Proving that $k|(n^k-n)$ for prime $k$ Prove that for any integer $n$,we have $(n^k)- n$ is divisible by $k$ for $k=3,5,7,11,13$
I tried using prime factorization but that does not work here

HINT: there are several ways to prove the congruence of the OP,  which corresponds to Fermat little theorem. A rather simple one: consider the sequence of integers $$n,2n,3n,…,(k−1)n$$
Among these integers, none is congruent modulo k to the others (this is not difficult to prove). So ${1,2,3,…,k−1}$ is the set of reduced residue system modulo k.
Taking the product, we have
$$n \cdot 2n \cdot 3n \cdot \dots \cdot (k-1)n \\ \equiv 1 \cdot 2 \cdot 3 \cdot \cdots \cdot (k-1) \pmod k
$$
and then
$$n^{k-1} \cdot (k-1)! \equiv (k-1)! \pmod k$$
Now you can easily complete the proof.

Agreed, this is just Fermat's little theorem restated, i.e. $n^{k-1 }\equiv 1\pmod k$ where$ k$ is prime. So $n^k\equiv n\pmod k$ , which is what you are trying to show.


How do you prove each $x$ for $a^x=y$ is unique? I'm not looking for limits or any calculus-related argument, I want to know how to prove uniqueness on a more fundamental level. I am at the point where I want to show $$a^x=a^y \implies x=y$$ but if I haven't yet proven the existence of a logarithm, how could it possibly be possible to show that $x=y$ ?

There is no way to get rid of that base $a$, but the fact that someone has already defined $\log_a(x)$ implies someone somehow did so some centuries ago.

In other uniqueness arguments like of rational functions, you can manipulate both sides with rational operations, but you can't do that here because you haven't proven a logarithm yet! So how could anyone possibly show $x=y$ for $x>0$, $y>0$.

Unless, can I use the properties of a logarithm after only making an argument for the existence of $x$ such that $a^{x}=y$ even if I haven't yet proven uniqueness and then use the existence to prove uniqueness for $a \neq 1$?

Without knowing about logarithms, the question need not even begin to make sense, since it is unclear just what $a^x$ might mean, even for positive $a$, when $x$ is not restricted.

You say in the comments that you would define the function $f(x)=a^x$ simply as a function with the property that
$$\begin{align*}
f(1) &= a\\
f(x+y) &= f(x)f(y).
\end{align*}$$

However, these properties do not suffice to conclude that $f$ is one-to-one (which is required in order to conclude that $f(x)=f(y)$ implies $x=y$), or that there is a unique $x$ such that $f(x)=y$ for a given $y$ (or at most one such $x$ if you don't want to assume surjectivity).

In particular, if we assume the Axiom of Choice, then there are functions that satisfy both $f(1)=a$ and $f(x+y)=f(x)f(y)$, but that are not one to one.

"Explicitly" (modulo the Axiom of Choice), let $\beta$ be a basis for $\mathbb{R}$ as a vector space over $\mathbb{Q}$ such that $1\in\beta$. Then any function $g\colon \beta\to\mathbb{R}$ can be extended to an additive function $g\colon\mathbb{R}\to\mathbb{R}$; that is, a function defined on all of $\mathbb{R}$, whose values at $\beta$ are as specified, and such that for all $x,y\in\mathbb{R}$, we have $g(x+y)=g(x)+g(y)$.

Now, define $g\colon \beta\to\mathbb{R}$ by letting $g(1) = 1$ and $g(r)=0$ for all $r\in\beta$, $r\neq 1$. Then define $f\colon\mathbb{R}\to\mathbb{R}$ by $f(x) = a^{g(x)}$.

Then $f(1) = a^{g(1)} = a^1= a$; and $f(x+y) = a^{g(x+y)} = a^{g(x)+g(y)} = a^{g(x)}a^{g(y)} = f(x)f(y)$. So this function $f$ satisfies the two given equations.

However, $\beta$ is uncountable, so pick $r\neq 1$ that is in $\beta$. Then $f(r) = a^{g(r)} = a^0 = 1$, and $f(0) = 1$ (since $g(0)=0$ must hold for $g$ to be additive). However, $r\neq 0$, since $0$ cannot be an element of $\beta$.

Thus, the two conditions $f(1)=a$ and $f(x+y)=f(x)f(y)$ do not suffice to show that $f$ is one-to-one.

Which means you need to specify a lot of other stuff; specifically, one need to know exactly what properties you are giving the function $f$.

(Yes, I know I'm using the exponential function to define this; but the point is that there are interpretations of the function $f$ that make all the assumptions true but the desired conclusion false, which means that one cannot prove the fact that $f$ is one-to-one using only the assumptions listed)

It is difficult to define either the exponential or the logarithm at the basic level of calculus-before-the-fundamental-theorem.

One can define the exponential function by first defining the functions $a\longmapsto a^n$ with $n$ a positive integer, inductively. Then for $n$ a negative integer using reciprocals. Then for $n$ the reciprocal of a positive integer using inverse functions. Then for $n$ a rational using $a^{p/q} = (a^p)^{1/q}$. Then prove that if $\frac{p}{q}=\frac{r}{s}$ with $p,q,r,s$ integers, $r,s\gt 0$, then we get $a^{p/q}=a^{r/s}$. Then define $a^x$ for arbitrary $x$ by letting $(q_n)$  be a sequence of rationals such that $q_n\to x$ as $n\to\infty$, and showing that the sequence $a^{q_n}$ is Cauchy and converges to a number we call $a^x$. Then showing that if $(q_n)$ and $(r_n)$ are two sequences of rationals that both converge to $x$, then $\lim_{n\to\infty} a^{q_n} = \lim_{n\to\infty} a^{r_n}$. And once all of this has been done, then one can show that the function is stricitly monotone when $a\gt 0$, $a\neq 1$, to deduce what you want (and hence that it has an inverse and logarithms exist).

Obviously, this requires a lot of work.

Or one can use integrals and define the natural logarithm by
$$\ln(x)  = \int_1^x \frac{1}{t}\,dt$$
for $x\gt 0$. Using the Fundamental Theorem of Calculus one can show that this function is continuous and differentiable; using the properties of the integral that it is strictly increasing, and so has an inverse. Call the inverse the exponential function $\exp(x)$; and then define $a^x = \exp(x\ln(a))$. And then prove that this function is strictly monotone when $a\gt 0$, $a\neq 1$.

This requires enough Calculus to prove the Fundamental Theorem of Calculus first. Again, a lot of work.

"Someone did it centuries ago"... The logarithm is a pretty recent "invention" as these things go, and it is closely connected with the development of calculus. Actual formal proofs of its properties (as well as actual formal proofs of the properties of the general exponential function) date from after the invention of calculus, and generally require some analysis or some calculus. I don't think you can really prove this via "elementary", non-analysis, non-calculus methods.

I would like to give a simple explanation.Let us consider a^x=a^y.
Which means aaa*-----(x times) = aaa*------- (y times).
It means aa   (x times)/aa  (y times) =1.
This is possible only when both numerator and denominator has same number of terms.
Which yields x=y.


"All math is useful eventually" We have all heard the argument : a lot of mathematics that was thought to be useless, abstract constructions with no links to the real world ended up being of use, like some arithmetic is useful in crypto. Some people say that the same applies to all of mathematics.

My question is : do you buy this ? More precisely, and to make this into a question that I hope fits the requirements of this website :

Do you have an example of a mathematical theory (i.e. not an isolated theorem, but a coherent set of mathematical concepts and theorems) that you believe will be of no use, ever, to let's say engineers or physicists or non-mathematician scientists in general ?

By of use I mean that it is a mathematical object so relevant to a field or a model that non-mathematicians have to think in terms of this mathematical theory, OR that it is a crucial ingredient to the mathematical proof of some other useful mathematical result. Giving one proof among other, more simple ones is not sufficient. And a theory which language can be used to describe certain models but gives no significant insight or power of prediction doesn't count.

A few examples :

hamiltonian systems are a good way to model most mecanical systems : useful
Fourier transforms are a useful tool in numerous calculations : useful
not being an expert in mathematical physics I can't give a precise example, but I would classify homology theory in useful because it is such a powerful mathematical tool that I'm sure it is a necessary ingredient to something with real-life applications, or will be
tilings : applications in chemistry, including recently one with Penrose's aperiodic tilings : useful
p-adic analysis : to my knowledge not useful.

Let me stress the "to my knowledge", as I know next to nothing about p-adic analysis and what may or may not be its applications.

I think "all math is eventually useful" must be false... There is certainly math that gets abandoned when better things come along, as with cylindrical algebras (as I understand their history). There are things like  first-order mereotopology which have seen little development outside philosophy departments (and not much in them either). Arguably (and I say this with much sadness), NF will never catch on as a serious set theory so much as a source of odd model theory; probably likewise with NFU (which is even less deserving of such a fate).

Not that I think uselessness is a bad thing, of course.

"Do you have an example of a mathematical theory (i.e. not an isolated theorem, but a coherent set of mathematical concepts and theorems) that you believe will be of no use, ever, to let's say engineers or physicists or non-mathematician scientists in general ?"

Take Laplace Transform (LT) and Fourier Transform (FT) for examples. Both are used widely in engineering and physics. Both use infinity. Infinity is not there in nature and in engineering. Therefore these theories are false and cannot or should not be used. A large number is never an approximation to infinity. If you replace infinity, by a large number, then the characteristics of both LT and FT will change. The result of applications will become false.

For example, finite LT will not have poles any more. Therefore by using LT we are corrupting engineering by introducing poles. Similarly, finite FT will change capacity theorem, and uncertainty principle (UP). Finite FT will remove uncertainty from quantum mechanics (QM). Take a look at chapter one on Truth and another chapter on QM in the book at the blog site https://theoryofsouls.wordpress.com/


Begin with one cell, which can die, do nothing, transform to 2 or 3 cells, with probability 1/4 respectively. How's the probability of extinction? A colony begins with a cell, which can die, do nothing, transform to two or three cells, with probability 1/4 for each case at next time point. Children cells share the same property described above. What's the probability of this colony's extinction?

I got two solutions, $1$ and $\sqrt{2}-1$, from a simple recursive equation, $p=\frac{1}{4}(1+p+p^2+p^3)$. But I've no idea which one is correct.
Thanks for your help!

You've done most of the work; what remains is to decide whether $1$ or $\sqrt2-1$ is the probability of eventual extinction. For that purpose, let $x_n$ be the probability of the event that the population goes extinct at or before the $n$-th time step.  Notice that these events form an increasing sequence with respect to $\subseteq$, so the probability of their union, the event of eventual extinction, is the supremum (and also the limit) of the increasing sequence of numbers $x_n$.  The issue is therefore whether this sequence ever gets above $\sqrt2-1$.

We have $x_0=0$ (since the population is initially a single cell, not extinct), and
$$
x_{n+1}=\frac14(1+x_n+{x_n}^2+{x_n}^3)
$$
(because of the rules for how the cells multiply or die or do nothing).  Notice that this equation says $x_{n+1}=f(x_n)$, where $f(x)=\frac14(1+x+x^2+x^3)$ is the function whose fixed points you already calculated. In particular, you know that $f(\sqrt2-1)=\sqrt2-1$.  But $f(x)$ is clearly an increasing function of $x$ as long as $x\geq0$.  So, since $x_0<\sqrt2-1$, we get, by induction on $n$, that $x_n<\sqrt2-1$ for all $n$.  Therefore, the probability of eventual extinction is not $1$ but $\sqrt2-1$.

I have translated this into a birt/death-process. Often one can set the master equations and derive the macroscopic relations from the microscopic. As there is no non-linearity (no liason of the cells) the rarction rates correspond to your probability of $1/4$ and we get

$$C \xrightarrow{k} D$$
$$C \xrightarrow{k} 2C$$
$$C \xrightarrow{k} 3C$$
$$C \xrightarrow{k} C$$

$$\frac{d C}{d t}=-k\,C+kC+k\,C+k\,C$$

$$\frac{d C}{d t}=2k\,C$$

Not sure if there is any threat to the colony to ever die out. $C$ in the equations represents the Expectation value (first moment) of the living cells over the population.

The above calculation holds if you start with a large population. If you start with one cell you will need to set up the master equations with the transition probailities and follow the progress of how your probability distribution changes over each time step. At the start it is $1/4$ then less and less. Larger the population gets less the probability that it dies out. Once large enough (law of large numbers) it will follow the above equations.


How does Cantor's diagonal argument work? I'm having trouble understanding Cantor's diagonal argument.  Specifically, I do not understand how it proves that something is "uncountable".  My understanding of the argument is that it takes the following form (modified slightly from the wikipedia article, assuming base 2, where the numbers must be from the set $ \lbrace 0,1 \rbrace $):

\begin{align}
s_1 &= (\mathbf{0},1,0,\dots)\\
s_2 &= (1,\mathbf{1},0,\dots)\\
s_3 &= (0,0,\mathbf{1},\dots)\\
\vdots &= (s_n \text{ continues})
\end{align}

In this case, the diagonal number is the bold diagonal numbers $(0, 1, 1)$, which when "flipped" is $(1,0,0)$, neither of which is $s_1$, $s_2$, or $s_3$.

My question, or misunderstanding, is: When there exists the possibility that more $s_n$ exist, as is the case in the example above, how does this "prove" anything?  For example:

\begin{align}
s_0 &= (1,0,0,\mathbf{0},\dots)\ \ \textrm{ (...the wikipedia flipped diagonal)}\\
s_1 &= (\mathbf{0},1,0,\dots)\\
s_2 &= (1,\mathbf{1},0,\dots)\\
s_3 &= (0,0,\mathbf{1},\dots)\\
s_4 &= (0,1,1,\mathbf{1},\dots)\\
s_4 &= (1,0,0,\mathbf{1},\dots)\ \ \textrm{ (...alternate, flipped } s_4\textrm{)}\\
s_5 &= (1,0,0,0,\dots)\\
s_6 &= (1,0,0,1,\dots)\\
\vdots &= (s_n \text{ continues})
\end{align}

In other words, as long as there is a $\dots \text{ continues}$ at the end, the very next number could be the "impossible diagonal number", with the caveat that it's not strictly identical to the "impossible diagonal number" as the wikipedia article defines it:

  For each $m$ and $n$ let $s_{n,m}$ be the $m^{th}$ element of the $n^{th}$ sequence on the list; so for each $n$,

  $$s_n = (s_{n,1}, s_{n,2}, s_{n,3}, s_{n,4}, \dots).$$

  ...snip...

  Otherwise, it would be possible by the above process to construct a sequence $s_0$ which would both be in $T$ (because it is a sequence of 0s and 1s which is by the definition of $T$ in $T$) and at the same time not in $T$ (because we can deliberately construct it not to be in the list). $T$, containing all such sequences, must contain $s_0$, which is just such a sequence. But since $s_0$ does not appear anywhere on the list, $T$ cannot contain $s_0$.

  Therefore $T$ cannot be placed in one-to-one correspondence with the natural numbers. In other words, it is uncountable.

I'm not sure this definition is correct, because if we assume that $m = (1, \dots)$, then this definition says that "$s_n$ is equal to itself"&mdadsh;there is no "diagonalization" in this particular description of the argument, nor does it incorporate the "flipping" part of the argument, never mind the fact that we have very clearly constructed just such an impossible $T$ list above.  An attempt to correct the "diagonalization" and "flipping" problem:

$$s_n = (\lnot s_{m,m}, \lnot s_{m,m}, \dots) \quad \text{where $m$ is the element index and} \quad\begin{equation}\lnot s_{m,m} = \begin{cases}0 & \mathrm{if\ } s_{m,m} = 1\\1 & \mathrm{if\ } s_{m,m} = 0\end{cases}\end{equation}$$

This definition doesn't quite work either, as we immediately run in to problems with just $s_1 = (0),$ which is impossible because by definition $s_1$ must be $ = (1)$ if $s_1 = (0)$, which would also be impossible because... it's turtles all the way down!?  Or more generally, with the revised definition there is a contradiction whenever $n = m$, which would seem to invalidate the revised formulation of the argument / proof.

Nothing about this argument / proof makes any sense to me, nor why it only applies to real numbers and makes them "uncountable". As near as I can tell it would seem to apply equal well to natural numbers, which are "countable".

What am I missing?

First, let me give you a proof of the following:

  Let $\mathbb{N}$ be the natural numbers, $\mathbb{N}=\{1,2,3,4,5,\ldots\}$, and let $2^{\mathbb{N}}$ be the set of all binary sequences (functions from $\mathbb{N}$ to $\{0,1\}$, which can be viewed as "infinite tuples" where each entry is either $0$ or $1$).

  If $f\colon\mathbb{N}\to 2^{\mathbb{N}}$ is a function, then $f$ is not surjective. That is, there exists some binary sequence $s_f$, which depends on $f$, such that $f(n)\neq s_f$ for all natural numbers $n$.

What I denote $2^{\mathbb{N}}$ is what Wikipedia calls $T$.

I will represent elements of $2^{\mathbb{N}}$ as tuples,
$$(a_1,a_2,a_3,\ldots,a_n,\ldots)$$
where each $a_i$ is either $0$ or $1$; these tuples are infinite; we think of the tuple as defining a function whose value at $n$ is $a_n$, so it really corresponds to a function $\mathbb{N}\to\{0,1\}$. Two tuples are equal if and only if they are identical: that is,
$$(a_1,a_2,a_3,\ldots,a_n,\ldots) = (b_1,b_2,b_3,\ldots,b_n,\ldots)\text{ if and only if } a_k=b_k\text{ for all }k.$$

Now, suppose that $f\colon\mathbb{N}\to 2^{\mathbb{N}}$ is a given function.
For each natural number $n$, $f(n)$ is a tuple. Denote this tuple by
$$f(n) = (a_{1n}, a_{2n}, a_{3n},\ldots,a_{kn},\ldots).$$
That is, $a_{ij}$ is the $i$th entry in $f(j)$.

I want to show that this function is not surjective. To that end, I will construct an element of $2^{\mathbb{N}}$ that is not in the image of $f$. Call this tuple $s_f = (s_1,s_2,s_3,\ldots,s_n,\ldots)$. I will now say what $s_k$ is. Define
$$s_k = \left\{\begin{array}{ll}
1 &\mbox{if $a_{nn}=0$;}\\
0 &\mbox{if $a_{nn}=1$.}
\end{array}\right.$$

This defines an element of $2^{\mathbb{N}}$, because it defines an infinite tuple of $0$s and $1$s; this element depends on the $f$ we start with: if we change the $f$, the resulting $s_f$ may change; that's fine.  (This is the "diagonal element").

Now, the question is whether $s_f = f(n)$ for some $n$. The answer is "no." To see this, let $n\in\mathbb{N}$ be any natural number. Then
$$f(n) = (a_{1n},a_{2n},a_{3n},\ldots,a_{nn},\ldots)$$
so the $n$th entry of $f(n)$ is $a_{nn}$. If the $n$th entry of $f(n)$ is $0$, then by construction the $n$th entry of $s_f$, $s_n$ is $1$, so $f(n)\neq s_f$. If the $n$th entry of $f(n)$ is $1$, then by construction the $n$th entry of $s_f$, $s_n$, is $0$. Then $f(n)\neq s_f$ again, because they don't agree on the $n$th entry.

This means that for every $n\in\mathbb{N}$, $s_f$ cannot equal $f(n)$, because they differ in the $n$th entry. So $s_f$ is not in the image of $f$.

What we have shown is that given a function $f\colon\mathbb{N}\to 2^{\mathbb{N}}$, there is some element of $2^{\mathbb{N}}$ that is not in the image of $f$. The element depends on what $f$ is, of course; different functions will have possibly different "witnesses" to the fact that they are not surjective.

Think of the function $f$ being hauled before a judged and accused of Being Surjective; to prove its innocence, $f$ produces a witness to verify its alibi that it's not surjective; this witness is $s_f$, who can swear to the fact that $f$ is not surjective because $s_f$ demonstrates that $f$ is not surjective: $s_f$ is not in $\mathrm{Im}(f)$; if the police hauls in some other function $g$ and accuse that function of being surjective, $g$ will also have to produce a witness to verify its alibi that it isn't surjective; but that witness does not have to be the same witness that $f$ produced. The "witness" we produce will depend on who the "accused" is.

The reason this is called the "diagonal argument" or the sequence $s_f$ the "diagonal element" is that just like one can represent a function $\mathbb{N}\to \{0,1\}$ as an infinite "tuple", so one can represent a function $\mathbb{N}\to 2^{\mathbb{N}}$ as an "infinite list", by listing the image of $1$, then the image of $2$, then the image of $3$, etc:
$$\begin{align*}
f(1) &= (a_{11}, a_{21}, a_{31}, \ldots, a_{k1},\ldots )\\
f(2) &= (a_{12}, a_{22}, a_{32}, \ldots, a_{k2},\ldots)\\
&\vdots\\
f(m) &= (a_{1m}, a_{2m}, a_{3m},\ldots, a_{km},\ldots)
\end{align*}$$
and if one imagines the function this way, then the way we construct $s_f$ is by "going down the main diagonal", looking at $a_{11}$, $a_{22}$, $a_{33}$, etc.

Now, remember the definition of "countable":

  Definition. A set $X$ is said to be countable if and only if there exists a function $f\colon\mathbb{N}\to X$ that is surjective. If no such function exists, then $X$ is said to be uncountable.

That means that the theorem we proved above shows that:

  Theorem. The set of all binary sequences, $2^{\mathbb{N}}$, is not countable.

Why? Because we showed that there are no surjective functions $\mathbb{N}\to 2^{\mathbb{N}}$, so it is not countable.

How does this relate to the real numbers? The real numbers are bijectable with the set $2^{\mathbb{N}}$. That is, there is a function $H\colon 2^{\mathbb{N}}$ to $\mathbb{R}$ that is both one-to-one and onto. If we had a surjection $\mathbb{N}\to\mathbb{R}$, then composing this surjection with $H$ we would get a surjection from $\mathbb{N}$ to $2^{\mathbb{N}}$, and no such surjection exists. So there can be no surjection from $\mathbb{N}$ to $\mathbb{R}$, so $\mathbb{R}$ is not countable (that is, it is uncountable).

Bijecting $\mathbb{R}$ with $2^{\mathbb{N}}$ is a bit tricky; you can first biject $\mathbb{R}$ with $[0,1]$; then you would want to use the binary representation (as in wikipedia's article), so that each sequence corresponds to a binary expansion, and each number in $[0,1]$ corresponds to a binary sequence (its digits when written in binary); the problem is that just like some numbers in decimal have two representations ($1$ and $0.999\ldots$ are equal), so do some numbers have two representations in binary (for example, $0.01$ and $0.0011111\ldots$ are equal). There is a way of fixing this problem, but it is a bit technical and may obscure the issue, so I would rather not get into it.

Instead, let me note that the set $2^{\mathbb{N}}$ can be mapped in a one-to-one way into $(0,1)$: simply take a binary sequence
$$(a_1,a_2,a_3,\ldots,a_n,\ldots)$$
and map it to the decimal number that has a $5$ in the $k$th decimal position if $a_k=0$, and has a $6$ in the $k$th decimal position if $a_k=1$. Using $5$ and $6$ ensures that each number has only one decimal representation, so the map is one-to-one. Call this map $h$. Define $H\colon\mathbb{R}\to 2^{\mathbb{N}}$ as follows: given a real number $x$, if $x$ is in the image of $h$, then define $H(x)$ to be the unique sequence $s$ such that $h(s)=x$. If $x$ is not in the image of $h$, then define $H(x)$ to be the sequence $(0,0,0,\ldots,0,\ldots)$. Notice that $H$ is surjective, because $h$ is defined on all of $2^{\mathbb{N}}$.

This is enough to show that there can be no surjection from $\mathbb{N}$ to $\mathbb{R}$: suppose that $f\colon\mathbb{N}\to\mathbb{R}$ is any function. Then the function $H\circ f\colon \mathbb{N}\stackrel{f}{\to}\mathbb{R}\stackrel{H}{\to}2^{\mathbb{N}}$ is a function from $\mathbb{N}$ to $2^{\mathbb{N}}$. Since any function from $\mathbb{N}$ to $2^{\mathbb{N}}$ is not surjective, there is some $s\in 2^{\mathbb{N}}$ that is not in the image of $H\circ f$. Since $s$ is in the image of $H$, there exists some $x\in\mathbb{R}$ such that $H(x)=s$. That means that $f(n)\neq x$ for all $n$ (since $H\circ f(n)\neq s$).

Since there can be no surjection from $\mathbb{N}$ to $\mathbb{R}$, that means that $\mathbb{R}$ is uncountable.

So, as to your questions. First, you should understand that the diagonal argument is applied to a given list. You already have all of $s_1$, $s_2$, $s_3$, etc., in front of you. Nobody is allowed to change them. You construct the "diagonal number" (my $s_f$ above) on the basis of that list. Yes, if you change the list then you can put the diagonal number $s_f$ in the new list; but $s_f$ is only a witness to the fact that the original list was not a list of all sequences. If you change to a different list, then I will have to produce a different witness. The witnesses depend on the given list. You know that $s_4$ is not equal to $s_f$ because $s_f$ is constructed precisely so that it disagrees with $s_4$ in the $4$th position, and one disagreement is enough to guarantee inequality.

Wikipedia's presentation seems to argue by contradiction; I don't like to introduce that into these discussions because the argument is difficult enough to "grok" without the added complication.  (The "Otherwise..." part is an argument by contradiction, arguing that if you could 'list' the elements of $T$, then you would apply the argument to show that this 'complete list' is not 'complete', etc). There's no need. Simply, there is no surjection from $\mathbb{N}$ to $T$, as discussed above.

Now, there is a common "first reaction" that this argument would apply "just as well" to the natural numbers: take a list of natural numbers listed in binary, and engineer an argument like the diagonal argument (say, by "reflecting them about the decimal point", so they go off with a tail of zeros to the right; or by writing them from left to right, with least significant digit first, instead of last) to produce a "number" not on the list. You can do that, but the problem is that natural numbers only corresponds to sequences that end with a tail of $0$s, and trying to do the diagonal argument will necessarily product a number that does not have a tail of $0$s, so that it cannot represent a natural number. The reason the diagonal argument works with binary sequences is that $s_f$ is certainly a binary sequence, as there are no restrictions on the binary sequences we are considering.

I hope this helps.

Several things that are taught, or implied by the way it is taught, about Cantor's Diagonalization are not correct. And they lead to misunderstandings like this one.

1) "Cantor wanted to prove that the real numbers are countable." No. Cantor wanted to prove that if we accept the existence of infinite sets, then the come in different sizes that he called "cardinality."

2) "Diagonalization was his first proof." No. His first proof was published 17 years earlier.

3) "The proof is about real numbers." No. The real numbers were the example he used in the first proof, but some other mathematicians objected to assumptions he made about them. So diagonalization explicitly did not. It used what I call Cantor Strings: infinite length strings of the characters "m" and "w." A similar proof can be used on real numbers, but it requires more steps that are often left out.

4) "It is a formal proof." No. It was an informal explanation of a more formal version, about power sets.

5) "Diagonalization is a proof by coontradiction." No, no, no. No. It is a proof by contraposition, which is similar but not the same. By proving "If A, then B" you also prove "If not B, then not A."

What diagonalization proves, is "If S is an infinite set of Cantor Strings that can be put into a 1:1 correspondence with the positive integers, then there is a Cantor string that is not in S." The contrapositive of this is "If there are no Cantor Strings that are not in the infinite set S, then S cannot be put into a 1:1 correspondence with the positive integers."


Solve the following parametric system of three equations and three unknowns This is my first time posting here so I'm exciting to join the community and gain as much knowledge as possible. My algebra is quite lacking and I'm a first year Physicist. Please could you help me solve the following three equations to find $I_3$ in terms of only $R$s and $\xi$ (i.e. no $I_1$ or $I_2$ terms). Thank you!

$$\begin{eqnarray}
\xi_1 - I_1 R_1 - (I_1 - I_2) R_4    &=& 0\\
- I_2 R_2 - (I_2 - I_3) R_5 - (I_2 - I_1) R_4  &=& 0\\
 - I_3 R_3- I_3 R_6 - (I_3 - I_2) R_5  &=& 0
\end{eqnarray}$$

So far, I get "augmented matrix"
$$
\left(
\begin{array}{ccc|c}
R_1 + R_4 & - R_4 & 0 & \xi_1 \\
-R_4 & R_2 + R_4 + R_5 & - R_5 & 0 \\
0 & - R_5 & R_3 + R_5 + R_6 & 0
\end{array}
\right)
$$

OOH, symmetric square part


Orienting Variables Matrix For the system of equations above, if I were to use matrices to solve for each of the variables, how is the variables matrix ordered? If all the variables in the equations are aligned vertically (as they already are), if I look at one of the equations, do the variables from the left to right correspond to the variables matrix's up and down? In other words, would the variables matrix then be ordered as such: the top variable is z, the middle variable is x, and the bottom variable is y? What is the general rule for orienting the variable matrix, because I can't seem to find one online?

Usually we choose the alphabetical order $x,y,z$ but of course we can also use any others order and in particular $z,x,y$.

In the latter case the matrix form will be

$$\begin{bmatrix}2&1&-1\\-1&2&1\\-3&-1&2\end{bmatrix}\begin{bmatrix}z\\x\\y\end{bmatrix}=\begin{bmatrix}-3\\0\\7\end{bmatrix}$$

which is completely equivalent to

$$\begin{bmatrix}1&-1&2\\2&1&-1\\-1&2&-3\end{bmatrix}\begin{bmatrix}x\\y\\z\end{bmatrix}=\begin{bmatrix}-3\\0\\7\end{bmatrix}$$


Finding the arc length of the parabola $y=x^2 \; from \; (0,0)\;to\;(1,1)$ As the title says, I need to find the arc length of that. This is what I have so far (I'm mostly stuck on the integration part): $${dy\over dx}=2x \Rightarrow L=\int_0^1 \sqrt{1+(2x)^2}dx$$
Substitute $$x=\tan\theta, \qquad dx=\sec^2\theta\,d\theta ,$$
giving $$\int_0^1 \sqrt{1+(2\tan\theta)^2}\sec^2\theta\,d\theta=\int_0^1 \sqrt{1+4\tan^2\theta}\sec^2\theta\,d\theta$$
That is where I'm stuck. Any help is appreciated, thank you.

Let $2x = \tan\theta$ instead. Then, the integral becomes $\displaystyle \int_0^{\arctan 2} \sqrt{1+\tan^2 \theta} \cdot \dfrac14\sec^2\theta \ \mathrm d\theta$ which is equal to $\displaystyle \frac14 \int_0^{\arctan2} \sec^3\theta \ \mathrm d\theta$.

You can make the substitution $2x=sh(t)$ it is simple ot obtain the result.


Fastest way to multiply numbers mentally? I'm wondering what is the fastest way to multiply numbers? For now, let's focus on 2-digit numbers and were one cannot use scrap paper.

I've come across 3 fast methods:

64x43

1)

    60x43 + 4x43         (note that 60x43 is actually a 1-dig. times 2-dig problem)
    2580  + 172  = 2752

2)

      64x43 = 60x40 + 4x3   =2412
             + 4x40  +3x60  = 340
                            =2752

3)

             64
            x43
           =-----
            (4x3)   (4 from 64)          12
      (6x3)+(4x4)   (cross multiply)    34
    (6x4)                              24
                                    ------------+
                                       2752

*I am aware to some multiplications can be done faster with specific tricks,
for example if we square a number, if one of the numbers ends with 5 or is equal to 11, or if they both end in the same number, or if one is close to a multiple of 10, or if one is a product of two 1 digit numbers, etc. But my question regards the general case.

Now practice makes perfect, but before I start training one method I would like to know which one is to be preferred. Is there someone who can say something useful about this?
I guess method 3 is not ideal, since the speed of doing the cross multiplication is partly because of the way it is denoted in this example (one above the other). So can someone say whether method 1 or 2 is to be preferred in sense of speed/simplicity?

Related: What is the fastest way to multiply two digit numbers?

What I usually do, which is probably not the fastest way, is try to look for patterns that I can resolve easily and split the numbers accordingly.

For example, in your case,
$$64\times43 = 64\times32+64\times11$$
I immediately recognize $64=2^6$ and $32=2^5$, and I know how to quickly multiply by $11$, so I get
$$64\cdot43 = 2^{11}+640+64=2048+704 = 2752$$

While strictly speaking this is not the "general case", I find that every time I have to multiply two digit numbers, a trick like this is possible - and I find it to be the most fun way to do it. This is probably hard to scale to more than two digits though.

may be my article will help you.
Here I have described how one can calculate multiplication of two digit numbers mentally.
http://mathsequation.com/how-to-memorize-multiplication-table-fast/

64*43= 64*40 +64*3
so basically problem breaks into multiplication of two digit number with 1 dgit
like 64*4, 64*3
which can be done mentally easily ,I have explained that too.
also once you calculate 64*40=2560 and 64*3=192 mentally.
you can add them  by my mental  addition trick written here.

http://mathsequation.com/mental-addition-trick-learn-within-5-minutes/

I have also explained some additional methods in special cases we can use.


Probability of getting 'k' heads with 'n' coins This is an interview question.( http://www.geeksforgeeks.org/directi-interview-set-1/)

Given $n$ biased coins, with each coin giving heads with probability $P_i$, find the probability that on tossing the $n$ coins you will obtain exactly $k$ heads. You have to write the formula for this (i.e. the expression that would give $P (n, k)$).

I can write a recurrence program for this, but how to write the general expression ?

You can use Dynamic Programming as $N$th turn's outcome is mutually independent to $N-1$th and there are two possible cases here :

$K$ heads already came in $N-1$ turns
$K-1$ heads already came in $N-1$ turns

$dp[i][j]$ : probability of getting $j$ heads in $i$ trials.

So, $dp[n][k] = dp[n - 1][k]\cdot (1 - P[n]) + dp[n - 1][k - 1]\cdot p[n]$

Let F(i,j) be the probability of finding exactly j heads when i biased coins are tossed. and P(i) be the probability of landing head when ith coin is tossed.
The recurrence relation for this problem would look like following :
 F(i, j) = F(i-1, j) * (1-P(i)) + F(i-1, j-1) * P(i)
With the base cases being as written here :- F(1,0) = 1-P(1) and F(1,1) = P(1)

So now the problem reduces to finding value F(n, k) which would give us the probability of finding exactly k heads in n tosses.
this can be extended to find at most k heads or atleast k heads also where we have to find the sum of solutions for all j in [0,k] and [k,n] respectively.


Can one work with any classes of numbers in a proof of number theory? Can one work with any classes of numbers, like natural, integer, rational, real and complex, in a proof of number theory, as long as the result tells something about the integers ? Or should the result be proven using integer-operations only (we can only divide one integer by another if the quotient is an integer, because otherwise we fall out of the default domain) ?

To be more specific, should we stay inside the group of integers when proving results about the integers ? If not, we are proving the result in a larger group containing the integers as a subgroup ?

Why on earth would you impose such restrictions? There's only one thing you should require of a proof: that it be logically valid. The end.

Not only is there absolutely no reason to apply a restriction like that, but modern number theory very frequently steps outside the realm of ordinary integers. Analaytic number theory is roughly speaking the application of calculus to number theory, while algebraic number theory uses systems of numbers that can include complex numbers (for example, the Gaussian integers $a+bi$ where $a,b$ are integers). I've often heard of something called ergodic theory used in contemporary number theory, although I have no idea how or even what exactly that is.

$$\text{Yes, only natural numbers!}$$
Says Kronecker.


Showing the countable direct product of $\mathbb{Z}$ is not projective I am trying to prove that the direct product $M = \mathbb{Z} \times \mathbb{Z}\times \cdots$ is not a projective $\mathbb Z$-module and I am stuck near the end of the proof because of the authors use of the word infinitely divisible.

I will list the sketch of the proof and try to explain where I am stuck.

We proceed by contradiction so suppose $M$ is contained in some free module $F$ with basis $B$.  Set $N = \mathbb{Z} \oplus \mathbb{Z} \oplus\cdots$ and observe that $N$ is a submodule of $M$.

Since $N \subset F$ there exists $B' \subset B$ such that $B'$ is a basis for $N$ and consider the free module $F' \subset F$ determined by $B'$.

Notice $F'+M \subset F$ gives $M/(M \cap F') \cong (F'+M)/F' \subset F/F'$ so we have $M/(M \cap F') $

The next step in the proof requires to consider sequences of signs so let $s = (s_1, s_2, \ldots)$ be a sequence of plus and minus signs and consider an element $m_s := (s_1 , 2 s_2, \ldots , k! s_k, \ldots) \in M$

The next point in the proof is what I don't understand, the notes I am using say $m_s +(M \cap F')$ is infinitely divisible in $F/F'$ and use this to show $M$ cannot be contained in any free $Z$ module.  My question is

  How do we show $m_s +(M \cap F')$ is infinitely divisible in $F/F'$ and how do translate the word infinitely divisible into definitions from Hungerford or Dummite and Foote?

First note that if $\mathbb{Z}^{\mathbb{N}}$ is free it has a basis $\beta$ with cardinality non-countable $\kappa$,  consider   as sets, then the cardinality of $Hom(\beta, \mathbb{Z})$ is $2^{\kappa}>\kappa$. Now the final argument is that if you take morphism $f$ from $\mathbb{Z}^{\mathbb{N}}$ in to $\mathbb{Z}$ and you compose it wiht the inclusion in the jth coordinate $i_j$ , $fi_j=0$ for almost all $j\in\mathbb{N}$. That is for almost all elements in the basis the morphism is zero, from here the cardinality of $Hom(\mathbb{Z}^{\mathbb{N}},\mathbb{Z})$ is $\kappa$, contradiction!!!!!, $Hom(\mathbb{Z}^{\mathbb{N}},\mathbb{Z})$, $ Hom(\beta, \mathbb{Z})$ must have the same cardinality.


Why is multiplication axiomatized? When we learned exponentiation, we first introduced it as an iterated multiplication (for integer powers), then using nth root we defined it for rational powers, and finally we took the limit to define it for arbitrary real power.

My question: why can't we do the same with multiplication? Why is multiplication defined with axioms, rather than as an iterated addition?

My understanding is that standard propositional logic, which has axioms for multiplication and addition, is complicated enough to be undecidable etc. (Gödel's theorems). However, an axiom set that uses only addition is decidable and is thus much less powerful. I don't understand why using multiplication is sufficient to get the full power of mathematics while addition is not sufficient, but that is the case logically.

Perhaps I should state explicitly that I am talking about the natural numbers, not the real numbers, but I think that the point of the power of multiplication (excuse the pun) is relevant.


How to find the total distance traveled, given the position function? A particle moves in a straight line according to the rule $x(t)=t^3-2t+5$, where $x(t)$ is given in meters and where $t$ is given in seconds. Determine the position, velocity, and acceleration of the particle at $t=0$ and $t=3$ seconds. How far has the particle moved during this $3$ second period?

Answer

\begin{align*}x(t)&=t^3-2t+5&x(0)&=5\,m&x(3)&=26\,m\\
v(t)&=3t^2-2&v(0)&=-2\,m/s&v(3)&=25\,m/s\\
a(t)&=6t&a(0)&=0&a(3)&=18\,m/s^2\end{align*}

Total distance traveled is $23.18$m.

My question concerns the total distance traveled. I know by definition distance is the total displacement (the net total distance, regardless of direction). But how do you get $23.18$ m from the equations?

You should integrate the absolute value of velocity from 0 to 3. Than you get the desired result.

Basically a particle will be moving in negative direction if its velocity is negative.As this type of motion is a straight line motion where $x$ is in terms of $t$ therefore total distance travelled =(distance travelled in $+v_e$ direction)+(mod of distance travelled in $-v_e$ direction)....

To solve for total distance travelled:

1.Find velocity vector by differentiating $x$ vector.

2.Find time intervals contained in the given time intervals where $v$ is $-v_e$

3.Integrate $v$ for time interval in which $v$ is $+v_e$ and add a '$-$' sign to those time time interval in which $v$ is $-v_e$ then integrate it for respective time in which $v$ is $-v_e$

4.Finally add the integrated values...


Seems like geometric programming, except equality constraints I have an optimization problem, which is quite similar to geometric programming, except that in equality constraints, I have posynomials instead of monomials. Is there any way to change it in to the form of GP?

My idea was to express $p(X)=1$ as $(1-\epsilon)\leq p(X)\leq 1$, and I was so happy that it leads me to an answer, but unfortunately I cannot handle the left-hand side inequality.
 Any help?

My group calls this a "Signomial Equality", and we've implemented a couple useful heuristics for solving this in http://gpkit.rtfd.org; it's also discussed in a 2013 paper by Xu et. al.


changing the order of the logical symbols $(\forall \epsilon) (\exists\delta)$ by $(\exists \delta)(\forall\epsilon) $ in limit definition Some time ago a professor told the class, which I was in, to analyze why this definition of limit is not good (or if it is a good definition to argument why):

  There exists a $\delta>0$ for all $\epsilon>0$ such that, for all x, whenever $ 0<|x-a|<\delta$ then $|f(x) - L|<\epsilon$

Or, in other words, changing the order of the logical symbols $(\forall \epsilon) (\exists\delta)$ by $(\exists \delta)(\forall\epsilon) $ in the epsilon-delta limit definition implies a bad functioning definition of limit?

Recently this question has returned from the universe of non answered question and it has brought another doubt. Is there some counterexample that can bring a contradiction to this bad definition?

When I try to respond this question I can't find a counterexample since everything I think is linked with the original limit definition. I am still trying to find sufficient small  $\delta$ for a arbitrary small $\epsilon$. But when I enunciate the above definition, intuitively something sounds different.

Can anyone help?

Let's analyse this logically, because what else do we do in mathematics?

  There exists $\delta$ for all $\epsilon$

This entails that for some $\delta$ we have that it is independed of our $\epsilon$. That means if $\epsilon=1$ or $\epsilon=10^8$ we have the same $\delta$, which is bad because it can be suddenly at any difference. For example

$$f(x)=\sin x$$
we can have $\epsilon=2$ and clearly there exists a $\delta$ satesfying the criteria but equally clearly is it that it does encompasses all all points, so if we let $x\to 0$ we have it converges to all values in $[-1,1]$.

The key here is that $(\exists\delta)(\forall\epsilon)$ implies that if $\delta$ is thought of as a function of $\epsilon$ then $\delta(\epsilon)=C$, that is with respect to $\epsilon$ it's constant for that given value.

As the $\epsilon$ bound is independent of the $\delta$, for any $\epsilon>0$,

$$|f(x)-L|<\epsilon$$

therefore $f(x)=L$ for all $0<|x-a|<\delta$. That is, $f(x)$ is a constant function on $0<|x-a|<\delta$.


Limit of sum of terms containing binomial coefficients $$\lim_{n \to \infty} \sum_{k=0}^n \frac{n \choose k}{k2^n+n}$$
The result is $0$. The $n$ from the denominator can be ignored.

If not for the $k$ at the denominator, the result would be $1$, but I can not find the right inequality.

$$\begin{eqnarray*} \sum_{k=0}^{n}\frac{\binom{n}{k}}{k 2^n+n}\leq \frac{1}{n}+\frac{1}{2^n}\sum_{k=1}^{n}\binom{n}{k}\frac{1}{k}&\stackrel{CS}{\leq}&\frac{1}{n}+\frac{1}{2^n}\sqrt{\zeta(2)\cdot\binom{2n}{n}}\\&\leq&\frac{1}{n}+\frac{1}{2^n}\sqrt{\frac{\zeta(2)\cdot 4^n}{\sqrt{\pi n}}}\\&=&O\left(n^{-1/4}\right).\end{eqnarray*}$$
"CS" stands for the Cauchy-Schwarz inequality and the inequality
$$ \sum_{k=0}^{n}\binom{n}{k}^2 = \binom{2n}{n}\leq\frac{4^n}{\sqrt{\pi n}} $$
is well-known. It comes from the usual manipulations about the Wallis product.

From this P. Stanica paper the approximation
\begin{align}
\binom{n}{k} \sim \sqrt{\frac{2}{n \, \pi}} \, 2^{n} \, e^{-\frac{(n-2k)^{2}}{2n}}
\end{align}
is given. Utilizing this result then the desired limit is as follows:
\begin{align}
\lim_{n \to \infty} \, \sum_{k=0}^{n} \frac{n \choose k}{k2^n+n} &= \lim_{n \to \infty} \, \sqrt{\frac{2}{n \, \pi}} \, \sum_{k=0}^{n} \frac{1}{k+ \frac{n}{2^{n}}} \, e^{-\frac{(n-2k)^{2}}{2n}} \to 0.
\end{align}


Proofs of AM-GM inequality The arithmetic - geometric mean inequality states that
$$\frac{x_1+ \ldots + x_n}{n} \geq \sqrt[n]{x_1 \cdots x_n}$$
I'm looking for some original proofs of this inequality. I can find the usual proofs on the internet but I was wondering if someone knew a proof that is unexpected in some way. e.g. can you link the theorem to some famous theorem, can you find a non-trivial geometric proof (I can find some of those), proofs that use theory that doesn't link to this inequality at first sight (e.g. differential equations …)?

Induction, backward induction, use of Jensen inequality, swapping terms, use of Lagrange multiplier, proof using thermodynamics (yeah, I know, it's rather some physical argument that this theorem might be true, not really a proof), convexity, … are some of the proofs I know.

Pólya's Proof:

Let $f(x) = e^{x-1}-x$. The first derivative is $f'(x)=e^{x-1}-1$ and the second derivative is $f''(x) = e^{x-1}$.

$f$ is convex everywhere because $f''(x) > 0$, and has a minimum at $x=1$. Therefore $x \le e^{x-1}$ for all $x$, and the equation is only equal when $x=1$.

Using this inequality we get

$$\frac{x_1}{a} \frac{x_2}{a} \cdots \frac{x_n}{a} \le e^{\frac{x_1}{a}-1} e^{\frac{x_2}{a}-1} \cdots e^{\frac{x_n}{a}-1}$$

with $a$ being the arithmetic mean. The right side simplifies

$$\exp \left(\frac{x_1}{a} -1 \ +\frac{x_1}{a} -1 \ + \cdots +  \frac{x_n}{a} -1  \right)$$

$$=\exp \left(\frac{x_1 + x_2 + \cdots + x_n}{a} - n \right) = \exp(n - n) = e^0 = 1$$

Going back to the first inequality

$$\frac{x_1x_2\cdots x_n}{a^n} \le 1$$

So we end with

$$\sqrt[n]{x_1x_2\cdots x_n} \le a$$

My idea is to use only the Cauchy Schwartz inequality by induction.

Applying the Cauchy_Schwartz inequality on the vectors $(\sqrt{a},\sqrt{a})^T$ and $(\sqrt{b},\sqrt{b})^T$ we get:
$$\sqrt{ab} +\sqrt{ab} \le (a+b)^{1/2}(a+b)^{1/2} $$
that is $$\sqrt{ab} \le \frac{a+b}{2}~~~~~~\color{red}{\text{AM-GM for $n =2$}}$$

  Hypothesis of induction: Assume that for all $p\in\{1,2,\cdots, n-1\}$ we have,
  \begin{align*}
\color{blue}{(a_1a_2\cdots a_p)^{1/p}\leq \frac{a_1+a_2+\cdots+a_p}{p}}
\end{align*}
  We want to prove that it is true for $n$. we will discuss two case: $n =2p$ and $n=2p-1$

If $n =2p$ Then we proceed as follows:
Using that case: $n=2$  and the case $n=p$, we have,

$$ \begin{align}\color{blue}{\frac{a_1+a_2+\cdots+a_n}{n}  }&= \frac{1}{2}\left(\overbrace{\frac{a_1+a_2+\cdots+a_p}{p}}^a+\overbrace{\frac{a_{p+1}+a_{p+2}+\cdots+a_n}{p}}^b  \right) \\
&\ge\sqrt{\left(\frac{a_1+a_2+\cdots+a_p}{p}\right) \left(\frac{a_{p+1}+a_{p+2}+\cdots+a_n}{p}\right)}\\
&\ge\sqrt{(a_1a_2\cdots a_p)^{1/p}(a_{p+1}a_{p+2}\cdots a_n)^{1/p}}\\
&=\sqrt{(a_1a_2\cdots a_n)^{1/p}}  = \color{blue}{(a_{p+1}a_{p+2}\cdots a_n)^{1/n} } ~~~~~~\color{red}{\text{AM-GM for $n =2p$}} \end{align}$$

If $n =2p-1$ (we use the case $n=2p$, as follows.we have,  $n+1 =2p $ and $p<n$ : Then taking $$ a_{n+1} = \frac{a_1+a_2+\cdots+a_n}{n}$$
in the previous case we obtain,

\begin{align}&\frac{a_1+a_2+\cdots+a_{n+1}}{n+1}   \ge (a_1a_2\cdots a_{n+1})^{\frac{1}{n+1}}\\
&\Longleftrightarrow\frac{1}{n+1}\left(a_1+a_2+\cdots+a_n+\overbrace{\frac{a_1+a_2+\cdots+a_n}{n}}^{a_{n+1}}  \right) \ge \left(a_1a_2\cdots a_{n}\left(\overbrace{\frac{a_1+a_2+\cdots+a_n}{n}}^{a_{n+1}}\right)  \right)^{\frac{1}{n+1}}\\
\\&\Longleftrightarrow\frac{a_1+a_2+\cdots+a_n}{n} \ge \left(a_1a_2\cdots a_{n}\left(\frac{a_1+a_2+\cdots+a_n}{n}\right)  \right)^{\frac{1}{n+1}}\\
&\Longleftrightarrow\left(\frac{a_1+a_2+\cdots+a_n}{n}\right)^{\frac{n}{n+1}} \ge \left(a_1a_2\cdots a_{n}\right)^{\frac{1}{n+1}}
\\&\Longleftrightarrow \color{blue}{\frac{a_1+a_2+\cdots+a_n}{n} \ge \left(a_1a_2\cdots a_{n}\right)^{\frac{1}{n}}}~~~~\color{red}{\text{AM-GM for $n =2p-1$}}\end{align}


Dirichlet characters mod $q$ of conductor $f_\chi|d|q$ Let $\hat{G}_q$ be the set of Dirichlet characters modulo $q$. If $d|q$, how many $\chi\in\hat{G}_q$ are there with $f_\chi|d$? ($f_\chi$ denotes the conductor of $\chi$)

And what if we count the $\chi\in\hat{G}_q$ with $f_\chi|d$ and $\chi(-1)=\pm1$? I expect the number to be $\frac12\phi(d)$ but don't know how to prove that.

Thanks alot!

There are $\phi(q)$ Dirichlet characters modulo $q$
If $\chi$ is a Dirichlet character modulo $d | q$ then $\tilde{\chi}(n)=\chi(n) 1_{gcd(n,q)=1}$ is a Dirichlet character modulo $q$.
Let $f(q)$ be the number of primitive Dirichlet characters modulo $q$ then

$$\phi(q) =\sum_{d | q} f(d) \quad \\ \implies \quad f(q) = \sum_{d | q} \mu(d) \phi(q/d) = \prod_{p^k \| q} f(p^k)= \prod_{p^k \| q} (\phi(p^k)-\phi(p^{k-1}))$$

For $q > 2$ : the map $\{ \chi \bmod q \} \to \{-1,1\}$,  $\chi  \mapsto \chi(-1)$ is a surjective group morphism with kernel the Dirichlet characters with $\chi(-1) = 1$. Thus there are $\phi_1(q) = \frac{\phi(q)}{2}$ Dirichlet characters modulo $q$ with $\chi(-1)= 1$ (excepted the special case $\phi_1(1)= \phi_1(2) =1$).

Also $\tilde{\chi}(-1) = \chi(-1)$ thus again (with $f_1(q)$ the number of primitive characters modulo $q$ with $\chi(-1)=1$)
$$\phi_1(q) =\sum_{d |q} f_1(q)  \implies f_1(q) = \sum_{d | q} \mu(d)\phi_1(q/d) = \frac{\mu(q)+\mu(q/2)+f(q)}{2}$$
and hence $$f_{-1}(q) = f(q)-f_1(q) = \frac{-\mu(q)-\mu(q/2)+f(q)}{2}$$

Ok, maybe I got it by myself: Let us define $\hat{P}_{d,\pm}:=\{\text{primitive Dirichlet characters $\tilde{\chi}$ with conductor dividing }d\text{ and }\tilde{\chi}(-1)=\pm1\}$ and $\hat{G}_{d,\pm}=\{\chi\in\hat{G}_q:f_\chi|d\text{ and }\chi(-1)=\pm1\}$. Let $\varphi:\hat{G}_{d,\pm}\longrightarrow \hat{P}_{d,\pm}$, $\chi\mapsto\chi'$, $\chi'$ the primitive character inducing $\chi$.

If $\tilde{\chi}\in\hat{P}_{d,\pm}$, set $\chi(n)=\tilde{\chi}(n)$ for all $(n,q)=1$ and $0$ otherwise. Thus, $\chi\in\hat{G}_q$ and $\tilde{\chi}$ induces $\chi$. So $\varphi(\chi)=\chi'$ and $\tilde{\chi}$ both are primitive inducing $\chi$, therefore $\varphi(\chi)=\tilde{\chi}$. But as $f_\chi=f_{\chi'}=f_{\tilde{\chi}}$ and $\chi(-1)=\chi(q-1)=\tilde{\chi}(q-1)=\pm1$ we got $\chi\in\hat{G}_{d,\pm}$. The uniqueness of $\chi$ is obvious. So we got $\hat{G}_{d,\pm}\cong\hat{P}_{d,\pm}$. Now, as $\hat{P}_d\cong\hat{G}_d$ it follows easily that $|\hat{P}_{d,\pm}|=\frac12\phi(d)$.


Finding $\lim\limits_{x\to 0} \frac{a^x-1}{x}$ without L'Hopital and series expansion. So we have to find $\lim\limits_{x\to 0} \frac{a^x-1}{x}$ without using any series expansions or the L'Hopital's rule.
I did it using both but I have no idea how to do it. I tried many substitutions but nothing worked. Please point me in the right direction.

In order to evaluate the limit $$\lim_{x \to 0}\frac{a^{x} - 1}{x}\tag{1}$$ for $a > 0$, it is necessary to have a proper definition of $a^{x}$. This has been pointed out in one of the comments to the question, but OP has perhaps not understood the necessity of such a comment. In case you want to evaluate the limit $$\lim_{x \to 1}\frac{\sqrt{x} - 1}{x - 1}$$ it is necessary to have a definition of the symbol $\sqrt{x}$ and it is easy to define it by saying that $\sqrt{x}$ is a non-negative real number which gives $x$ when squared.

The unfortunate part here is that there is no definition of $a^{x}$ which is as simple as the definition of $\sqrt{x}$ and most introductory calculus texts try their best not to define this symbol in proper manner. The real sin is committed when they are able to convince the students that there is no necessity of such a definition. It is one thing to avoid a complicated definition but totally another matter to convince that such a thing does not exist.

There are multiple approaches to define $a^{x}$ and almost all of them are hard enough for a beginner in calculus (see these posts here in case you are interested).

I prefer the following approach which is suitable for a beginner in calculus:

Theorem: For a given real number $a > 0$ there exists a unique function $F(x)$ defined and continuous for all real $x$ such that $F(0) = 1, F(1) = a$ and $F(x + y) = F(x)F(y)$ for all $x, y$. Moreover if $x$ is a rational number (say equal to $r \in \mathbb{Q}$) then $F(x) = F(r) = a^{r}$.

This function is called the general exponential function and denoted by symbol $a^{x}$. And then we come to the limit in question:

Theorem: For any real number $a > 0$ the limit $$\lim_{x \to 0}\frac{a^{x} - 1}{x}$$ exists and hence defines a function of $a$ (say $L(a)$) and this function $L(a)$ satisfies the following properties: $$L(1) = 0, L(xy) = L(x) + L(y)$$ Further $L(x)$ is continuous for all $x > 0$ and $$\lim_{x \to 0}\frac{L(1 + x)}{x} = 1\tag{2}$$ The function $L$ is traditionally denoted by $\log x$ and is called the natural logarithm (or just the logarithm) of $x$.

Essentially the above approach gives two standard limits $$\lim_{x \to 0}\frac{a^{x} - 1}{x} = \log a,\,\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1$$ and defers their proof (to advanced courses). But it is more honest than supplying vague ideas and intuition about exponential and logarithmic functions which I consider more of a hand waving.

Convert it into standard limit as write $a^x$   as $e^{x\ln a}$ and
$\lim_{x \to 0} \dfrac{e^{x-1}}{x}=1$


Approximation of a $L^1$ function by a dominated sequence of continuous functions Consider $\mathbb{T}=\{z\in\mathbb{C}\mid |z|=1\}$ and the Lebesgue measure on it. Denote by $L^1(\mathbb{T})$ the set of integrable functions on $\mathbb{T}$ and by $C(\mathbb{T})$ the set of continuous functions on $\mathbb{T}$. I wonder whether or not the following proposition is true.

Proposition. Let $0\leq f\in L^1(\mathbb{T})$. Then there exist a sequence $\{g_n\}_{n\in\mathbb{N}}\subset C(\mathbb{T})$ and a function $g\in L^1(\mathbb{T})$ such that

$g_n\to f$ a.e.
$|g_n(t)|\leq g(t)$ a.e. for all $t\in\mathbb{T}$ and all $n\in\mathbb{N}$.

Just to add some context, I came up with this question while trying to find an alternative proof for a Fejér's-like theorem. I tried the usual techniques (density of continuous function in $L^1(\mathbb{T})$, Luzin's Theorem, etc) but I didn't succeed.

Thank you in advanced.
MD

Yes, your proposition is true.

By density of the continuous functions, there is a sequence of continuous $g_n$ with $g_n \to f$ in $L^1$.  Replacing $g_n$ with $g_n^+$, we can assume $g_n \ge 0$.  Passing to a subsequence, we can also assume $g_n \to f$ almost everywhere.  Passing to a further subsequence, we can assume $\|g_n - f\|_{L^1} \le 2^{-n}$.

(Actually, the first "pass to a subsequence" is unnecessary. As soon as $\sum_n \|g_n - f\|_{L^1} < \infty$, a Borel-Cantelli argument ensures that $g_n \to f$ almost everywhere.)

I claim $g_n$ is the desired sequence.  It remains to construct the dominating function $g$.  Let $h_n = (g_n - f)^+$.  Then $h_n$ is measurable, $\|h_n\|_{L^1} \le \|g_n -f\|_{L^1} \le 2^{-n}$, and we have $h_n \ge 0$ and $g_n \le f + h_n$.  Set $g = f + \sum_{n=1}^\infty h_n$.  Now by monotone convergence
$$\int \sum_{n=1}^\infty h_n = \sum_{n=1}^\infty \int h_n \le \sum_{n=1}^\infty 2^{-n} = 1$$
so $g$ is integrable.  And for each $n$ we have $g_n \le f + h_n \le g$.

This argument didn't use any topology.  Indeed, it still goes through if we replace $\mathbb{T}$ by any measure space $(X,\mu)$, and replace $C(\mathbb{T})$ by any dense subset $E \subset L^1(X,\mu)$ which is closed under the "positive part" operation.  Maybe that latter condition can be weakened even further.

If $f$ is continuous, the result is a consequence of Arzela-Ascoli theorem. Because the sequence is non-negative, there is a sequence of measurable functions in $L^1$ that converges to $f$. Besides, thanks to the density of the continuous functions in $L^1$, we can assume that this functions are continuous. Let $F$ be such sequence of functions, i.e. $F=\{f_{n}:n\in \mathbb{N} \}⊂C(\mathbb{T})$. Any sub-sequence of functions in $F$ converges in $F\cup\{f\}⊂C(\mathbb{T})$. Arzela-Ascoli implies that $F$ is uniformly bounded, i.e. that the sequence satifies 2. For the general case, since $f$ is $L^1$, you can take a sequence of continuous functions converging to $f$. You can prove that there is a bound that holds for any case.


Reducing 12 modulo 9 This example is from a text on Modular Arithmetic

We want to reduce 12 modulo 9. Here are several results which are
correct according to the definition:
1) 12 ≡ 3 mod 9,     3 is a valid remainder since 9|(12−3)
2) 12 ≡ 21 mod 9,    21 is a valid remainder since 9|(21−3)
3) 12 ≡ −6 mod 9,    −6 is a valid remainder since 9|(−6−3)

I understand the first one.
For a ≡ r mod m, we can write m | (a - r)
So in the first example, a = 12 r = 3 m = 9.
So from that we get 9 | (12 - 3)
But for the second, a similar substitution would be
9 | (12 - 21) - But in the example, they have (21 - 3) instead of (12 - 21)
Likewise in the $3^{rd}$, they have (-6 -3) instead of (12 + 6)
How do they arrive at this in the $2^{nd}$ & $3^{rd}$ example?

If $a \equiv b \pmod m$ and $b \equiv c \pmod m$, then $a \equiv c \pmod m$. This is called transitivity and you should prove that it is true.
For your second example: The author in your text uses $12 \equiv 3 \pmod 9$ and $3 \equiv 21 \pmod 9$, so $$ 12 \equiv 3 \equiv 21 \pmod 9.$$ Of course, you can also just check $9 \mid 12 - 21$, as you said.

I am answering my own question. I think the author is trying to find all answers which form an equivalence class for mod 9.
3, 21 & -6 are all part of the equivalence class for mod 9.
{ ...-15, -6, 3, 12, 21, ... } is one of the 9 equivalence classes for mod 9.
That's how he gets to (2) & (3) from (1)


Checking understanding on proving uniqueness of identity and inverse elements of a group. Sorry for such a trivial question, but just wanted to check my understanding.

When proving a statement, for example, that the inverse of a group element is unique (in elementary group theory) one starts by supposing that there exists two inverses $h$ and $k$ for a given element $g \in G$, where $G$ is some group with binary operation $\ast:G\times G \rightarrow G$, such that $h\ast g = g\ast h = e$  and $k\ast g = g\ast k = e$ where $e \in G$ is the unique identity for the group $G$. From this, one can show that $h=k$ in the following manner: $$ h=h\ast e =h\ast\left( g\ast k\right) = \left(h\ast g\right)\ast k = e\ast k =k$$ and as such $h=k$.
Now, is the reason we can from this state that the inverse of an element $g\in G$ is unique because $h$ and $k$ were chosen arbitrarily, apart from the requirement that they are inverses of $g$, and as such, if we know the value of one of the inverses, say $h$, then we know that for any other value $k$ to be an inverse of $g$ it must be equivalent to the known inverse $h$, and thus $h$ is the unique inverse of $g$. Is this reasoning correct?

Sorry for the wordiness of this question, but just wanted to check my understanding explicitly. Also, although I understand that, logically, I should have proven this first (but I'd already written out the inverses part before thinking of this, so apologies for that), but is the reasoning the same for arguing that the identity element of a group is unique? (i.e. If we assume that there are two identity elements $e,g \in G$ and subsequently show that $e=g$, then this implies that if we know the form of one of the identities, say $e$, then for any other value $g$ to be an identity of the group $G$ it must be equivalent to $e$ and thus the identity element of a group is unique).

Group axioms tell you that an identity element must exist, and also that every element has an inverse. They don't tell you that there's only one identity element, and they don't tell you that an element can have only one inverse: these are things that you have to prove.

So, if for example you want to prove that there is only one identity element, suppose instead that there are more than one. If so, you can choose two of them that are not the same element: yet as the proof goes on you see that those elements are instead the same one, as you know. This means that supposing that there are lots of identity elements yields a contradiction: hence, since you can't have more than one of them, and since at least one has to exist because of the group axioms, you see that the only possible option is that there's exactly only one identity element.

The same happens when you're trying to show that every element of the group has only one inverse.

Suppose $x\in G$ has two inverses $y$ and $z$ then $$zx=yx\Rightarrow (zx)y=(yx)y\Rightarrow z(xy)=y(xy)\Rightarrow ze=ye$$ So $y=z.$


Good book on abstract algebra I am an undergraduate student - just started to learn about group theory, ring theory and module theory. I am self studying from the book "Basic Abstract Algebra" by P.B.Bhattacharya, S.K.Jain, S.R.Nagpaul. Prof said that this book is written in an uncomplicated manner but i found the statements in it with more hints rather than complete proofs. What are your thoughts about this book? Our prof also suggested Abstract Algebra by Dummit and Foote - but i found it quite bulky. Please enlighten me with your thoughts as i am very much lost - which one to follow. Thank you.

Dummit and Foote. Its bulkiness means it'll serve you well for years. It's detailed, patient, logical, and comprehensive. Some of its later material isn't great in my opinion, like its treatment of homological algebra with basically no motivation or punchline, but few people make it there anyway. Its character theory is actually quite good too.

Lang's Algebra is a classic,  or near classic,  and there's no reason not to start consulting it, even though it is advanced.
Meanwhile,  to address the question,  Fraleigh's A First Course in Abstract Algebra would surely not have been chosen by professor John Stallings, who was surely one of the best group theorists of his generation,  as the textbook for his undergrad course at Berkeley,  if it were anything but a lovely introduction to the subject.   And that's exactly what I found it to be.


Integer solutions of an equation that is set to a number How many integer solutions for $a$ and $b$ in $(ab)/(a+b)=3600$?

My attempt:

$(ab)/(a+b)=3600$

= $ab=3600(a+b)$

= $ab=3600a+3600b$

=$ab=3600a=3600b$

Dividing $3600b$ on both sides

=$a(1-3600)/3600$

I am not really sure if this is correct. I was just doing this for fun after I was reading this in some mathematical book. It just explained about a few things not much. Can someone please help me with this? I was trying this for a while now and wanted to know what it would look like.

HINT : Note that
$$\begin{align}\frac{ab}{a+b}=3600&\iff ab=3600(a+b)\ \ \text{and}\ \ a+b\not=0\\&\iff (a-3600)(b-3600)=3600^2\ \ \text{and}\ \ a+b\not=0.\end{align}$$
Can you take it from here?

We get $ab=3600(a+b)$ or $-3600a+ab-3600b=0$. We can factor $-3600a+ab$ as $a(-3600+b)$. We get $$a(-3600+b)-3600b=0$$. Adding $3600^2$ to both sides gives $$a(-3600+b)+3600(3600)-3600b=3600^2$$ or equivalently, $$a(-3600+b)-3600(-3600)-3600b=3600^2$$.

This factors as $a(-3600+b)-3600(-3600+b)$ or equivanlently, $$(a-3600)(-3600+b)=3600^2$$ .

The prime factorization of $3600^2$ is $(2^4 \cdot 3^2 \cdot 5^2)^2$ or $2^8 \cdot 3^4 \cdot 5^4$. There are $225$ factors of $3600^2$.

$(a-3600)(-3600+b)$ can each be factors of $3600^2$ and since there are $225$ factors, there are 225 integer solutions for positive $a$ and $b$.

There are also $225$ integer solutions for negative $a$ and $b$.

Now we must deal with the condition where $a+b \neq 0$. We can let $a=-b$ and plug this into our equation to find our wrong solutions.

We get $$(-b-3600)(b-3600)=3600^2$$.

$(-b-3600)$ must equal $\pm 3600$. $3600$ factors as $2^4 \cdot 3^2 \cdot 5^2$ so there are $45$ solutions for positive $b$ and $45$ solutions for negative $b$. So there are $90$ wrong solutions.

In total, we have $225+225-90=\boxed{360}$ integer values for $a$ and $b$.


Evaluate the double integral $\int_{-1}^{0} \int_{1}^{2}(x^2y^2 + xy^3)dydx$ $\int_{-1}^{0} \int_{1}^{2}(x^2y^2 + xy^3)dydx$

\begin{align}
\int_{-1}^{0} \int_{1}^{2}(x^2y^2 + xy^3)dydx = \int_{-1}^{0} \bigg[\frac{x^2}{3}y^3 + \frac{x}{4}y^4\bigg]_{1}^{2}dx = \\ \int_{-1}^{0} \bigg[ \frac{8x^2}{3} + \frac{16x}{4}\bigg] - \bigg[\frac{x^2}{3} - \frac{x}{4} \bigg]dx = \int_{-1}^{0}\bigg( \frac{7x^2}{3} + \frac{17x}{4}\bigg)dx \\= \bigg[\frac{7}{9}x^3 + \frac{17}{8}x^2 \bigg]_{0}^{-1} = 0 - \bigg[ \frac{-7}{9} - \frac{17}{8}\bigg] = \frac{209}{72}
\end{align}

This is my work, but the textbook says the answer is $-\frac{79}{72}$. I double checked my work and everything seems to be checking out for me on my end, wasn't sure if the textbook made a mistake or if there is a super small mistake I am missing here.

Here had been the first out of two sign errors:
$$\bigg[ \frac{8x^2}{3} + \frac{16x}{4}\bigg] - \bigg[\frac{x^2}{3}
+ \frac{x}{4} \bigg]$$
See the + before the $\frac{x}{4}$. At the end it should be:
$$\bigg[\frac{7}{9}x^3 + \frac{15}{8}x^2 \bigg]_{-1}^{0} = 0 - \bigg[ \frac{-7}{9} + \frac{15}{8}\bigg] = -\frac{79}{72}$$

Note the + instead of the - before the $\frac{15}{8}$.


Prove that $n^2(n^2+1)(n^2-1)$ is a multiple of $5$ for any integer $n$. Prove that $n^2(n^2+1)(n^2-1)$ is a multiple of $5$ for any integer $n$.

I was thinking of using induction, but wasn't really sure how to do it.

Lemma:

For any polynomial of integer coefficients $P(n)$, $n=0,1,2,3,4$ exhaust the possible values of $P(n)\bmod 5$.

Indeed for any $a$, $(n+a)\bmod 5=(n\bmod5+a)\bmod5$, and $(na)\bmod5=((n\bmod5)a)\bmod5$, so that $P(n)\bmod5=P(n\bmod 5)\bmod5$. QED.

Now,
$$0^2(0^4-1)\bmod5=0\\
1^2(1^4-1)\bmod5=1\cdot0\bmod5=0\\
2^2(2^4-1)\bmod5=4\cdot15\bmod5=0\\
3^2(3^4-1)\bmod5=9\cdot80\bmod5=0\\
4^2(4^4-1)\bmod5=16\cdot255\bmod5=0\\$$


Uniform continuity implies existence of increasing continuous function In the book that I’m reading, the author makes the following assertion, which I was not able to prove:
If $c:\mathbb R \times \mathbb R\to \mathbb R$ is a continuous function on a compact set (i.e $c$ restricted to a compact set) and hence uniformly continuous, then there exists an increasing continuous function $w:\mathbb R_+ \to \mathbb R_+$, with $w(0)=0$ and such that
$$
\mid c(x,y) - c(x’,y’) \mid \leq w(d(x,x’) + d(y,y’))
$$
Can anyone prove that this is in fact true?
P.s: Note that $d$ is a metric.

A friend of  mine showed me the solution, so here it is:
I'll sove this for $\mathbb R$ instead of $\mathbb R^2$ for sake of simplicity, because the proof is the same. So, I'll change $c:\mathbb R \times \mathbb R \to \mathbb R$ for $f:\mathbb R\to \mathbb R$, also continuous.
Let $K$ be the compact and
$$ w(z) := \sup_{x,y:d(x,y)\leq z} |f(x) - f(y)|$$
Hence, $|f(x) - f(y)|\leq w(d(x,y))$. Also, it's clear that $w$ is an increasing function. We have to prove that $w$ is continuous, and $w(0)=0$.
It's clear that $w(0) = 0$, since $d(x,y) =0 \iff x=y$.
The only thing left to prove is the continuity, which we'll prove for $w$ at 0, and a similar proof can be done to other points.
Note that since $f$ is continuous on a compact set, it is then uniformly continuous. Therefore,
$$\forall \epsilon>0, \exists  \delta>0, \quad \text{s.t} \quad  d(x,y) \leq \delta \implies |f(x) - f(y)| \leq \epsilon$$
Taking the $\sup$ , we obtain that:
$$
\sup_{x,y:d(x,y)\leq \delta}|f(x) - f(y)| = w(\delta) \leq \epsilon
$$
This means that $d(x,y)\leq \delta \implies w(\delta) \leq \epsilon$, hence,
$\lim_{\delta \to 0}w(\delta) = 0$.
Which proves that $w(0) =0$m and that $w$ is continuous on $0$. The only thing left to do is to prove the continuity for the rest of the points, which I could not do actually.

This is not true in general. Consider $c(x,y) = x^2$ and the points $(x, y), (x', y')$ such that $d(x, x') = d(y, y') =a$ in which case the RHS is 0. But $x^2 - x'^2$ is not always 0.


Show that there are polynomials $q(x)$ and $r(x)$ with integer coefficients such that $f(x)=g(x)q(x)+r(x)$ and $\deg(r)<\deg(g)$. Let $f(x), g(x)$ be polynomials with coefficients in $\mathbb Z$. Suppose that $\deg(g)≥1$ and that the leading coefficient of $g$ is $1$. Show that there are polynomials $q(x)$ and $r(x)$ with integer coefficients such that $f(x)=g(x)q(x)+r(x)$ and $\deg(r)<\deg(g)$.

By letting $q=0, r=f$, we see that there are polynomials $q,r$ with integer coefficients such that $f=qg+r$. Among all such pairs of polynomials, pick one that minimizes $\deg r$. I claim that $\deg r<\deg g$. Indeed, if we assume that $\deg r\ge \deg g$ is possible where $g(x)=x^n+a_{n-1}x^{n-1}+\ldots +a_0$ and $r(x)=b_mx^m+\ldots +b_0$ with $m\ge n$ and $b_m\ne 0$, say, then for $q^*(x):=q(x)+b_mx^{m-n}g(x)$ and $r^*(x):=r(x)-x^{m-n}g$ we have $f=q^*g+r^*$ and $\deg r^*<m=\deg r$, contradiction.

You can get $q(x)$ by doing polynomial division of $\frac{f(x)}{g(x)}$. Then you will find out that all coefficients of $r(x)$ with degree$\ge \deg(g)$ is $0$. Therefore $\deg(r)<\deg(g)$ or $r(x)=0$

It's called division algorithm for polynomials.


Circumsphere of a tetrahedron undefined? I am trying to find 3D alpha shapes from my data-set. In doing so, I am keeping only those tetrahedra that have circumradius below a certain threshold. However, while finding the circumradius of the tetrahedra making up the delaunay triangulation of the data, I am finding that for some cases, it becomes imaginary. I am using the standard formulae, given for example, in http://mathworld.wolfram.com/Circumsphere.html.
Are these formualae always valid for any (non-degenerate) tetrahedron ?
An example of the vertices of a tetrahedra for which it fails:

1.0e+03 *

-0.882361572000000   1.832846680000000   8.039920898000000

-0.871205933000000   2.190948975000000   7.713502440999999

-0.874571533000000   1.637495972000000   7.953884766000000

-0.945120239000000   1.753712891000000   8.093748535000000

I am using Matlab here, along with its delaunayn function. Essentially in the formula
$r = \frac{ \sqrt{D_x^2 + D_y^2 + D_z^2 - 4 a c} }{2 |a| }$, the discriminant under the square root becomes negative. My computed values are:

[$D_x \, D_y \, D_z \, a \, c$] =

1.0e+14 *

0.000152327124454

0.000160507051388

-0.000890797019744

-0.000000057932589

-3.705507881061755

If this was StackOverflow they would say "show me your code." It looks like you've at least done something with the minus sign in calculating Dy. Here's my code where the four vertices of the tetrahedron are stored in the 4-by-3 matrix xyz:

w = ones(4,1);
s = sum(xyz.^2,2);
Dx = det([s xyz(:,2:3) w])
Dy = -det([s xyz(:,[1 3]) w])
Dz = det([s xyz(:,1:2) w])
a = det([xyz w])
c = det([s xyz])
r = 0.5*sqrt(Dx^2+Dy^2+Dz^2-4*a*c)/abs(c)

The resulting radius is a positive real value: about 8.72e-6. Yes, the negative Dy goes away when it gets squared (and could be omitted in your case) – did you maybe put the negative inside of the square root?

I can't answer your your more mathematical question, but I'd hazard to guess that this formula is fine. I'd think that numerical issues would potentially be a bigger issue. There's potential for overflow in your case if the numbers get bigger.

In my opinion the radius should be:

$r = 0.5\times\sqrt{Dx^2+Dy^2+Dz^2-4ac}/|a|\approx 0.5577$

But thanks, you helped me a lot.


Examples of Diophantine equations with a large finite number of solutions I wonder, if there are examples of Diophantine equations (or systems of such equations) with integer coefficients fitting on a few lines that have been proven to have a finite, but really huge number of solutions?

Are there ones with so large number of solutions that we cannot write any explicit upper bound for this number using Conway chained arrow notation?

Update: I am also interested in equations with few solutions but where a value in a solution is very large itself.

I think the rough answer to your question is that "there are no such natural equations." Let me try to justify this heuristic.

 Caporaso, Harris, Mazur; and Baker's method: Consider diophantine equations corresponding to curves,  for example equations of the form:

$$C: f(x,y) = 0$$

where $f(x,y)$ is a polynomial with integer coefficients. One knows that the number of solutions (with $x$ and $y$ in $\mathbf{Z}$ or $\mathbf{Q}$) is --- to some extent --- determined by the complex geometry of $C$. If the genus $g$ of $C$ is greater than two, then Faltings proved that $C$ has  only finitely many rational points. However, much more (may) be true. Assuming a conjecture of Lang, it was proved by Caporaso, Harris, and Mazur (J. Amer. Math. Soc. 10 (1997), 1-35) that the number of rational points $\#C(\mathbf{Q})$ is bounded by a function $A(g)$ which only depends on $g$, not on $C$. This doesn't prove that $A(2)$ [for example] is not enormous, but all known lower bounds in $g \ge 2$ on the number of rational points are at most polynomial. For example, I think the largest number of rational points on a genus two curve that has ever been found has at most a few hundred rational points. Since the genus is controlled by the degree of $f$, this suggests that it is highly unlikely to find an equation with small coefficients and degree which has absolutely massive solutions, even if one makes the coefficients big. Note that this is only a heuristic, but it strongly suggests that there are no such equations. It certainly says that no such equations are currently known.

One is in slightly better shape (or worse, depending on what you're trying to do) if one restricts to integral points of $C$. For certain classes of equations (say $F(x,y) = m$ for homogenous $F$) there are explicit bounds on the number of integral solutions in terms of the coefficients coming from Baker's theorem (on linear forms in logarithms). The bounds are not great (perhaps super-super exponential in the coefficients), but they certainly preclude numbers of the size you are interested arising from anything that one can sensibly write down. Moreover, one conjecturally expects that Baker's bounds are not optimal.

For genus 1 and 0, the situation is similar. There's difference here again between whether one wants to consider only integral solutions or rational solutions, but the result (in either case) is that if one insists that there are only finitely many solutions (in either integers or rationals), then there is a bound for the largest such solution in terms of the coefficients (which, using Baker's method for integral points, might be quite large, but is still tiny compared to the numbers you are discussing).

Moral: If one assumes Lang's conjecture (and we certainly don't have any evidence that it is false), there is still heuristic evidence to suggest that the the types of equations  you are looking for do not exist, either for curves or higher dimensional varieties: this is very speculative, but certainly means that writing down any such equations is either impossible or very hard.

Exponential Diophantine Equations:
There are other flavours of Diophantine equations besides algebraic varieties. For example, you could consider exponential Diophantine equations, e.g.: $2^n = x^2 + 23$. In this case, one can often convert solutions to these equations into subsets of polynomial equations of genus at least two. For example, the previous equation gives solutions to one of the five genus three curves:
$$Ay^5 = x^2 + 23, \qquad A = 1,2,4,8,16.$$

Thus one can often "reduce" to the case of curves. Alternatively, Baker's method can often be used directly on exponential equations to give upper bounds.

The phenomena of a big "smallest" solution:
There are some equations which have infinitely many solutions of which the "smallest" is quite large, for example, the first non-trivial solution to Pell's equation $x^2 - 61 y^2 = 1$ has
$$[x,y] = [1766319049, 226153980].$$
 However, in these cases (and in similar cases coming from elliptic curves) one at least expects (and knows for Pell) that there exists a bound on the
 regulator which is polynomial in the coefficients, where the regulator involves the logarithm of the entries. So this gives upper bounds for primitive solutions which are exponential in a polynomial in the coefficients, still enough to prevent super-huge numbers arising as the smallest interesting solution.

Summary: There are many other flavours of Diophantine equation one can write down, but I think the summary is that the conjectures of Lang suggest, on some heuristic level, that the type of phenomena you are looking for simply does not occur naturally. There may be a way of embedding Ackerman's function in to some system of equations (but not into polynomial equations), but I think you would consider that cheating.

Let $b$ is a non-zero integer, and let $n$ is a positive integer.
The equation $y(x-b)=x^n$ has only finitely many integer solutions.
The first solution is:
$x=b+b^n$ and $y=(1+b^{n-1})^n$.
The second solution is:
$x=b-b^n$ and $y=-(1-b^{n-1})^n$,

cf. [1, page 7, Theorem 9] and [2, page 709, Theorem 2].

The number of integer solutions to $(y(x-2)-x^{\textstyle 2^n})^2+(x^{\textstyle 2^n}-s^2-t^2-u^2)^2=0$
grows quickly with $n$, see [3].

References

[1] A. Tyszka, A conjecture on rational arithmetic
which allows us to compute an upper bound for the
heights of rational solutions of a Diophantine
equation with a finite number of solutions,

http://arxiv.org/abs/1511.06689

[2] A. Tyszka, A hypothetical way to compute an upper
bound for the heights of solutions of a Diophantine
equation with a finite number of solutions,
Proceedings of the 2015 Federated Conference on
Computer Science and Information Systems
(eds. M. Ganzha, L. Maciaszek, M. Paprzycki),
Annals of Computer Science and Information Systems, vol. 5, 709-716,
IEEE Computer Society Press, 2015.

[3] A. Tyszka, On systems of Diophantine equations with a large number of integer solutions,

http://arxiv.org/abs/1511.04004


For Orthogonal Unit Vectors, Prove that Let $A$ be an $n \times n$ matrix. Prove that for $n\geq 2$ there exist $n$ orthogonal unit vectors $u_1,\ldots,u_n$ in $\mathbb{R^n}$ such that $Au_1,\ldots,Au_n$ are also orthogonal.
As I recall this is a theorem but couldnt find its proof. Can someone help me please.

This seems just a direct consequence of the Singular Value Decomposition for any matrix. In fact, $A=UDV$, where $D$ is diagonal and $U,V$ are orthogonal. As a consequence, $AV^T = UD$, but the columns of both $V^T$ and $UD$ are orthogonal.


Why isn't several complex variables as fundamental as multivariable calculus? One typically studies analysis in $\mathbb{R}^n$ after studying analysis in $\mathbb{R}$.  Why can't the same be said of $\mathbb{C}$?

I would say that the main reason that several complex variables is rarely seen in the undergraduate curriculum (and even not that often in the graduate curriculum unless the department has some specialists in SCV) is that you can't get very far without lots of prerequisites.

You can for example start by proving Cauchy's integral formula for a polydisc, and from that Liouville's theorem and a few other well known results from one complex variable quickly follow.

From Cauchy's integral forumla, it also follows that holomorphic functions of several variables admit power series expansions (but the domain of convergence is not usually a ball in $\mathbb{C}^n$: compare $\sum_{j,k} z^j w^k$, $\sum_{k} (z+w)^k$ and $\sum_{k} z^k w^k$ for a few examples of what might happen). From here you can go on and study logarithmically convex Reinhardt domains.

Note, however, that the mere definition of a holomorphic function in several variables is a little problematic. You want to say that a function is holomorphic if is is holomorphic in each variable separately, but to show that this is equivalent to other plausible definitions (without assuming that the function is for example locally bounded or jointly continuous) is surprisingly difficult.

You may even get as far as showing a version of Hartogs' extension theorem: If $\Omega$ is a domain in $\mathbb{C^n}$ and $K$ is a compact subset such that $\Omega \setminus K$ is connected, every holomorphic function on $\Omega\setminus K$ extends to $\Omega$. (Here $n > 1$, of course.)

I think this is about how far you can get without bringing in tools from PDE, potential theory, algebra (sheaf theory), functional analysis, differential geometry, distribution theory and probably a few more fields.

The big highlight in a first course in several complex variables is usually to solve the Levi problem, i.e. to characterize the domains of existence for holomorphic functions. (Hartogs' extension theorem shows that some domains are unnatural to study, since all holomorphic functions extend to a bigger domain.) This is usually done with Hörmanders $L^2$-solution of the $\bar\partial$-equation. (Or via sheaf theory a la Oka.) While it's not strictly necessary to have a modern PDE course as a prerequisite, it's certainly valuable. At the very least you need to know some functional analysis (and preferably some potential theory as well), including some exposure to unbounded linear operators on Hilbert spaces to be able to understand the Hörmander solution. (For the sheaf theory solution, you need a healthy background in algebra instead.)

Similarly, you need some differential geometry (at least familiarity with differential forms and tangent bundles) to understand the more complicated integral formulas such as Bochner-Martnielli's formula and the geometric aspects of pseduoconvexity, which is central for a deeper understanding of SCV. In fact, the interplay between the complex geometry of the domain and the corresponding function theory is a reoccuring theme in SCV. Function theory in strictly pseudoconvex domains, for example, look rather different from function theory in weakly pseudoconvex domains. (Many finer points concerning weakly pseudoconvex domains are still open problems.)

Summing up, to do a real meaningful course in SCV, you really need more background than what is reasonable to expect from an undergraduate. After all SCV is really a 20th century field of mathematics! The Levi problem for example wasn't solved until the early 50's (Hörmander's solution is as late as 1965).

You need to know concepts from analysis on $\mathbb R$ and $\mathbb R^2$ including power series, derivatives, line integrals, functions, analytic geometry, and basic topology of those real (metric) spaces before you talk about analytic functions from $\mathbb C$ to $\mathbb C$. That is the main reason why you see analysis on real spaces before you get to analysis on $\mathbb C$. $\mathbb R^2$ with cartesian and polar coordinates is $\mathbb C^1$.  And of course you do $\mathbb C^1$ before analysis on higher dimensional $\mathbb C$-spaces.


Proper definition use in Stoke's theorem Let the curve C be a piecewise smooth and simple closed curve enclosing a region, D.

Some sources asserts Stoke's theorem to be:

$$\oint_{C} F.dr = \iint_{R}\nabla \times FdS$$

Whereas, some claims it to be

$$\oint_{C} F.dr = \iint_{R}\nabla \times F.n.dS$$

Could someone clear the air as to which of the above definition used is correct?

Thanks in advance.

From the first of the statements, the surface integral is written as $$\iint_R \nabla\times \vec F \cdot d\vec S$$ where I have added overarrows to clarify vector quantities. In particular, $d\vec S$ means integrating over the surface in the direction of the unit normal.

In the second statement the surface integral is written as

$$\iint_R \nabla\times\vec F \cdot \hat n \ dS$$

where $\hat n$ is the unit normal and the integral now is over the surface treated as a 'scalar'. Loosely speaking, $d\vec S = \hat n \, dS$.

In other words, the two statements properly understood are equivalent.


Give two integers $m$ and $n$ such that $n^2$ is a multiple of $m$, $n$ is not a multiple of $m$ and $n > m$ Give two integers $m$ and $n$ such that $n^2$ is a multiple of $m$, $n$ is not a multiple of $m$ and $n > m.$

I think that such two integers don't exist, but even if that were true, I'm having trouble with proving that. I have found clear examples when I neglect the last condition ($n > m$).

Any help on this problem would be much appreciated. The observation that I did make is the fact that since $n^2$ is a multiple of $m$ we can write it as $k * m$, which means that $k$ has to be larger than $n$.

I think that could be a starting point of some sort, but I wasn't capable of going any further.

In general pick $n=pq$, where $p$ is a prime and let $m=p^2$, where $q$ is also a prime such that $p<q$. This satisfies all your requirements.

Given a squarefree $a$ and $c>b>1$ such that $b$ is not a factor of $c,$ then we can define $m=ab^2$ need $n=abc.$

Then $n=abc>ab^2=m,$ and $n^2=a^2b^2c^2$ is divisible by $m=ab^2$, and $\frac{n}{m}=\frac{abc}{ab^2}=\frac{c}{b}$ is not an integer since $c$ is not divisible by $b.$

On the other hand, given any two $m,n$ satisfying these conditions, we can find $a,b,c.$

If $m$ is square-free, then $m\mid n^2$ implies $m\mid n.$ (Why?) So $m$ is not square-free.

This means that we can write (uniquely) $m=ab^2$ where $a$ is square-free and $b>1.$

Now, since $ab^2\mid n^2,$ with $a$ square-free, then then $ab\mid n.$ (Why?)

Writing $n=abc$ for some $c,$ then we then see $n>m$ means $c>b.$ We also see that that $m=ab^2$ is a factor of $n=abc$ if and only if $b$ is a factor of $c.$

So this construction gives all $m,n$ uniquely: Start with $a$ square-free, and $c>b>1$  with $c$ not a multiple of $b.$ Then take $m=ab^2,n=abc.$

Simple cases: When $n=1$ and $b>1$ we can choose $c=b+1$ then we get $m=b^2$ and $n=b(b+1).$


Prove by induction that $(n+1)^2 + (n+2)^2 + ... + (2n)^2 = \frac{n(2n+1)(7n+1)}{6}$ Prove by induction that $$(n+1)^2 + (n+2)^2 + ... + (2n)^2 = \frac{n(2n+1)(7n+1)}{6}.$$
I got up to:
$n=1$ is true, and assuming $n=k$ prove for $n=k+1$.

Prove...

$$\frac{k(2k+1)(7k+1)+6(2(k+1))^2}{6} = \frac{(k+1)(2k+3)(7k+8)}{6}$$

I keep trying to expand to $6(2k+2)^2$ and factorising but I end up being short on one factor, e.g., I end up with $\frac{(k+1)(2k+3)(7k+6)}{6}$.

Alternative route (also with induction)

You can start proving inductively that: $$\sum_{k=1}^{n}k^{2}=\frac{1}{6}n\left(n+1\right)\left(2n+1\right)$$
Then have a look at: $$\sum_{k=n+1}^{2n}k^{2}=\sum_{k=1}^{2n}k^{2}-\sum_{k=1}^{n}k^{2}$$


Limit vs interior definition of continuity Suppose I have two topological spaces $X$ and $Y$ whose topologies are defined by interior operators $\text{int}_X$ and $\text{int}_Y$ respectively, as well as a function $f$ with domain $I$ (for input) and codomain $O$ (for output).

Here's where I hit my quandary: I see two potential routes for defining continuity. I could say the function $f: I \rightarrow O$ is continuous at the point $\tilde{x}$ iff $\lim_{x \rightarrow \tilde{x}} f(x) = f(\tilde{x})$, then say it's continuous everywhere in $I$ if it's continuous at every point $\tilde{x}$ in $I$.

Or, I could say the function $f: I \rightarrow O$ is continuous iff $f^{1-}\left[\text{int}_X(A)\right]$ is a subset of $\text{int}_Y \left( f^{1-}[A] \right)$ for any $A \subseteq O$. My question is: are these two definitions equivalent? Could someone give me some insight into their equivalence or lack thereof? (I'm assuming the limits are defined with respect to the topologies of $\text{int}_X$ and $\text{int}_Y$.)

Following article is giving the proof you're looking for: Continuous functions and convergent sequences


Uniform convergence and sequence of functions I was asked to prove that $f_n(x) = (1-x)^{\alpha}x^n$ is uniformly convergent on $[0,1]$ where $\alpha>0$ I did this part then I was asked if wether the corresponding series of functions converges uniformly on the given interval ? And if it does for what alpha it should be ?
   Hints are really appreciated ? Thanks

Let $S_n(x)=\sum _{k=0}^n(1-x)^\alpha x^k=(1-x)^\alpha \sum_{k=0}^nx^k=(1-x)^\alpha\dfrac{1-x^n}{1-x}=(1-x)^{\alpha-1}(1-x^n)$

$S(x)=\lim_{n\to\infty}S_n(x)=(1-x)^{\alpha-1}(1-x^n)$ which can be checked to be discontinuous irrespective of $\alpha$


Differentiability: Partially Defined Functions Reference

These ideas came to my mind while reading Lee's Introduction to Smooth Manifolds (cf. p. 45).

Definition

Let $E$ and $F$ be two Banach spaces together with a plain subset $A\subseteq E$.

Here, a partially defined function $f:A\to F$ is called differentiable at $a\in A$ if it admits an extension $\bar{f}_a:E\to F$ differentiable at $a$.

Remarks

Note the dependence of the extension on the point under consideration: $\bar{f}_a$

Also a function $f:U\to F$ with open domain $U$ is differentiable at $u\in U$ in the definition given above iff it is differentiable there in the ordinary sense.

The leading principle of this approach to differentiability is that a linear approximation foots on linear spaces. Plain subsets or opens in general aren't!

Problems

(Riesz-Dunford Functional Calculus) (resolved!)
Let a function $f:A\to F$ be (continuously) differentiable in $A$ in the definition given above.
Does it necessarily admit an extension $\bar{f}:E\to F$ that happens to be (continuously) differentiable on some whole neighborhood $U_A$ of $A$ rather than merely on $A$?
(Manifolds with Boundary)
Let a function $f:A\to F$ be (continuously) differentiable in $A$ in the definition given above.
Does it necessarily admit an extension $\bar{f}:A\to F$ that happens to be (continuously) differentiable at every point $a\in A$ simultaneously rather than for every point a separate extension $\bar{f}_a:E\to F$?

Explanation

(Riesz-Dunford Functional Calculus)
The Riesz-Dunford Calculus applies only to functions that happen to be holomorphic on some neighborhood of the spectrum of an operator. A positive result here would pin the problem to holomorphic functions on the spectrum precisely.
(Manifolds with Boundary)
On manifolds a map is differentiable on the boundary iff its coordinate expression has one-sided directional derivatives within half space. A negative result here would complicate the situation alot.
Moreover, the definition given in Lee's book for differentiability of partially defined functions slightly varies from the one given above to the extend that it requires the existence of a common extension. The lack, however, there is that though differentiability is a local property it is defined pointwise. So from a structural point the definition given above shows consistency while for practical purposes the definition given in Lee's book is favourable. A positive result here would unveil them as equivalent and therefore justify the approach.

Attempts

(Riesz-Dunford Functional Calculus)
(Manifolds with Boundary)
For some function on half space $f:\mathbb{H}^m\to\mathbb{R}^n$ to be differentiable in the sense given above it must hold that locally at specific points it extends infinitesimally as:
$$F_E(a_0+v):=2F(a_0)-F(a_0-v),v\notin \mathbb{H}^n$$
while globally at all points it extends infinitesimally as:
$$F_E(a-n):=2F(a)-F(a+n),n\bot\partial\mathbb{H}^n$$
These guiding constructions seem to clash. But this still requires a rigorous counterexample.

1.(Riesz-Dunford Functional Calculus)

Consider the function $f(z):=|z|^2$ defined on the real and imaginary axis only. Then around every point it has an extension to a continuously differentiable function within some neighborhood. But that extension is confined to the Cauchy Riemann equations and therefore it must be $f(z)=+z^2$ and $f(z)=-z^2$ simultaneously in every neighborhood of zero which is impossible. So the answer to the first problem is: No, in general there won't be an extension continuously differentiable in a whole neighborhood.

2.(Manifolds with Boundary)

Besides this example still doesn't resolve(!) the second problem as one can choose the following extension:


Inverse Fourier Transform of Fourier Transform Given the Fourier Tranform defined as
$$\color{red}{\mathcal{F}\{}\color{green}{f(}t\color{green}{)}\color{red}{\}}(\xi):=\int_{-\infty}^{+\infty}\color{green}{f(}t\color{green}{)}e^{-2\pi i \xi t}dt=\color{#F0F}{\hat{f}(}\xi\color{#F0F}{)}$$
They say inverse Fourier Transform is given by
$$\color{blue}{\mathcal{F}^{-1}\{}\color{#F0F}{\hat{f}(}\xi\color{#F0F}{)}\color{blue}{\}}(t)=\int_{-\infty}^{+\infty}\color{#F0F}{\hat{f}(}\xi\color{#F0F}{)}e^{2\pi i \xi t}d\xi=\color{green}{f(}t\color{green}{)}$$
I'm expecting that $$\color{blue}{\mathcal{F}^{-1}\{}\color{red}{\mathcal{F}\{}\color{green}{f(}t\color{green}{)}\color{red}{\}}(\xi)\color{blue}{\}}(t)=\color{green}{f(}t\color{green}{)}$$
But actually
$$\color{blue}{\mathcal{F}^{-1}\{}\color{red}{\mathcal{F}\{}\color{green}{f(}t\color{green}{)}\color{red}{\}}(\xi)\color{blue}{\}}(t)=\int_{-\infty}^{+\infty}(\int_{-\infty}^{+\infty}\color{green}{f(}t\color{green}{)}e^{-2\pi i \xi t}dt)e^{2\pi i \xi t}d\xi=$$$$=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}\color{green}{f(}t\color{green}{)}dtd\xi=\int_{-\infty}^{+\infty}(\int_{-\infty}^{+\infty}d\xi)\color{green}{f(}t\color{green}{)}dt=\int_{-\infty}^{+\infty}0\cdot \color{green}{f(}t\color{green}{)}dt=0 \text{ ??}$$

Let $f \in L^1(\mathbb{R})$ so that
$\hat{f}(\xi) = \int_{-\infty}^{\infty} f(x) \, e^{-ix\xi} \, dx$
is defined. Then
$$
|\hat{f}(\xi)| = \left|\int_{-\infty}^{\infty} f(x) \, e^{-ix\xi} dx\right|
\leq \int_{-\infty}^{\infty} \left|f(x) \, e^{-ix\xi}\right| \, dx
= \int_{-\infty}^{\infty} \left|f(x) \right| \, dx
< \infty
$$
so $\hat{f}\in L^\infty(\mathbb{R})$ but we cannot be sure that $\hat{f} \in L^1(\mathbb{R}).$ Therefore, let us define $\hat{f_\epsilon}(\xi) := e^{-\epsilon \xi^2/2}\hat{f}(\xi)$ so that $\hat{f_\epsilon} \in L^1(\mathbb{R}),$ and at the end take the limit $\epsilon\to 0.$ We then have
\begin{align}
\int_{-\infty}^{\infty} \hat{f_\epsilon}(\xi) \, e^{iy\xi} \, d\xi
&= \int_{-\infty}^{\infty} e^{-\epsilon \xi^2/2}\hat{f}(\xi) \, e^{iy\xi} \, d\xi \\
&= \int_{-\infty}^{\infty} e^{-\epsilon \xi^2/2} \left( \int_{-\infty}^{\infty} f(x) \, e^{-ix\xi} \, dx \right) \, e^{iy\xi} \, d\xi \\
\\
&= \int_{-\infty}^{\infty} f(x) \left( \int_{-\infty}^{\infty}  e^{-\epsilon \xi^2/2} \, e^{-ix\xi} \, e^{iy\xi} \, d\xi \right) \, dx \\
&= \int_{-\infty}^{\infty} f(x) \sqrt{\frac{2\pi}{\epsilon}} e^{-(x-y)^2/(2\epsilon)} \, dx \\
\\
&\overset{\{ x=y+z\sqrt{2\epsilon} \}}{=}
 \int_{-\infty}^{\infty} f(y+z\sqrt{2\epsilon}) \sqrt{\frac{2\pi}{\epsilon}} e^{-z^2} \, \sqrt{2\epsilon} \, dz \\
&= 2\sqrt{\pi} \int_{-\infty}^{\infty} f(y+z\sqrt{2\epsilon}) \, e^{-z^2} \, dz \\
&\to 2\sqrt{\pi} \int_{-\infty}^{\infty} f(y) \, e^{-z^2} \, dz
= 2\sqrt{\pi} \int_{-\infty}^{\infty} \, e^{-z^2} \, dz \, f(y)
= 2\pi \, f(y).
\end{align}

The first integral should be
\begin{align}
I(u) &= \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(t)e^{-i2\pi\xi t}e^{i2\pi\xi u}dtd\xi\\
&=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f(t)e^{-i2\pi\xi t+i2\pi\xi u}dtd\xi\\
&=\int_{-\infty}^{\infty}f(t)\left(\int_{-\infty}^{\infty}e^{i2\pi\xi (u-t)}d\xi\right)dt\\
\end{align}
We have
$$\delta(u-t) = \frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i(u-t)x}dx = \int_{-\infty}^{\infty}e^{i2\pi(u-t)\xi}d\xi
$$
Thereby it becomes
\begin{align}
I(u) &= \int_{-\infty}^{\infty}f(t)\delta(u-t)dt\\
&= f(u)
\end{align}
The function as mapping is returned, doesn't matter how its argument is denoted


Product rule for partial derivatives I am going through the solution for a problem (1.7 from Goldstein's Classical Mechanics) where it says:

I don't understand why the right-hand side of the second line only contains 4 terms when there should be 5. The very last term on line 1 has been expanded into 1 term on line 2 using the product rule, but according to the product rule there should be 2 terms.

$q$, $\dot{q}$ and $\ddot{q}$ are being treated here as separate variables, so
$\dfrac{\partial \ddot{q}}{\partial \dot{q}} = 0$.

I think this results from:
$$ {\frac {\partial {\dot q}} {\partial {\dot q}}} = 1
$$
Hence:
$$
\frac {\partial {\ddot {q}  }} {\partial \dot q}  =
\frac {\frac {\partial {\dot q}} {\partial t}} { \partial {\dot q}  } = \frac {\partial( \frac {\partial {\dot q}} {\partial q}) }{\partial t} = \frac {\partial 1} {\partial t} = 0
$$


Asymptotic expansion for $\frac{1}{2\zeta(3)}\int_x^\infty \frac{u^2}{e^u - 1} du$? Is there an asymptotic expansion for the function:
\begin{equation}
g(x)=\frac{1}{2\zeta(3)}\int_x^\infty \frac{u^2}{e^u - 1} du,
\end{equation}
over the domain $x\in [0,\infty)$ in terms of elementary functions? Here $\zeta$ is the Riemann zeta function and $1/2\zeta(3)$ is a normalization factor included to ensure $g(0)=1$.

We have $$g(x) = \dfrac1{2 \zeta(3)}\underbrace{\int_x^{\infty} \dfrac{u^2}{e^u-1} du}_{I(x)}$$
We will now obtain a series expansion for $I(x)$. We have $$I(x) = \int_x^{\infty} \dfrac{u^2}{e^u-1} du = \int_x^{\infty} \dfrac{u^2 e^{-u}}{1-e^{-u}} du = \int_x^{\infty} \sum_{k=0}^{\infty} u^2 e^{-(k+1)u} du = \sum_{k=1}^{\infty} \int_x^{\infty} u^2 e^{-ku} du$$
We now have that
$$\int_x^{\infty} u^2 e^{-ku} du = \dfrac{e^{-kx} \left(k^2 x^2 + 2kx + 2\right)}{k^3}$$
We have have
$$I(x) = \sum_{k=1}^{\infty} \dfrac{e^{-kx} \left(k^2 x^2 + 2kx + 2\right)}{k^3}\tag{$\star$}$$
For a given $x$, truncating $(\star)$ will give you an exponentially converging approximation. Note that $$I(0) = \displaystyle \sum_{k=1}^{\infty} \dfrac2{k^3} = 2 \zeta(3)$$

yes is equal to the incomplete gamma function http://mathworld.wolfram.com/IncompleteGammaFunction.html

$$ \Gamma (3,x) $$


Physics Vector algebra question If $\vec A=2\vec i+4\vec j-\vec k$, $\vec B=2\vec i-3\vec j+\vec k$ and $\vec C=- \vec i+3\vec j$, then unit vector in direction of $\vec A+\vec B+\vec C$. I tried to find unit vector of all vectors and to add all of them. But I didn't get correct answer. It's a competitive question. I am unable to get the correct answer. I added unit vector of all vectors then got unit vector.

$$\frac1{\sqrt{21}}\vec A=\frac1{\sqrt{21}}(2\vec i+4\vec j-\vec k)$$
$$\frac1{\sqrt{14}}\vec B=\frac1{\sqrt{21}}(2\vec i-3\vec j+\vec k)$$
$$\frac1{\sqrt{10}}\vec C=\frac1{\sqrt{21}}(-\vec i+3\vec j)$$
Now what I should do to get the correct answer?

You should first add the vectors $\vec  A, \vec  B$ and $\vec C$, and then calculate the unit vector.

$$\vec{A} + \vec B + \vec C = 3\hat\imath + 4\hat\jmath$$
$$\text{Unit vector }= (3/5)\hat\imath + (4/5)\hat\jmath$$