{n},\frac{f\left(\frac{i}{n+1}\right)}{n+1}\right\}
\end{align*}

We want to find a tight lower bound for $F(n)$ in terms of $n$, knowing that $\int_0^{1} f(x) dx =1 $.

For instance, I am able to prove that $F(1)\geq \frac{1}{2}$, $F(2)\geq \frac{2}{3}$, $F(3)\geq \frac{1}{3}+\frac{3}{8}$ and see that they are tight for the function $f(x)=2x$. Also, I can prove that $\displaystyle\lim_{n\rightarrow \infty} F(n) \geq 3/4$, and again observe that it is tight for the function $f(x)=2x$.
 Thanks for your time and attention!

$f(x)$ is non-decreasing. For given $n$, split the range of the sum at $n^*$,
where
$$
\frac{f(n^* / (n + 1))}{n + 1} = \frac{1}{n}
$$
Then the sum is:
$$
F(n) = \frac{1}{n + 1} \sum_{1 \le i \le n^*} f(i / (n + 1)) + (n - n^*) / n
$$
We want to minimize this, and this means keeping as much in the first range (which collects the small values) as we can. By the relation defining $n^*$ we see that it is enough to make sure that $f(x) < \frac{n + 1}{n}$ to have $n^* = n$ for all $n$. By the condition that $\int_0^1 f(x) dx = 1$, this means $f(x) = 1$, and the sum is just $F(n) = \frac{n}{n + 1}$.


Intermediate value property problem and continuous function How to prove that a strictly increasing function $f:[a, b]\rightarrow \mathbb{R}$ which has the intermediate value property is continuous on $[a, b]$.

Let $c$ be a point with $a<c<b$, and suppose given a small $\epsilon > 0$ (small enough that $c+\epsilon<d$ and otherwise arbitrarily small). Use the intermediate value property (and strict increasingness of $f$) to pick $d$ with $c<d<b$ for which $f(d)=f(c)+\epsilon.$  Since $f$ is strictly increasing, for all $x$ in $[c,d]$ we have $$f(c) \le f(x) \le f(d).$$ And $f(d)-f(c)=\epsilon.$ [Note: In the $\epsilon, \delta$ continuity definition, we have chosen $\epsilon$ first, and then found the point $d$, and are defining $\delta=d-c$.]  This given "half" of continuity at $c$, and the left half is similar to show.

At either endpoint $a$ or $b$ we only need the above argument one way.

You cannot prove that because it is not true, there are functions that are not continuous that have this property. Just think of a function that is strictly increasing but with discontinuities and that is defined in the hole interval that you will notice that it can still obey the property.


How does it make sense to rearrange equations like this to find a function Like for example I had $\dfrac{(g(x)-1)}{g(x)}=\dfrac x{(x+1)}$ and rearranged to get $g(x)=x+1$ involving multiplying through by $g(x)$.

Clearly $g(x)$ can equal $0$ but it doesn't seem to make sense to me to multiply through by $g(x)$ to get a solution when $g(x)$ can equal $0$ so how does it work and get the right solution. Thanks

edit: Sorry I didn't word this very well. It is the composition of two functions f(x) and g(x) is the solution I showed x/(x+1) which is f(g(x)) however I knew f(x) which was (x-1)/x and needed to find g(x) which is what I was trying to do. So then g(x)=x+1 would satisfy this. And g(-1) would be defined in this case. It is the 4th row down in this image link gyazo.com/adbd45de4445ac8600c6f9695d66fd29

Your solution is partially correct. If someone writes $(g(x)-1)/g(x)=x/(x+1)$, they are assuming $g(x)$ is non-zero and $x\neq -1$, so you must specify the domain has changed for $g$, that is, $g(x)=x+1, x\neq -1$. From an algebraic point of view, you can also just solve this formally.

I see two possible interpretations:

either the first identity is guaranteed to hold for all $x\ne -1$, then implicitly $g(x)\ne0$ and $g(x)=x+1$,
or the first identity holds when it is defined, then $g(x)$ is either $0$ or $x+1$.


Maximum Perimeter of a triangle inside a circle How would i find/prove the maximum perimeter of an equilateral triangle inside a circle:

$$x^2+y^2=4$$

ps. sorry for my bad english and bad question making

If you really need to use Calculus, try proposing three points $P_1,P_2,P_3$. This points have coordinates $(x_i,y_i)$, which satisfy the relation $x^2+y^2=2^2$.

You need to find the perimeter of the triangle formed by those points, so you want $Perimeter = d(P_1,P_2)+d(P_1,P_3)+d(P_3,P_3)$. Expanded is:

$$
Perimeter =\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\ +
\sqrt{(x_1-x_3)^2+(y_1-y_3)^2}\ +
\sqrt{(x_2-x_3)^2+(y_2-y_3)^2}.
$$

You can set $P_1=(2,0)$ without loss of generality, so above is simplified to:

$$
Perimeter =\sqrt{(2-x_2)^2+y_2^2}\ +
\sqrt{(2-x_3)^2+y_3^2}\ +
\sqrt{(x_2-x_3)^2+(y_2-y_3)^2}.
$$

You also know that both $P_2$ and $P_3$ need to satisfy the relation. Try first with $y_i^2 = 2^2-x_i^2$. Set it again in the above equation and rewrite:

$$
Perimeter =\sqrt{(2-x_2)^2+2^2-x_2^2}\ +
\sqrt{(2-x_3)^2+2^2-x_3^2}\ +
\sqrt{(x_2-x_3)^2+(2^2-x_2^2+2^2-x_3^2)+2\sqrt{(2^2-x_2^2)(2^2-x_3^2)}}.
$$

Now to try a long-shot of intuition, set $x_2=x_3=z$. Then:

$$
Perimeter =2\sqrt{(2-z)^2+2^2-z^2}\ +
\sqrt{2(2^2-z^2)+2\sqrt{(2^2-z^2)(2^2-z^2)}}
$$

You might simplify further if needed. Then differentiate with respect to $z$. You should get $z=-1$. This is the easiest Calculus way I can think of.

The equation $x^2+y^2=4$ describes a circumference of radius $\rho=2$ centered at $(0,0)$. If you have a equilateral triangle inscribed in this circle (which is unique up to rotations) then this means that $\rho=2r$, where $r$ is the inradius of your triangle; i.e. $r=1$. As in the following picture (borrowed from http://mathworld.wolfram.com/EquilateralTriangle.html)

The length $a$ of each side happens to be $\frac{6}{\sqrt 3}r$ so your answer is
$$
perimeter=3a=6\sqrt3.
$$
Notice that the up-to-rotation-uniqueness implies that every equilateral triangle incribed in this circumference will have the same perimeter.

If you don't restrict yourself to the inscribed case and consider a triangle inside the circle then you can maximize the perimeter maximizing the function
$p(r)=6\sqrt3r$, which will tell you that the biggest value is attained when $r$ is maximun; i.e. when $r=1$.


Show that $u$ and $v$ are solutions of the wave equation Suppose that $u = u(x, t)$ and $v = v(x, t)$ have partial derivatives as follows:
$$u_t=-v_x \quad\text{and}\quad v_t=-u_x.$$
Show that $u$ and $v$ are solutions of the wave equation:
$$u_{tt}=u_{xx}.$$

My solution effort:
taking derivative of $v_t=-u_x$ w.r.t $x$, $v_{xt}=-u_{xx}$, we get $u_{xx}=-v_{xt}$. Similiarly; $u_{tt}=-v_{xt}$. Then $u_{tt}=u_{xx}$.

Is this right?

If $v$ is a $C^2$ function, then $v_{xt}=v_{tx}$ (Schwaz's theorem) and you are right.


If $\sum a_n$ converges does it imply $\sum a_n^3$ converges? As title says does the converges of $\sum_{i=0}^\infty a_i$ imply that $\sum_{i=0}^\infty (a_i)^3$ converge?

If we assume $a_i\leq 0$ or $a_i\geq 0$ from some point on the statement is obviously true.
If we remove the crucial assumption about the sign, it might not. Let
$$ a_n = \frac{1}{\log^2(n+2)}\cdot\left\{\begin{array}{lcl}1&\text{if}&n\equiv 1\pmod{3}\\1&\text{if}&n\equiv 2\pmod{3}\\-2&\text{if}&n\equiv 0\pmod{3}\end{array}\right.$$
It it simple to check that $\sum_{n\geq 0}a_n$ is convergent, but $\sum_{n\geq 0}a_n^3$ is not, essentially by Kronecker's lemma (summation by parts).

$\textbf{Hint: }$ If
$$\sum_{n=1}^{\infty}a_n$$
converge, then $a_n\to 0$ as $n\to \infty$. Which means, there exists some $N\in \mathbb{N}$ such that $|a_n|<1$ for all $n\ge N$. Then $|a_n|^3=|a_n^3|<|a_n|$, and hence
$$\sum_{n=N}^\infty a_n^3$$
converge (the absolute convergence is necessary).


Properties of Equilateral Triangles in Circles If there is an equilateral triangle in a circle, would the midpoint of any of the 3 sides be half the radius?

e.g if the radius was 6 and at the midpoint of the triangle (call it B) would center to B be 3 and then B to circle be 3 as well?

Yes. If you're familiar with construction using compass and straight edge, one of the easiest ways to construct an equilateral triangle is to draw two circles where each circle's centre lies on the other circle's edge. Drawing a line between the two intersection points and then from each intersection point to the point on one circle farthest from the other creates an equilateral triangle. You can see from this construction that the side of the equilateral triangle between intersection points is equidistant from each centre, proving that the side is halfway between the circle's centre and its edge.

No. In that case, the angle of triangle must be $45^{\circ}$ because is isosceles.


What makes a theorem "fundamental"? I've studied three so-called "fundamental" theorems so far (FT of Algebra, Arithmetic and Calculus) and I'm still unsure about what precisely makes them fundamental (or moreso than other theorems).

Wikipedia claims:

  The fundamental theorem of a field of mathematics is the theorem considered central to that field. The naming of such a theorem is not necessarily based on how often it is used or the difficulty of its proofs.

So I am left wondering: what is the main criterion for a theorem to be considered fundamental?

Why, for instance, is there no "Fundamental Theorem of Complex Analysis" (say, de Moivre's formula, for example), or "Fundamental Theorem of Euclidean Geometry" (e.g. Pythagoras' theorem)?

Does anyone have a source of a more-rigourous definition of "fundamental", in this context?

Let's examine some of these so-called "fundamental" theorems.
($1$) The Fundamental Theorem of Arithmetic.

"Every natural number greater than $1$ has a unique representation as a product of primes."

This theorem about factorization establishes the primes as fundamental building blocks for studying numbers. This idea and the obsession with these numbers (who have an entire theorem named after them) which are the core building blocks of all integers has sparked multiple entire fields of research. Almost all problems about integers in Diophantine equations makes use of this theorem, because it makes proving an enormous body of results vastly simpler and in many cases makes them possible where direct proof is impossible. Most fields in mathematics would probably be crippled without this theorem. And because of the immense advantage of unique factorization (which not all settings have) number theorists were led to develop the field of algebraic number theory where this theorem finds its ultimate generalization in the context of Dedekind domains. In fact, to underscore just how good unique factorization is:  if it were true that every number ring had unique factorization, there are old proofs of Fermat's last theorem which would have given us the result ages ago, instead of making us wait 350 years!

($2$) The Fundamental Theorem of Algebra.

"Every polynomial of degree $n$ with coefficients in the complex numbers has exactly $n$ complex roots (counting multiplicity)."

Algebra has its roots in the study of polynomials (no pun intended). You learn about other topics in a high school course named "algebra," but good old polynomials were some of the first objects considered by mathematicians in this field. Ring theory uses polynomials extensively, group theory was created to deal with the problem of understanding solutions to polynomials. Field theory would not be a thing without polynomials. And the fact that a polynomials is essentially uniquely determined by its roots (in the classical case of $\Bbb C$ certainly) is--like the FTA from ($1$)--essential in huge numbers of proofs in the subject, and without which, again, it probably would not be so much of an object of study, and certainly would be hundreds of years behind in results proven without it.

($3$) The Fundamental Theorem of Calculus.

"If $f$ is a continuous, $\Bbb R$-valued function on an interval $[a,b]$ and for $a<x<b$ define
$$F(x) = \int_a^x f(t)\,\mathrm{d}t.$$
Then $F(x)$ is differentiable on $(a,b)$ and $F'(x) = f(x)$."

This theorem links the two biggest concepts in calculus:  derivatives, which measure rates of change, and integrals, which measure areas underneath curves--which have huge numbers of physical interpretations other than just the naive notion of "area." Before this theorem, computing integrals was a task of doing Riemann sums directly using hard limit arguments and a lot of work. This theorem revolutionized the field because it related integration to differentiation, a process which is much easier to perform because--for example--any function expressible in terms of elementary functions has an elementary function styled derivative. It destroyed excess computation time, and made the use of both tools much richer and formed the basis for the study of calculus in greater generality and in more applications. The FTC allows one to prove the change of variables formula and integration by-parts. The former, which is basically the chain rule in its derivative form, is a basic tool in proving many identities, which are the basic things in any mathematician's toolbox. The latter--Leibniz' product rule in derivative form--gives rise to the notion of a weak derivative and--for some very general notions about integration--allows one to rigorously explore ideas in areas such as partial differential equations, which are basically what describe most everything in the physical world. This idea founded several fields as well, including differential geometry and the idea of differential forms through Stokes' theorem, which is something that has helped us understand huge areas in mathematical physics and high-dimensional analysis, and has advanced society itself an enormous amount.

($4$) The Fundamental Theorem of Galois Theory.

Let $k$ be a field and $\overline{k}$ be its separable closure. Denote by $G_k=\text{Gal}\left(\overline{k}/k\right)$ the group of continuous field automorphisms of $\overline{k}$ which fix $k$ pointwise. Then there is an inclusion reversing bijection between subfields $k\subseteq K\subseteq\overline{k}$ and closed subgroups $H\le G_k$."

This is another theorem that launched a huge enterprise and--along with its applications--made Galois theory a topic which today finds applications in number theory, algebra, and beyond. The field itself is mostly settled now, but this, its main theorem has the staying power to make powerful advances in all the applications it has found, and has a central role in the Langland's program, a current area of research which is massive and global in scale.

Addendum (non "fundamental" fundamental theorems)
Here I'll note that there are fundamental theorems in all areas, not just the ones which have theorems that possess the word "fundamental" in the title. This also means that this is just works of my opinion, and not necessarily what everyone else might consider the "most fundamental" theorem to the topic.
($1'$) Demorgan's laws. (Fundamental Theorem of Logic)

"If we have two logical statements, $P,Q$. Then the negation of $P\land Q$, is $\lnot P\lor\lnot Q$, and the negation of $P\lor Q$ is $\lnot P\land\lnot Q$."

This set of relations, which fundamentally links logic to set theory--replace statements with sets, $\land$ with $\cap$, $\lor$ with $\cup$ and $\lnot$ with set complement--allows for a greater exploration of how one can do mathematics in a way which utilizes mathematics itself. It is one of the first theorems in logic, and provides a basis also for the idea of quantifiers and the duality which states $\lnot (\forall x : P)\iff \exists x : \lnot P$, which is an idea used in mathematical proofs as a technique time and again. Never mind content, which the other theorems thus-far address:  without this, we'd have trouble proving anything!

($2'$) The law of cosines. (Fundamental Theorem of [Analytical] Geometry)

"If three sides of a triangle have lengths $a,b,c$ and the angles opposite those sides are $A,B,C$ respectively, then the relation
$$c^2=a^2+b^2-2ab\cos(C)$$
holds."

I choose this one, because the Pythagorean theorem, while excellent, is not the one that best generalizes. This notion of using this extra term, $2ab\cos(C)$ leads to the idea of an inner-product, which is an idea exploited in functional analysis, and is a foundation for the study of linear algebra, a subject without which no professional mathematician would function. It is essential in areas such as functional analysis, applied mathematics, mathematical physics, and much more. It's a revolutionary idea which allows us to change from pure geometry to the idea of analytical geometry.

($3'$) The Central Limit Theorem and Law of Large Numbers. (Fundamental Theorems of Statistics)

(CLT) "If we look at a bunch of averages of samples from our population of the same size, then their distribution looks like a bell curve."
(LLN) "If we take the average of some observations, then--so long as we make enough of them--we will get closer to what the real average of all the numbers is."

Here I do not think either can supersede the other. These two theorems together, form the basis for statistical analysis. Without the Central limit theorem, it would be difficult to impossible to analyze averages of random variables in any sort of efficient way. It takes very general objects one could care about (usually interpreted as samples of data from a population) and allows us to make inferences with good confidence on our errors to better understand the world around us. Without it, civilization would not exist in this information age of the internet. Information is the basis for the kind of advanced understanding we have of objects and physical laws, and has allowed us to develop things like the internet, smart phones, cars, spaceships, and so much more. The LLN is what allows us to be confident in our use of data. That is:  without know much of anything about data we're dealing with (i.e. how it's distributed, which can drastically affect what we observe and what happens) we can still make good guesses about what will happen on average, which allows us safety to build on this information and achieve the same things as in the CLT.

($4'$) Tychonoff's Theorem and Urysohn's Lemma. (Fundamental Theorems of [Point Set] Topology

(Tychonoff) "A product of any collection of compact spaces is compact."
(Urysohn) "A topological space is normal if and only if two closed sets can be separated by a continuous function."

These theorems are two of the most fundamental and often-applied theorems from classical topology. The applications of Tychonoff's theorem top all other topological theorems by a huge margin. While not necessarily central to topology itself, the field as a whole is built more horizontally than vertically, and Tychonoff's theorem is one of the results which realizes the immense utility and ubiquity of topological methods. With the advent of the modern definition of compact as finite subcovers of open covers replacing the old notion from metric spaces, this result stands at a turning-point in the history of topology. Along with the modern definition of continuity, compact spaces are the new ABCs of modern mathematics as finite sets were in the classical days. Almost nowhere in modern mathematics will one see a proof about an object where topological concepts do not occur, and compact sets being the "nicest" sorts of sets, knowing that your favorite object is compact (or locally compact et cetera) is an advantage in proving results that cannot be understated. Furthermore, the idea behind using a product topology instead of the box topology originated in Tychonoff's work, and from a very general standpoint, this idea is somehow the most "right" way to think about topological products, so that--aside from the theorem itself--the ideas of its proof are also fundamental.
Urysohn's lemma, while less applied overall, is more central within the field of topology itself. Topology came as a refinement to the old notion of a metric space, but naturally metric spaces already had a great deal of theorems proven about them, so it was a worthwhile enterprise to find topological descriptions of a space which implies ones favorite space is a metric space and hence allowed for the use of all the hard work everyone had put into metric spaces to be applied to one's favorite spaces in this new context. This is the basic idea behind metrization theorems. Urysohn's lemma is a foundational result which helps one to prove many of these theorems and hence is very useful withing the field itself. While not carrying as much utility overall as Tychonoff, it is used to prove more theorems in topology than is Tychonoff, which is more useful for proving theorems with topology.

($5'$) The Three Isomorphism Theorems. (Fundamental Theorems of Group and Ring Theory)

Again there is a lot since there are three theorems and each is slightly different for groups vs rings, so I'll leave the link for those interested in reading about them.

These three results constitute the bedrock of group and ring theories--areas in modern algebra which study symmetry and algebraic relations in their most general forms. Group and rings are both primarily understood by examining sub and quotient groups/rings. The idea of a homomorphism is quite possibly the backbone of all modern mathematics. It's a notion of a function which preserves structure, in other words it's a notion of how similar two objects are while taking into account only the relevant properties we care about. In the case of groups, we care if we preserve the main bit of structure in a group, the group multiplication. In the case of rings, they need to preserve multiplication and addition, the two ring operations. The same holds for fields, boolean algebras, this list can go on ad nauseum, so I'll stop here before I write an entire paragraph for the list of things we use homomorphisms for. The isomorphism theorems are essential building blocks towards understanding the fundamental essence of homomorphisms and how to think about the objects that naturally arise in their study, namely subobjects (i.e. subgroups or rings) and quotient objects (i.e. quotient groups and rings). Thanks to the idea of a composition series and the celebrated Jordan-Hölder theorem, it tells us that a thorough understanding of homomorphisms can give us essentially unlimited insight into the structure of groups.

($6'$) Cauchy's Integral Formula. (Fundamental Theorem of Complex Analysis)

"Let $f(z)$ be holomorphic function on the open set $U\subseteq\Bbb C$. Let $z\in U$ and $\gamma=\gamma_z: [0,1]\to U$ be a simple, closed curve around $z$, oriented positively. Then
$$f(z)=\oint_{\gamma}{f(\xi)\over \xi-z}\,\mathrm{d}\xi.$$

Cauchy's integral formula deserves to be called "fundamental" in complex analysis if anything does. It proves the residue theorem which is hugely important in complex analysis, algebraic geometry, and analytic number theory--where it connects to the idea of a $\zeta$-function and distributions of primes in a very general setting. It proves the argument principle which led to the idea of branch cuts and multi-valued functions. This allowed us to more concretely understand the complex logarithm and root functions in sufficient generality in such a genius way that it basically jump started the field of topology through the notion of a Riemann surface. Finally, this result establishes that a single complex derivative ensures infinite differentiability and a local power series everywhere on an open set where a function is holomorphic, completely abolishing the notion that complex-differentiable and real-differentiable functions are behave similarly. This is what makes classical complex analysis such an immensely beautiful subject, the rigidity of the functions gives them amazing properties real functions cannot compete with for the very mild condition of being differentiable.

($7'$) The Artin Reciprocity Law. (Fundamental Theorem of Class Field Theory)

"Let $K/k$ be an abelian extension of fields and $\mathfrak{c}$ be an admissible cycle for this extension. Then the Artin map
$$I(\mathfrak{c})\to \text{Gal}(K/k)$$
is surjective with kernel $P_{\mathfrak{c}}\mathfrak{N}(\mathfrak{c}$) (described in the articles cited, omitted for brevity)."

Artin's reciprocity law is the ultimate generalization of quadratic reciprocity a theorem first proved by Gauß and lauded by him as the "golden theorem." Combined with the Hasse principle, it describes an incredible fact about quadratic forms and quadratic fields. These forms are heavily studied objects. Even today interesting theorems from many areas--such as the dynamical systems approach of Margulis to problems in fields such as Diophantine approximation, and $\vartheta$-functions and hence modular forms--are proven about them. Alongside the work of Hecke and Tate these forms are of huge importance. In the general case, it establishes norms as very special objects, and connects the idèle class group to the Galois group of a field extension in a beautiful and deep way. This is the foundation of modern ideas related to the Hilbert symbol, the Kronecker-Weber theorem, and the Langland's program (before mentioned).

($8'$) Parseval's Formula and the Plancherel Theorem. (Fundamental Theorems of Fourier Analysis)

(Parseval) "Let $f,g$ be functions of period $2\pi$ with Fourier series
$$f(x)\sim \sum_{n\in\Bbb Z} a_ne^{inx},\quad g(x)\sim\sum_{n\in\Bbb Z}b_ne^{inx}$$
Then the inner-product of the $\ell^2(\Bbb Z)$ functions, $(a_n),(b_n)$ is equal to
$${1\over\pi}\int_{-\pi}^\pi f(x)\overline{g(x)}\,\mathrm{d}x."$$
(Plancherel) "Let $\mathcal{F}$ be the Fourier transform on $(L^1\cap L^2)(\Bbb R)$ with normalization
$$\mathcal{F}(f)(\xi)=\int_{\Bbb R}f(x)e^{-2\pi i x\xi}\,\mathrm{d}x.$$
Then the continuous extension of $\mathcal{F}$ to $L^2(\Bbb R)$ is an isometry (in fact a unitary operator)."

The study of Fourier analysis goes back hundreds of years (and prototypes date back much farther). The Parseval formula transforms questions about periodic functions--which are a class of what are arguably one of the most important in all of mathematics--into those about its Fourier series, where techniques involved with sequences. Together with its generalizations, this greatly simplifies the study of functions to understanding their Fourier series (more on this after Plancherel). The Plancherel theorem establishes a truly remarkable duality between functions and their Fourier transforms. Because of the myriad properties of this transform, one can transfer very hard problems into much easier ones. In differential equations, one can turn differentiation (a very hard process to deal with in general) to questions about polynomials:  objects studied since antiquity. In applied mathematics, one uses Fourier analysis and things like the sampling theorem to reconstruct signals from sampled data. This is what powers digital music, your ability to hear people over wireless phones, and a nearly infinite list of other things. Indeed, to do justice to the sheer magnitude and variety of uses of Fourier analysis and this duality established by Plancherel, would take three answers as long as this one to do a true modicum of justice. It is used in topology, representation theory, all forms of number theory (in things like the Poisson summation formula), and on and on and on. Parseval embodies analysis on the circle group--the prototype for general periodic functions--an idea which has proven to have a lot of mileage.
The two of them together underpin the works of Fourier analysis, a subject which has generalized to locally compact groups and their characters by the miracle of the Haar measure. This allows this fruitful and powerful field reign in very general problems, and is the source of entire industries of and careers in mathematics.

The Punchline There are a lot of things which make a theorem "fundamental" to a field.
The $4$ with "official" status have a variety of reasons for their names. ($1$) is extremely old (due to Euclid). It is also powerful and useful in proving things in arithmetic (i.e. number theory). ($2$) gets the name as a product of history. To see how much, one considers that Cardano's method for solving cubics was first used to win some of the old math contests from the 1500s. Algebra has branched off a lot since, but for a long time it was much simpler in scope, and unique factorization of polynomials found itself as a long-standing central interest and has kept its name due to this history. ($3$), like ($1$), richly deserves its name:  the huge conceptual leap to connect the two different ideas of differentiation and integration is perhaps the single result which it is most satisfying to call "fundamental." Finally ($4$) stands at the top of its theory. It qualitatively embodies the philosophy of Galois theory in the capstone of the subject. No other result is as reflects the culmination of a field better than the FTGT. Notice, that it is not used a great deal to prove more results within Galois theory itself, here the result simply stands at the end of a long road, rather than near the beginning.
There are myriad reasons to lack a "fundamental theorem." In ($1'$), it would be strange to have a single, fundamental result. If the entire field of logic depended on one result, all of mathematics would be as well, which would either make this result unreasonably powerful or mathematics as a whole ridiculously simple.
For ($2'$), the field--as in ($4'$)--the results build more horizontally, so because of the diversity it makes less sense to name something "fundamental."
In ($3'$) and ($5'$), one notes that: (i) they have multiple results which are fundamental, no single one takes center stage and (ii) the chosen names are just better. Calling something "the fundamental theorem of [blah]" underscores--and can slightly over-inflate--a result's importance, but it makes it a lot less descriptive. The LLN and CLT are best with the name they have, as are the isomorphism theorems.
Finally ($6'$), ($7'$), and ($8'$) share the distinction that their theorems are named after the geniuses who proved them. This is more and more common in modern days, especially. Lang calls Artin Reciprocity the Fundamental Theorem of Class Field Theory, and most would probably agree. However, its official name is still "Artin reciprocity."
On the whole:  what makes the theorems we have "fundamental," in name is often a combination of the amount of history the subject and result have. For ($1$) and ($2$) especially, the fields are extremely old, whereas subjects like analysis, group/ring theory, and topology don't date back nearly as far. What makes all these results more "fundamental" in spirit is their incredible power and flexibility. These are some of the results which have helped shape the fields they correspond to. Some motivate a great deal of further research, some provide the key ingredients for an absurd number of proofs, some act as ambassadors between subjects in a way that links them much more strongly than others.
This multiplicity of reasons means that it's fundamentally a flawed idea to attach particular importance just to the official name "fundamental," or hope for a very insightful and "right" definition of what that means. If you look at the two lists, the "unofficial" list ended up being twice as long as the official one, and some of the unofficial ones--such as ($6'$)-($8'$)--are perhaps more "fundamental" to their subjects than ($2$). So any definition tailored to fit the current state-of-affairs which also did justice to the name "fundamental" would probably need to include a few of the results in the unofficial list and could not be a good definition since it is clear that it is not always used.

Just a little addendum after @Adam Hughes' nice answer.
Let $(\Omega, \mathcal F, P)$ be a probability space. The information $I(A)$ of an event $A$ (any element of the $\sigma$-algebra $\mathcal F$) is given by $I(A)=f(P(A))$, denoting by $f$ a real function on $[0,1]$.
Then one has the

Fundamental theorem of information (FTI)
The Shannon entropy $f(x):=x\ln \frac{1}{x}+(1-x)\ln\frac{1}{1-x}$
is the unique continuous solution of the fundamental equation of information (FEI)
$$f(x)+(1-x)f\left(\frac{y}{1-x}\right)=f(y)+(1-y)f\left(\frac{x}{1-y}\right) ~~(*)$$
on  $D:=\{(x,y)~|~ x\in [0,1), y\in[0,1), x+y\leq 1\}$.

The Shannon entropy plays a fundamental role in information theory, information geometry (it even generates Kullback Leibler divergences), and in many applications. There exists even an analogue of the FEI in number theory; for the whole exposition I refer to
Kontsevich, The $1\frac{1}{2}$-logarithm. Composition Math. 130 (2002).
Apart from applications et similia, I believe that it is the formulation of the FTI via functional equations that makes it interesting: after all, many important functions in mathematics are defined as solutions of functional equations (in primis the exponential function): the Shannon entropy becomes one of them.


Solution of a differential equation having a singularity (not everywhere defined) Remind me about ordinary differential equations, whose solutions are not everywhere defined (have a singularity).

I want to remember the exact definition of a solution with singularity, which I studied in a university long time ago.

And please, an example.

The easiest example is the `finite time blow-up' ODE:

$$
y^\prime =y^2,\qquad y(0)=1
$$

This equation has the unique solution

$$
y(t)=\frac{1}{1-t}
$$ which is not defined when $t=1$.  So technically speaking, $y(t)$ isn't a solution to $y^\prime=y^2$ for all time - we have to choose an interval where the solution makes sense.  The rigorous approach would go as follows:

We say that $y(t)$ is a (classical) solution to the following ODE

$$
y^\prime=f(t,y),\qquad y(0)=y_0,\qquad a<t<b
$$ If

$y^\prime(t)$ exists for all $a<t<b$ (note this requires that $y(t)$ itself exists for all $a<t<b$)
$y^\prime(t)=f(t,y)$ for all $a<t<b$
$y(0)=y_0$

Any solution which fails to satisfy these conditions at some point could be called a `singular' solution.  The example $y^\prime=y^2$ violates the first condition: $y^\prime$ doesn't exist when $t=1$.  However, if we restrict the problem to read as follows:

$$
y^\prime=y^2,\qquad y(0)=1,\qquad 0\leq t<1
$$ Then $y(t)=(1-t)^{-1}$ is a classical solution.  Note that I have included the left endpoint assuming the appropriate definition of the derivative at $t=0$.

There is much more - I recommend picking up a book on ODE theory e.g. Hartman or Arnold.

A solution $x(t)$ of the ODE is called maximal if it is defined on an
  open interval and cannot be extended to any larger open interval.

From "Ordinary Differential Equation". Alexander Grigorian. University of Bielefeld. Lecture Notes, April - July 2008

So the things I search for are maximal solutions.

The question remains: What is the best generalization of this for PDE?


How to prove this Tree/Set claim? If $T$ is a tree and $T_1, T_2, \ldots , T_k$ are all subtrees of $T$ such that the vertices of $T_i$ are $V_i$. Suppose $V_i\cap V_j$ is non-empty for all $i$ and $j$. Show that there is a vertex that exists in all $T_i$.

I'm honestly not even sure what the question is asking. This is just a completion grade, but I honestly do want to know how I should go about proving this. What method to use?

Let $F = \{V_1, \ldots, V_k\}$. We prove by induction on $j \in \{2, \ldots, k\}$ that:

  Any $j$ vertex sets chosen from $F$ all share some vertex $v$ in common.

This is immediate for $j = 2$. For the inductive step, choose any $j$ vertex sets from $F$, say $W_1, \ldots, W_j$. By the inductive hypothesis, we know that for each $i \in \{1, \ldots, j\}$, there is some vertex $v_i \in W_1 \cap \cdots \cap W_{i-1} \cap W_{i + 1} \cap \cdots \cap W_j$. We claim that in the sequence of vertices $(v_1, \ldots, v_j)$, some two of them must repeat.

Otherwise, if all $j$ vertices are distinct, then we can paste the following $j$ paths together to form a cycle, a contradiction:

The $(v_1, v_2)$-path in $T_3$ (which is guaranteed to exist since trees are connected).
The $(v_2, v_3)$-path in $T_4$.
$\cdots$
The $(v_{j-2}, v_{j-1})$-path in $T_j$.
The $(v_{j-1}, v_j)$-path in $T_1$.
The $(v_j, v_1)$-path in $T_2$.

Hence, we know that $v_a = v_b$ for some $a < b$. But then this vertex belongs to $W_1, \ldots, W_j$ as desired.

I can give a sketch of an inductive proof for this which can be expanded into a full proof.

The base case $ k = 1 $ is simple.  Since all $V_1$ are in the tree $T_1$.

Now assume that this is true for $k$.  There exists a $V_{bottleneck}$ in every $T_1,T_2,..$.

$T_{k+1}$ must have a shared note with each of $T_1,T_2,..$ however because trees are acyclic to get from any vertex in $T_1$ to a vertex in $T_2,T_3,...$ you have to travel through $V_{bottlenet}$.


Find the magnitude of the vertex angle of an isosceles triangle of the given area $A$ Find the magnitude of the vertex angle $\alpha$ of an isosceles triangle with the given area $A$ such that the radius $r$ of the circle inscribed into the triangle is maximal.

My attempt:

We'll prove that for all triangle
$$r\leq\sqrt{\frac{A}{3\sqrt3}}.$$

Indeed, let $AB=c$, $AC=b$ and $BC=a$.

Hence, $A=\frac{1}{2}ra+\frac{1}{2}rb+\frac{1}{2}rc$, which gives $r=\frac{2A}{a+b+c}$.

Thus, we need to prove that
$$\frac{2A}{a+b+c}\leq\sqrt{\frac{A}{3\sqrt3}}$$ or
$$12\sqrt3A\leq(a+b+c)^2.$$
But by Heron formula we have
$$A=\sqrt{p(p-a)(p-b)(p-c)},$$
where $p=\frac{a+b+c}{2}$.

Thus, we need to prove that
$$(a+b+c)^2\geq12\sqrt3\sqrt{\frac{a+b+c}{2}\cdot\frac{a+b-c}{2}\cdot\frac{a+c-b}{2}\cdot\frac{b+c-a}{2}}$$ or
$$(a+b+c)^3\geq27(a+b-c)(a+c-b)(b+c-a)$$ or
$$\frac{(a+b-c)+(a+c-b)+(b+c-a)}{3}\geq\sqrt[3]{(a+b-c)(a+c-b)(b+c-a)},$$
which is AM-GM.

The equality occurs for $$a+b-c=a+c-b=b+c-a$$ or
$$a=b=c,$$
which says that $r$ gets a maximal value, when our triangle is an equilateral triangle,

which says that $\measuredangle BAC=60^{\circ}$.

Done!

  If the following is obvious for you, then I am ready to delete it.

The equality occurring in our AM-GM just for $a+b-c=a+c-b=b+c-a$ we can understand by the following way.

Let $a+b-c=x^3$, $a+c-b=y^3$ and $b+c-a=z^3$.

Hence, $x$, $y$ and $z$ are positives and the equality case gives
$$\frac{x^3+y^3+z^3}{3}=xyz$$ or
$$x^3+y^3+z^3-3xyz=0$$ or
$$x^3+3x^2y+3xy^2+y^3+z^3-3x^2y-3xy^2-3xyz=0$$ or
$$(x+y)^3+z^3-3xy(x+y+z)=0$$ or
$$(x+y+z)((x+y)^2-(x+y)z+z^2)-3xy(x+y+z)$$ or
$$(x+y)^2-(x+y)z+z^2-3xy=0$$ or
$$x^2+y^2+z^2-xy-xz-yz=0$$ or
$$2x^2+2y^2+2z^2-2xy-2xz-2yz=0$$ or
$$(x-y)^2+(x-z)^2+(y-z)^2=0$$ or
$$x=y=z,$$
which gives $a=b=c$.

It is well known (and easy to prove via the convexity of $\tan$) that a triangle of smallest area circumscribed to a circle $C$ of radius $r>0$ has to be equilateral. A fortiori an isosceles triangle of smallest area circumscribed to $C$ has to be equilateral. It follows that ${{\rm area}(\triangle_{\rm iso})\over r^2}$ is minimal for an equilateral triangle; hence ${r^2\over {\rm area}(\triangle_{\rm iso})}$ is maximal when the angle at the vertex is $60^{\circ}$.


How to show that this set is bounded? Given set $\{3 - x \mid x < 0\}$, find if it is bounded.

So, from intuition it should be bounded from below, but how to show it?

If $x<0$ then $3-x>3$, so the set in question is the open interval $(3,\infty)$. Hence it is bounded below (by $3$ for instance) but not above.


Trigonometry system - complex conjugate I have the following function
$$e_n(t) = e^{2\pi int}, t \in R, n \in Z$$
Could anyone explain how one can go from this:
$$e_m(t) \bar e_n(t) $$
to
$$e^{-2\pi i(m - n)t}$$

Shouldn't it be $e^{2\pi i(m - n)t}$ instead of $e^{-2\pi i(m - n)t}$?

If $e_n(t)=e^{2πint}$, the conjugate is $\bar{e}_n(t)=e^{-2πint}$

Then

$e_m(t)\bar{e}_n(t)=e^{2πimt}e^{-2πint}=e^{-2πi(n-m)t}.$


How to evaluate $\lim\limits_{h \to 0} \frac {3^h-1} {h}=\ln3$? How is

$$\lim_{h \to 0} \frac {3^h-1} {h}=\ln3$$

evaluated?

There are at least two ways of doing this: Either you can use de l'Hôpital's rule, and as I pointed out in the comments the third example on Wikipedia gives the details.

I think a better way of doing this (and Jonas seems to agree, as I saw after posting) is to write $f(h) = 3^{h} = e^{\log{3}\cdot h}$ and write the limit as $$\lim_{h \to 0} \frac{f(h) - f(0)}{h}$$ and recall the definition of a derivative. What comes out is $f'(0) = \log{3}$.

Using the formula
$$\lim_{x \to 0} \frac {a^x-1} {x}=\ln a,$$
we have for $a = 3$
$$\lim_{h \to 0} \frac {3^h-1} {h}=\ln 3.$$


Möbius tranformation taking reals to reals can be written with real coefficients I'm working on this one (from Ahlfors' Complex Analysis):

A Fractional Linear Transformation of form $\displaystyle T(z) = \frac{a z + b}{c z + d}$ which takes the real numbers into the real numbers can be written in a way where all the coefficients are real.

I'm pretty sure I know a way to get the answer.

First, the thing is a constant exactly when $ad - bc = 0$. Then you're done. If not, some coefficient is not zero. in that case, you can change variables and stuff around in a way so that you can assume WLOG $a \ne 0$.

Then, you pick three different real-valued inputs and show that all the three remaining coefficients $b, c, d$ are real-valued.

My question is:

Is there an easier way to do this, without using more advanced complex analysis (which I don't know yet)

It turns out that this is not an answer to the OP's question. I will leave this answer nevertheless, perhaps someone might find it useful for something. This was my fault for not reading the question more carefully

I will start the answer with an exercise. We will assume throughout that $a\neq 0$, and $ad-bc\neq 0$. Throughout, $L$ will denote a linear fractional transformation that sends extended reals to the extended reals, that satisfies the conditions above.

Exercise: Let $(z_1,z_2,z_3)$, and $(w_1,w_2,w_3)$ be two triples of extended complex numbers (the complex numbers with the point at infinity include). Then their is a unique linear fractional transformation, $L$ such that $L(z_i)=w_i,i=1,2,3.$

Now to take on the literal statement of the question, if a linear fractional transformation takes the reals to the reals, then it is a linear function (and takes infinity to infinity), but this is not so interesting. More interesting (the one I assume that you want to prove) is that a LFT that takes the extended reals to the extended reals (the reals with infinity included) mat be written to have all real coefficients. Since by your remark, that we may assume $a\neq 0$, multiply the numerator and denominator by $\frac{1}{a}$, so now without loss of generality, we may assume that $a=1$.

We will now invoke our exercise. If we say where we send $\{0,1,\infty\}$, we completely determine the LFT. Again, if we send infinity to infinity, we have a linear function. Let us assume that we send infinity to some real number. If $L(z)=\frac{z+b}{cz+d}$, then $L(\infty)=\frac{1}{c}$. If we assume that $L$ sends extended reals to extended reals, the, $\frac{1}{c}$ is real and therefore ,so is $c$. Likewise $L(0)=\frac{b}{d}$, which is also real if $d\neq 0$(we will also need to deal with the case when $d=0$.
Since our linear fractional transformation does not send infinity to infinity, the number, $-\frac{d}{c}$ is real, since $L(-\frac{d}{c})=\infty$. Since $c$ is real, so is $d$. Since $\frac{b}{d}$ is real, so is $b$, since $d$ is real. Now suppose that $d=0$. Then the LFT is of the form, $\frac{z+b}{cz}=\frac{1}{c}+\frac{b}{cz}$. We still know that $c$ is real, the previous argument for this did not require $d\neq 0$. Now pick $z=1$ and plug it into $\frac{1}{c}+\frac{b}{cz}$, which concludes this argument.


Why isn't finitism nonsense? This is a by product of this recent question, where the concept of ultrafinitism came up. I was under the impression that finitism was just "some ancient philosophical movement" in mathematics, only followed by one or two nowadays, so It sounded like a joke to me.

But then I got curious and, after reading a bit, It seems to me that the only arguments against infinite mathematics that finitists seem to have are that "there are numbers so big that we couldn't computate in a lifetime" or the naive set theory paradoxes. The former doesn't seem like a serious argument, and the latter is not a problem now that mathematics relies on consistent axioms.

Are there some (maybe arguably) good mathematical reason to deny the existence of $\infty$ or is it just a philosophical attitude? The concept of unboundedness seems pretty natural to me, so what could be a reason to avoid it? Does this attitude even make any sense?

In short, why today-finitists have a problem with $\infty$?

Edit: First of all, thank you so much for your answers (and comments), they have been enormously illuminating. :)

I didn't know that "finitism vs. infinitism" was such a polemic topic. Now I myself agree this question might look as primarily opinion-based. However, It was not my intention to open a debate about "which posture is better"; I was just meaning to ask about what specific mathematical reason (argued and not-primarily opinion based) do finitists have to reject the "infinitists" use of infinity.

Based on the two excellent answers I've already had (thank you again :) It is my understanding that their main problem with the use of $\infty$ is that it leads to mathematical results (like the Banach-Tarski paradox) which they don't recognize as true when looked through the glasses of our real-world experience.

Final edit: After reading every answer and comment (specially Asaf Karagila's) I've came to the conclussion that there are not strictly mathematical reason to avoid the use of infinite. That my specified question on the last edit has no answer, and that the motivation to stick to finitists or infinitists view of mathematics relies on how much one expects mathematics to describe each one's "real world". As Wildcard's answer is the one that clarifies that matter best to me, I am accepting it. Thank you all again for your answers and comments!

It's not known that modern set theory is consistent; in fact, by the Incompleteness Theorem, we can't ever have a system of axioms that we can prove is consistent. Which means that the only condition we can rely on for determining whether a set of axioms is "right" is whether or not it produces absurd results.

Under $ZFC$, we have different sizes of infinity - there are sets which are larger than the set of natural numbers in a precise sense. We also have a lot of weirdness involving the Axiom of Choice - for example, with the Axiom of Choice, a theorem of Banach and Tarski states that a hollow sphere can be disassembled into five pieces and then reassembled (without stretching, tearing, or otherwise deforming the pieces) into two spheres that are both identical to the first one in both size and shape. But the Axiom of Choice simply states that given a set of sets, we can "choose" one element from each set - which seems intuitively true.

A finitist's perspective on $ZFC$ is often that results like the hierarchy of infinite cardinals and the Banach-Tarski paradox are absurd - that they should count as contradictions, because they patently disagree with the intuitive picture of mathematics. The sensible conclusion is that one of the axioms of $ZFC$ is wrong. Most of them are intuitively obvious, because we can demonstrate them with finite sets - the only one we can't is Infinity, which states that there exists an infinite set. So a finitist's conclusion is to reject the Axiom of Infinity. Without that axiom, $ZFC$ becomes purely finitistic.

Now, many finitists are happy to stop here. But some are bothered by the fact that we still have an infinite collection of natural numbers; the infinite still "exists", in a sense, and gives the opportunity for the above weirdnesses to arise in the same way. So some people (including some mathematicians) subscribe to ultrafinitism and insist that there are only finitely many numbers at all. One ultrafinitist mathematician I know defines the largest integer to be the largest integer that will ever be referenced by humans.

Among mathematicians, ultrafinitists are much rarer than simple finitists. Finitists generally agree with you that "unboundedness" is a natural idea - it's essential, for example, in the definition of a limit. But they would go on to insist that this is just a formalism - that a limit, for example, is just a statement of eventual behavior, involving only finite numbers. So $\infty$ isn't an object, it's just a shorthand. This is (at least to my mind) more mathematically defensible than ultrafinitism.

EDIT: Since a lot of people seem to be having a hard time with my first sentence, I thought I'd clarify. The Incompleteness Theorem states that we cannot have a set of axioms powerful enough to express arithmetic and still be able to prove its consistency within the system. The reason I didn't include this phrase above is because it unnecessarily weakens the point. Any axiom system intended to codify all of mathematics defines the idea of "proof"; if, say, $T$ is intended to underlie all of mathematics, then by "proof" we must mean "proof in $T$". With such a $T$, we can say that $T$ cannot be proven consistent at all; because by Incompleteness, any proof of the consistency of $T$ would not be a proof from inside $T$, but $T$ is supposed to be powerful enough that all proofs are proofs from inside $T$. Thus: we can't ever have a system of axioms for mathematics that we can prove is consistent - full stop.

Many quality responses here, but ill add my perspective.

I agree that many defenses of finitism are silly or incomplete, but there is an underlying point to it that is quite valid and important.

The way I would put it, is that infinity as a concept should not be given the same epistemic status as other mathematical ideas. As has been noted elsewhere, the concept fundamentally cannot be defended in an empirical manner. A pure mathematician may shrug as to that argument; but even within the confines of mathematics, there are clear conceptual differences. Math involving infinity cannot be described in a system that is proven to be free of contradiction; incompleteness is a feature of math involving infinity; not of math or formal systems per se.

Personally, I am wholly unimpressed by the so called modern set theoretic 'solutions' to the properties of infinity known since ancient times. 'naively' speaking, youd think that 1-to-1 correspondence, and the ability to construct one set as a subset of the other, would be equally good axioms to decide on the relative size of sets. Both are observably true; and we should be willing to embrace all the conclusions that can be derived from either axiom, as truths about the real world. The 'modern' approach is to observe that you can just drop one of these axioms, thereby get rid of some obvious contradictions involving infinity, and be so enamoured with this to jump to the conclusion that therefore this must be a good idea, without much further consideration.

The Banach-Tarski 'paradox' can be seen as a rather direct result of this hasty philosophical choice. Its not a paradox of modern math of course; yet it puzzles a lot of people; because most people, physicists especially, do like to be able to assume their math has anything whatsoever to do with the real world.

The way physics should work, is that you try and identify what axioms must be true, and then try and reason about what other implications those givens have, and see if that actually holds in the real world. What infact we have is our best physicists (Feynman specifically) mocking the Banach-Tarski paradox as silly, rather than going out and trying to set up an experiment to reproduce it (and the latter goes for all physicists I know). Thats a big epistemological breakdown. Now I think Feynmans common sense is as usual spot-on; but if there are some conclusions from math that we should take less seriously than others, id like to know which one's those are; to codify that intuition, if you will.

Personally, I find the paradoxes of infinity hard to get around. If I divide a number by two an infinite number of times, is it zero? No, because non-zero-ness is conserved by halving; yet it has to be, since if it is nonzero, I divided a finite amount of times. Are there more natural numbers than even numbers? I could go on, and I find the arguments for and against usually equally compelling. Sure you can 'solve' these problems by 'clever' redefinition of terms and other trickery; but invariably this seems to involve driving a wedge between what you take to be axiomatically true, and any empirical obviousness of such axioms.

Which isn't only bad for the physical/empirical users of math; but generally lends a 'flavor' to modern mathematics that is off-putting to many. Bertrand Russel, who gave a big impetus to the idea that if we just boil down mathematics to as few and abstract axioms as possible, all will be great, actually disavowed the wisdom of doing so in his last years. And I imagine he would agree with the later ideas of Quine on epistemological holism. That is, the quality of an axiom should be decided on its interaction with the entirety of our web of beliefs; not on what pure mathematician or logicians thinks is 'elegant' or looks pretty on paper.

Personally, I am more in favor of a *para-consistent approach to infinity (*I guess thats what its called but it seems to mean many things to many people). That is, Id rather take an approach where I embrace the contradictions in infinity. Which doesn't mean you cannot do math at all; plenty of people knew how to do math throughout history while believing in contradictions of infinity, and it wasn't because they were 'naive', but because they had common sense. In fact, the principle of explosion is nothing but an artefact of so called 'classical' logic (specifically, the fact that it does not specify a deterministic rule for symbol-lookup). Its really not that hard in the 21th century to define your logical language to be robust to contradictions, while merrily continuing to perform deductions on your non-contradicting clauses.

What that means for the 'existence' of infinity, I dont know, but I am happy with that gray status until someone can show me any of these alleged paradoxes empirically. Personally, I am inclined to believe that the infinite divisibility of space is probably as much a figment of our imagination as the infinite divisibility of matter still was well into the 19th century.

This is mostly an academic debate with little practical implications in the end. I don't think I (or anyone with finitist inclination) should be expected to come to different conclusions on any key point. They just come to fewer conclusions. And I think research into say the real number continuum is quite valuable; even if we disagree as to the parity of the last digit of the square root of two. I say the answer is both even and non-even; you say that it's not a rational number and you think sidestepping the question like that makes you less 'naive'. Soit.

But if I was the world's science Tsar, I would definitely defund any research into the continuum hypothesis, or anything involving hierarchies of infinities. Id tell these people to stop chasing the implications of Cantors hasty assumptions, and start doing something comparatively useful for society, like solving Sudoku's for instance.


Stuck on question with asymptotes Ive been stuck on this quest from textbook and I can't find an answer for it..

Question is, what may the equation of a rational function with below conditions look like? Explain how you got the equation.

$\displaystyle f(1)=0$

$\displaystyle \lim_{x \to 0}  \; f(x)= - \infty$

$\displaystyle \lim_{x\to2^+} f(x)= -\infty$  (from right side)

$\displaystyle  \lim_{ x \to 2^-} f(x)= +\infty$  (from left side)

$\displaystyle   \lim_{x \to \pm \infty} f(x)= 0$

For the activity around x=2 you need $(2-x)$ raised to an odd power in the denominator. To make $f(1)=0$ you need $(x-1)$ in the numerator. And for the activity at plus and minus infinity you need a power of $x$ in the denominator. Thus I would suggest
$$f(x)=\frac{(x-1)}{(2-x)^3x}.$$
Edit: Following comments below, a better suggestion is
$$f(x)=\frac{(x-1)}{(2-x)^3x^2}.$$


help solving an integral $\int\left(\frac {x-1}{3-x}\right)^\frac{1}{2} dx$ $\displaystyle\int \left(\frac {x-1}{3-x}\right)^\frac{1}{2}\,\rm dx$

I am stuck on this part:

Let $u=\dfrac{x-1}{3-x}~\longrightarrow$  $~~\rm du=\dfrac {2}{(x-3)^2}\,\rm dx,$ which can be represented as

$\rm du=\dfrac{1}{3-x} - \dfrac{1-x}{(3-x)^2}\,\rm dx$

I cannot "see" how to get to this $2$ $\displaystyle\int \: \frac{(u)^\frac{1}{2}}{(u+1)^2} \:\rm dx$

after this part I know how to solve it;
I just wish someone would show me "step by step" this part
It seems it involves some sort of "leap" of thought; or is there a systematic way doing this using basic algebra?

Thanks.

The integrand is so seductively pointing to the use of complex variables which reduces it to a trivial problem. Beware we have a pole within a branch cut!


Factorial question: number of trailing zeroes in 125! How many zeros are after the last nonzero digit of 125! ?

The answer is 31, but how do you solve it?

Well, we know that to have a zero at the end then $10$ must be a factor, which means $5$ and $2$ must be factors. However, every other factor is even, so there are far more factors of $2$ than $5$ - As such, we have to count the number of factors divisible by $5$. The number of factors divisible by $5$ less than or equal to $125$ is $25$ (we just do $\frac{125}{5}$), so the answer appears to be $25$, but then we remember that $25 = 5\cdot5$, so we must count double for each factor divisible by $25$, of which there are $5$; accordingly, our new answer is $30$. BUT WAIT... $125 = 5^3$, so we have to count it too, giving us a final answer of $31$

I just count them.

125! = 188267717688892609974376770249160085759540364871492425887598231508353156331613598866882932889495923133646405445930057740630161919341380597818883457558547055524326375565007131770880000000000000000000000000000000

Have a Apple computer? Compute the value yourself. Below is the applescript. Paste into the Script Editor application and click on the "Run the Script" button.

on ids(digits as text)
    if digits is "" or digits is not "0" and digits starts with "0" then error "bad integer " & digits
    set x to (id of digits) as list
    repeat with digit in x
        if digit < 48 or digit > 57 then error "bad integer"
    end repeat
    return x
end ids

on add(x as text, y as text)
    set {a, b, n, carry} to {ids(x), ids(y), (length of x) - (length of y), 0}
    if n < 0 then set {a, b, n} to {b, a, -n}
    repeat with i from length of a to 1 by -1
        set {carry, sum, j} to {0, carry + (item i of a), i - n}
        if j > 0 then set sum to sum + (item j of b) - 48
        if sum > 57 then set {carry, sum} to {1, sum - 10}
        set item i of a to sum
    end repeat
    if carry > 0 then set beginning of a to carry + 48
    return character id a
end add

on multiply(x as text, y as text)
    set {a, b, c, sum} to {ids(x), ids(y), {}, "0"}
    repeat with m from length of b to 1 by -1
        set multiplier to (item m of b) - 48
        if multiplier is not 0 and x is not "0" then
            set carry to 0
            repeat with i from length of a to 1 by -1
                set product to carry + ((item i of a) - 48) * multiplier
                set carry to product div 10
                set beginning of c to product mod 10 + 48
            end repeat
            if carry > 0 then set beginning of c to carry + 48
            set sum to add(sum, character id c)
        end if
        set end of c to 48
        set c to items (m - (length of b) - 1) thru end of c
    end repeat
    return sum
 end multiply

 set x to 1
 repeat with y from 1 to 125
    set x to multiply(x, y)
 end repeat


How to solve $x \times 0 = n$ where $n$ does not equal $0$? So recently I was watching some videos on complex numbers. I was surprised to learn the uses of $i$ in different problems. For example, I never thought it would make sense that $x^2 + 1 = 0$ has any solutions at all. But apparently it does when you consider that this equation actually has two solutions that aren't visible by graphing this equation on your calculator.

However, my question is somewhat related yet somewhat different. What if we had the following equation:

$$x \times 0 = 1$$

Now, from my knowledge, there's no solutions to this problem because no number that I know multiplied by zero would be equal to anything but zero. But would it be possible to create your own solutions to this problem?

I noticed when I tried to solve this problem I came across two main issues:

$$x \times 0 \, / \, 0 = 1 \, / \, 0$$

First off, what is $0 \, / \, 0$? Does $0 \, / \, 0$ "cancel out?" Secondly, isn't $1 \, / \, 0$ undefined? Not to mention, what if I have another problem like this one:

$$x \times 0 = 5$$

How would the solutions, if any, differ from this problem and the first problem?

Keep in mind, I'm not a mathematics genius or anything, so I hope you won't be disappointed with me for suggesting a problem like this because I'm simply just curious. This is not meant to stump anyone, nor is it meant to waste your time. Any responses would be greatly appreciated.

You can define a new (extended) number system with the property that there is a solution to $0x=1$, and call that solution (or one of the solutions if there are more) $k$. You just have to be willing to accept the consequences. For instance, if you still want the property $0+0=0$, and the property of multiplication distributing over parentheses, then you get
$$
1=0k=(0+0)k=0k+0k=1+1=2
$$
so suddenly $1=2$ in this new number system. That's probably not desirable. Therefore you have to give up at least one of the three assumptions I've made above. I prefer to give up the existence of $k$.

The way I see it your question is this:

$x^2 + 1 = 0$ means $x^2 = -1$.  If $x \ge 0$ then $x^2 \ge 0$.  If $x < 0$ then $x^2 > 1$.  So $x^2 < 1$ is impossible.  So there should be no answer.  But mathematicians say:  "Oh, that's okay.  It's no problem.  We just introduce a new number called $i$ where $i^2 = -1$ and we can say $i=\sqrt{-1}$.  That's easy why do you have a problem with it? You must be stupid to be stuck in your old world-view that you can't accept something new."

But if we did $x\times 0 = 1$ there is no solution.  But what if I said "Let $\mu$ be a new type of number where $\mu \times 0 =1$ and we can say $\mu = \frac 10$."  But when you say this mathematicians say:  "What?  Are you crazy?  We can't define $\frac 10$! $x\times 0 = 0$ and nothing else ever.  That's a fact.  Man, you sure are stupid, if you think you can just make up answer just by inventing numbers. "

So why is it okay to "make up" a number $i = \sqrt{-1}$ but it's not okay to "make up" a number $\mu \times 0 = 1$?  It seems like both of them should not be allowed or both should be all.  (And why are these hypothetical mathematician so dang rude?)

.....

Okay.  The idea of "making up" a new kind of number is called "extending" a number system, and it's allowed as long as the results are consistent.

For instance the first number system we have is $\mathbb N = \{1,2,3,4....\}$.  This is good.  Our basic two rules are that if you have two different numbers $a$ and $b$ they are ... different; They can't be equal to each other.  And that if you have $a+ b$ then $a + b = c$ which is also a number as well as $a \times b = d$ is also a number.  (We don't have two numbers that we can't add together.) Those are basic rules.

And there are some not so basic rules.  $1$ is the smallest number.  Every number has a number that comes immediately after it.  Etc.  Those are interesting rules but they are not fundamental.

If I asked you to solve $x + a = a$... there is not solution.  So we say, let's make up a new type of number called $0$ where $0 + a = a$ for all $a$.

This is consistent with our idea that all numbers are distinct and that we can add any two numbers together.

To make it consistent if $a \times b = b + b+ b....$ then $0\times a = nothing$ and $a*0 = 0+ 0+0+.... = 0$ so that's a new rule.  $0*a = a*0 = 0$.

Some of a rules change.  $1$ isn't the smallest number; $0$ is.

The next  extension: Can we solve $x + a = 0$.  We "make up" negative numbers.  They are consistent with our most important rules. But we no longer have the rule that there is a smallest number.

The next extension: Can we solve $nk = m$.  We have to make up the rationals.  We no longer have the rules that every number has a number that comes immediately after it.  But the must important one is that numbers are distinct.  We have to make an exception that $0k = m$ has not solution. As that would mean $m=0*k = 0+0+0... = 0$ and we have two different numbers equaling the same.

We also extended and discovered that $\frac km*\frac jn$ doesn't mean adding $\frac jn$ a total of $\frac km$ times.  We are no longer doing whole numbers but that's okay.  $\frac km *\frac jn = \frac {kj}{mn}$ works and is consistent.

So.....

Upshot.  Allowing $x^2 + 1 =0$ by "making up" $i^2 = -1$ makes things weird BUT consistent.  We still have multiplication and addition are close operations.  (We always add or multiply any two numbers together) and numbers are distinct.

We lose some basic rules.  The most strange is we can no longer say numbers are bigger or smaller than each other!  $i \not < 0; i\ne 0$ and $i \not  > 0$.  $i$ simply... goes off at a right a right angle.  We just can't say that one complex number is bigger than another!

But otherwise things are consistent.

Not so if we try to "make up"  $\mu\times 0 = 1$ and $\mu = \frac 10$.  We can simply get $(a*\mu)*0 = a*(\frac 10\times 0) = a*1 = a$ but $a*0= 0$ so $a*0^2 = 0$ so $0 = a*0^2\mu = 0*\mu = 1$.  .... we just can not make consistent sense of this.

.....

So we have:  $\mathbb N\subset \mathbb Z \subset \mathbb Q \subset \mathbb R \subset\mathbb C$ are all consistent exstensions.

$x + a = a$ means $x \in \mathbb N$ is not possible but $x \in \mathbb Z$ is.  $\frac x7 = 4$ means $x \not \in \mathbb N$ but $x \in \mathbb Q$ is possible.  And $x^2 = 2$ means $x \not \in \mathbb Q$ but $x \in \mathbb R$ is possible. And $x^2 + 1=0$ means $x\not \in \mathbb R$ but $x \in \mathbb C$ is possible.  All that is consistent.

But $x\time 0 = 1$ is simply not consistent and can't happen in any extension, on which we have basic rules of arithmetic.