Applying the Lagrangian function to find critical points So I have the following function

$$ f(x,y) = x^2+y^2 $$

subject to

$$ g(x,y) = x+y-1 = 0. $$

And I have to use the Lagrangian to find the critical points, and determine wether they are constrained minima, constrained maxima or neither.

I am starting on this guy right now. I have no clue, and any hints would be great (not necessarily an answer). Im on wikipedia, which is making me more confused right now!

For your better reading, I'll follow the notation used on wikipedia.

Let $(x,y)\in \Bbb R^2$.

Start of by defining $\Lambda(x,y,\lambda):=f(x,y)+\lambda g(x,y)$.

Then find $\dfrac{\partial\Lambda }{\partial x}(x,y,\lambda), \dfrac{\partial\Lambda }{\partial y}(x,y,\lambda)$ and $\dfrac{\partial\Lambda }{\partial \lambda}(x,y,\lambda)$:

$$\begin{align} \dfrac{\partial\Lambda }{\partial x}(x,y,\lambda)&=2x+\lambda\\
\dfrac{\partial\Lambda }{\partial y}(x,y,\lambda)&=2y+\lambda\\
\dfrac{\partial\Lambda }{\partial \lambda}(x,y,\lambda)&=x+y-1 .\end{align}$$

Next solve the system:

$$\begin{align} \begin{cases}\dfrac{\partial\Lambda }{\partial x}(x,y,\lambda)&=0\\
\dfrac{\partial\Lambda }{\partial y}(x,y,\lambda)&=0\\
\dfrac{\partial\Lambda }{\partial \lambda}(x,y,\lambda)&=0  \end{cases}&\iff \begin{cases}\\2x+\lambda=0\\
2y+\lambda=0\\
x+y-1=0 \end{cases}\\
&\iff  \begin{cases}\\2x=-\lambda\\
2y-2x=0\\
x+y-1=0 \end{cases}\\
&\iff \begin{cases}\\2x=-\lambda\\
x=y\\
2x=1 \end{cases}\\
&\iff x=\dfrac 1 2=y \land \lambda =-1.\end{align}$$

Thus finding the set of critical points: $\left\{(x,y)\in \Bbb R^2\colon x=\dfrac 1 2 = y\right\}=\left\{\left(\dfrac 1 2, \dfrac 1 2\right)\right\}$.

From here I'll let you make the final conclusions.

In order to verify the conclusions you find, note that $x+y-1=0\iff y=1-x$, so you can just replace $y$ with $1-x$ in $f$ and you have reduced the problem to a single variable which you should know how to solve.


Can every proof by contradiction also be shown without contradiction? Are there some proofs that can only be shown by contradiction or can everything that can be shown by contradiction also be shown without contradiction? What are the advantages/disadvantages of proving by contradiction?

As an aside, how is proving by contradiction viewed in general by 'advanced' mathematicians. Is it a bit of an 'easy way out' when it comes to trying to show something or is it perfectly fine? I ask because one of our tutors said something to that effect and said that he isn't fond of proof by contradiction.

To determine what can and cannot be proved by contradiction, we have to formalize a notion of proof.  As a piece of notation, we let $\bot$ represent an identically false proposition. Then $\lnot A$, the negation of $A$, is equivalent to $A \to \bot$, and we take the latter to be the definition of the former in terms of $\bot$.

There are two key logical principles that express different parts of what we call "proof by contradiction":

The principle of explosion: for any statement $A$, we can take "$\bot$ implies $A$" as an axiom.  This is also called ex falso quodlibet.
The law of the excluded middle: for any statement $A$, we can take "$A$ or $\lnot A$" as an axiom.

In proof theory, there are three well known systems:

Minimal logic has neither of the two principles above, but it has basic proof rules for manipulating logical connectives (other than negation) and quantifiers. This system corresponds most closely to "direct proof", because it does not let us leverage a negation for any purpose.
Intuitionistic logic includes minimal logic and the principle of explosion
Classical logic includes intuitionistic logic and the law of the excluded middle

It is known that there are statements that are provable in intuitionistic logic but not in minimal logic, and there are statements that are provable in classical logic that are not provable in intuitionistic logic. In this sense, the principle of explosion allows us to prove things that would not be provable without it, and the law of the excluded middle allows us to prove things we could not prove even with the principle of explosion.  So there are statements that are provable by contradiction that are not provable directly.

The scheme "If $A$ implies a contradiction, then $\lnot A$ must hold" is true even in intuitionistic logic, because $\lnot A$ is just an abbreviation for $A \to \bot$, and so that scheme just says "if $A \to \bot$ then $A \to \bot$". But in intuitionistic logic, if we prove $\lnot A \to \bot$, this only shows that $\lnot \lnot A$ holds. The extra strength in classical logic is that the law of the excluded middle shows that $\lnot \lnot A$ implies $A$, which means that in classical logic if we can prove $\lnot A$ implies a contradiction then we know that $A$ holds. In other words: even in intuitionistic logic, if a statement implies a contradiction then the negation of the statement is true, but in classical logic we also have that if the negation of a statement implies a contradiction then the original statement is true, and the latter is not provable in intuitionistic logic, and in particular is not provable directly.

Following Carl Mummert in considering the three main systems of propositional logic, let's re-interpret the question once again as

  Does there exist a proof by contradiction that is valid in Classical Logic, yet invalid in Minimal Logic (resp. Intuitionistic Logic)?

The systems of Minimal Logic, Intuitionistic Logic and Classical Logic are three systems of propositional logic of strictly increasing strength. (I will be using the textbook 'Foundations of logic and mathematics' by Nievergelt as a reference, especially Sections 1.1 and 4.1.) To begin to answer this question, we first need to formalise what 'proof by contradiction' is, as a logical principle. Let us look at two examples.

Take first the usual proof of the infinitude of primes: Suppose $p_1, \ldots, p_N$ is the list of all primes. Then the smallest prime factor of $p_1 \cdots p_N + 1$ is larger than $p_N$. Thus there are infinitely many primes. The underlying logical principle applied at the word 'thus' is the so-called Law of Reductio Ad Absurdum:
$$(P \to Q) \to ((P \to \neg Q) \to \neg P),$$
where $P$ is '$p_1, \ldots, p_N$ is the list of all primes' and Q is 'the smallest prime factor of $p_1 \cdots p_N + 1$ is a prime larger than $p_N$'. So if the Law of Reductio Ad Absurdum is valid, then the proper conclusion of the above proof is that $p_1, \ldots, p_N$ is not the list of all primes, which is reasonable as a possible definition of the infinitude of primes (where the prefix 'in-' means 'not', so that 'in-finite' means 'not-listable').

There is another kind of proof of contradiction, namely of the Pigeonhole Principle: Given $n$ holes, if $n + 1$ pigeons are put into them, then there must be some hole with at least two pigeons. So the proof goes: Were there no hole having at least two pigeons, then at most $n$ pigeons were put into the $n$ holes. Thus, if $n + 1$ pigeons were put into the $n$ holes, then there is some hole with at least two pigeons. And the underlying logical principle at the word 'thus' is now the so-called Converse Law of Contraposition:
$$(\neg P \to \neg Q) \to (Q \to P),$$
where $P$ is 'there is some hole with at least two pigeons' and $Q$ is '$n + 1$ pigeons are put into the holes'.

By these two examples, I hope that the reader sees and is convinced that what is generically regarded as 'proof by contradiction' is formalisable as either the the Law of Reductio Ad Absurdum or the Converse Law of Contraposition, which are two separate laws distinct from each other.

The subtlety now arises that in fact

the Law of Reductio Ad Absurdum is valid in Minimal Logic, in Intuitionistic Logic and in Classical Logic.
The Converse Law of Contraposition is not valid in Minimal Logic (resp. Intuitionistic Logic). However, adding the Converse Law of Contraposition to Minimal Logic (resp. Intuitionistic Logic) gives a logic equivalent to the full Classical Logic (see Appendix).

So finally, we can arrive at an answer to the re-interpreted question in the yellow box. For our first example, the proof of the infinitude of primes uses 'proof by contradiction' in the sense of the Law of Reductio Ad Absurdum. This proof is valid in Classical Logic, but by (1), is also valid in Minimal Logic and in Intuitionistic Logic. However, for our second example, the proof of the Pigeonhole Principle uses 'proof by contradiction' in the sense of the Converse Law of Contraposition. Although this proof is valid in Classical Logic, by (2), it is not valid in Minimal Logic nor in Intuitionistic Logic. So we must be careful not to reject as non-intuitionistic or non-minimalistic those proofs in classical mathematics that uses the Law of Reductio Ad Absurdum, and inspect carefully whether it is this law or the Converse Law of Contraposition that is being employed.

Appendix:

For the convenience of the reader, I write down the axioms of these three systems of logic, as taken from Nievergelt's book. One of the purposes of writing this down is that in @Carl Mummert's answer, he uses a constant symbol $\bot$ to denote the falsum. However, it is possible avoid the falsum and to write down the axioms of Minimal Logic, Intuitionistic Logic and Classical Logic completely over the language $\{\neg, \to, \vee, \wedge\}$, with the symbol $\neg$ for negation, the symbol $\to$ for implication, the symbol $\vee$ for disjunction and the symbol $\wedge$ for conjunction. In this language, the use of a constant symbol $\bot$ for the falsum is avoided.

To give the details, let $CL^-$ be the system consists of the following two axiom schemas:

$P \to (Q \to P)$
$(P \to (Q \to R)) \to ((P \to Q) \to (P \to R))$

Then Classical Logic (CL) is $CL^-$ together with the Converse Law of Contraposition (p.58).

Let $T$ denote $CL^{-}$ together with the additional five axiom schemas:

$(P \wedge Q) \to P$
$(P \wedge Q) \to Q$
$P \to (Q \to (P \wedge Q)$
$P \to (P \vee Q)$
$Q \to (P \vee Q)$
$(P \to R) \to ((Q \to R) \to ((P\vee Q) \to R))$

Then Minimal Logic (ML) is $T$ plus the Law Of Reductio Ad Absurdum (p.228). And, Intuitionistic Logic (IL) is $T$ plus the Special Law of Reductio Ad Absurdum -- $$(P \to \neg P) \to \neg P$$
and also plus the Law of Denial Of the Antecedent
$$\neg P \to (P \to Q).$$

The facts are $ML + \text{Law of Denial Of the Antecedent} \Leftrightarrow IL$ (Exercise 755, p.231), $ML + \text{Law of Double Negation} \Leftrightarrow CL$ (Theorem 653, p.229) and $IL + \text{Law of Double Negation} \Leftrightarrow CL$ (Exercise 754, p.231). Since the Law of Reductio Ad Absurdum is an axiom of ML, hence it is valid in both $IL$ and $CL$. Next, $ML$ is strictly weaker than $IL$ since the Law of Denial of the Antecedent is not valid in $ML$. And also $IL$ is strictly weaker than $CL$ since the Law of Double Negation is not valid in $IL$. Hence, the Law of Double Negation being the special case of the Converse Law of Contraposition by taking $Q = \top$ as the verum, the Converse Law of Contraposition is not valid in $ML$ nor in $IL$.


Every homomorphic image of abelian group is abelian but converse need not be true. Every  homomorphic image  of abelian  group is  abelian but converse  need  not be  true.

My  exercise is  to give an counterexample  that  converse  need not be true.

My attempt  :  I tried to construct a  mapping  $f : S_3 \rightarrow  \frac{S_3}{A_3}$ defined by $f(x) = x$ where $ x \in S_3$

I thinks  this  will  be  counter example  but here $S_3$  is non- abelian.

Is  jt true?

I would have written $f(x)=xA_3$, as that's what the elements of $S_3/A_3$ look like. But otherwise, yeah, that's a good example. This $f$ (and all other homomorphisms constructed from a group to a quotient of that group in this way) is called the canonical quotient homomorphism. Because it's the most obvious way to map a group into a quotient.

For an even simpler example, take any non-abelian group and map it to the trivial group (or to the identity element of your favourite group).

Another  counter example  from trancelocation answer . $$\det:  GL(n,R) \rightarrow \{-1,1\}$$


Proof that $n^2 < 2^n$ How do I prove the following statement by induction?

$$n^2 \lt 2^n$$

$P(n)$ is the statement $n^2 \lt 2^n$

Claim: For all $n \gt k$, where $k$ is any integer, $P(n)$

(since $k$ is any integer, I assume I have to prove this for positive and negative integers)

So let base case be $P(1)$, and I have to prove $P(n)$ for $n \ge 1$ and $n \lt 1$

$P(1)$ is $1^2 \lt 2^1$ which is clearly true.

Induction hypothesis: $k^2 \lt 2^k$

Inductive step $(k+1)^2 \lt 2^{(k+1)}$

$(k+1)^2 = k^2 + 2k + 1$

$k^2 + 2k + 1 \lt 2^k + 2k + 1$ by inductive hypothesis

Not sure how to proceed. Is my previous intuition that I have to prove this for $n \ge 1$ and $n \lt 1$ correct?

Hint only: For $n \geq 3$ you have $n^2 > 2n + 1$ (this should not be hard to see) so if $n^2 < 2^n$ then consider
$$
2^{n+1} = 2\cdot2^n > 2n^2 > n^2 + 2n+1 = (n+1)^2.
$$
Now this means that the induction step "works" when ever $n\geq 3$. However to start the induction you need something greater than three. By trial an error you can find the smallest $n(\geq0)$ such that $n^2 < 2^n$.

As a fun fact another way to see that this inequality is true eventually is to compare the sum of the numbers in the $n-1$ 'st row of pascal's triangle to the $n$'th triangular number, which appears in the $n+1$ 'st row and occurs along one of the diagonal's of pascal's triangle.
See the link for more details, http://mathforum.org/workshops/usi/pascal/pascal_triangular.html.  The fact presented on that page plus the fact the sum of the row of pascal's triangle is a power of two can be used to prove the desired inequality.

This amounts to proving that $\frac{n(n+1)}2 \lt 2^\left(n-1\right)$ which is stronger than the original inequality, since $n^2 \lt n(n+1)$ for $n \gt 0$ .


Is every linear representation of a group $G$ on $k[x_1,\dots,x_n]$ a dual representation? Let $\rho\colon G\to GL(V)$ be a linear representation of $G$ on a $k$-vector space $V$. The dual representation is

$$G\to GL(V^*),\quad g\mapsto(\varphi\mapsto\varphi\circ\rho(g^{-1})).$$

By the same rule, we get a linear representation on the coordinate ring $k[V]$ of $V$, which is the $k$-algebra generated by $V^*$, with pointwise multiplication. Now if $V$ is finite-dimensional, $e_1,\dots,e_n$ the standard basis, and $x_1,\dots,x_n$ the dual basis, then $k[V]\cong k[x_1,\dots,x_n]$. Hence there is a linear representation $\bar\rho$ of $G$ on $k[x_1,\dots,x_n]$.

My question is: Is there a way to go "back"? That is, given a linear action of a group $G$ on $k[x_1,\dots,x_n]$, does it come from a linear representation of $G$ on an $n$-dimensional vector space $V$?

More generally, given a linear representation $\bar\rho:G\to GL(k[V])$, does it come from a representation $G\to GL(V)$?

Edit: As an example, consider $G\subseteq GL_n(k)$ acting on $k[x_1,\dots,x_n]$ via $(A,f)\mapsto f(A\cdot x)$. Does this action come from a linear representation of $G$ on $k^n$? I could consider $G$ acting on $k^n$ by multiplication, but the action on the coordinate ring corresponding to this should be $(A,f)\mapsto f(A^{-1}\cdot x)$, or am I mistaken here?

This is motivated by the fact that in invariant theory, one studied actions of subgroups $G\subseteq GL_n(k)$ on the polynomial ring $k[x_1,\dots,x_n]$ as above. But many theorems or facts in invariant theory are formulated in the language of representations, and the invariant ring there is $k[V]^G$. Now if I know that under certain circumstances $k[V]^G$ is finitely generated, I wanted to "come back" to the initial problem of linear actions on $k[x_1,\dots,x_n]$. But I can't really connect these two yet. As written above, any representation of $G$ on some f.d. vector space $V$ induces an action on the coordinate ring $k[V]$, but why do these include the above actions on the polynomial ring? And if they do, how does the initial action on $k^n$ look, given $\rho:G\to GL(k[V])$?

Edit5: I finally found a way to hopefully make clearer what I'd like. Let $G\subseteq GL_n(k)$. Can every linear $G$-action on $k[x_1,\dots,x_n]$ be written as $\bar\rho:G\to GL(k[V])$, coming from some $\rho:G\to GL(V)$? So that the $k[V]^G$ really are a "generalization" of the invariant ring $k[x_1,\dots,x_n]^G$.

If I understand your question correctly, the answer is no, not every representation of a group $G$ on the polynomial algebra $k[V]$ comes from an action of $G$ on $V$.

One way to construct examples to see this is by using the fact that the polynomial ring $k[V]$ is graded:
$$
k[V] = \bigoplus_{m=0}^\infty k[V]_m,
$$
where $k[V]_m$ consists of the polynomials of degree $m$.

Then a group $G$ can act on just one component of this direct sum without acting on the others.  For instance, let $G = \mathbb{Z}/2\mathbb{Z}$, written multiplicatively with generator $\alpha$ (so $\alpha^2 = e$), and for $f \in k[V]_m$, define
$$
\alpha \cdot f =
\begin{cases}
-f & \text{   if } m=0\\
f  & \text{    if } m>0
\end{cases}
$$
Essentially, $G$ acts on a polynomial by changing the sign of the constant term and leaving everything else alone.
This clearly doesn't come from an action of $G$ on $V$.

Yes.  This is related to Peter Weyl Theorem that every representation has a dual representation.

\[ L^2(G) \simeq \bigoplus_{\mathrm{Irr}(G)} R \otimes \overline{R}  \]

Physicists have a shorthand for this using "bra-ket" notation: the "ket" or left-action $ g \,|v \rangle $ is dual to "bra" right-action $ \langle v | g$.

In these notes on representation theory, in section 1.3, they talk about left- and right-modules.


Exchanging expectation and limits on a finite probability space Suppose I have an infinite sequence of random variables $X_n \rightarrow X $ a.s.

Suppose that the sample space $\Omega$ is finite.

Does it follow that E[$\textrm{lim}_{n\rightarrow\infty} X_n] = \textrm{lim}_{n\rightarrow\infty} E[X_n]$ ?

It seems to me that it should follow, since given that $\Omega$ is finite, there should be a dominating r.v. and thus the dominated convergence theorem should apply. Is this correct?

(Apologies if this is a dumb question, I am not a mathematician by training.)

Thanks!

case your sum is diverging :

a simple counter exmple :
let $\Omega$ = {-1;1} and define $(\epsilon_i)_i$ an infinite sequence of rademacher variables i.e. $\mathbb{P}(\epsilon_i=-1)=1/2$ then :
$\sum_{i=1}^{\infty}\mathbb{E}[\epsilon_i]=0$
But we know that $S_\infty=\sum_{i=1}^{\infty}\epsilon_i$ is divergent so it can't have a finite expectation

Indeed if $\mathbb{E}[\sum_{i=1}^{\infty}\epsilon_i]<+\infty$ then it would imply imediatly that $\sum_{i=1}^{\infty}\epsilon_i < \infty$ a.s wich is known to be false (infinity will be visited i.o. it's a known proprety of the random walk, for a little more intuition on this see : https://mathoverflow.net/questions/60417/will-a-random-walk-on-0-inf-tend-to-infinity)

case your sum is converging :

Here you have two options either your sum is made of positiv random variables and then Fubini-Tonelli applies without problem whatever the cardianality of $\Omega$
But if your sum is not made of positiv r.v. then it must satisfy absolute convergence to apply Fubini-Lebesgue theorem and so it still does not depend on the cardinality of $\Omega$

https://en.wikipedia.org/wiki/Fubini%27s_theorem


Why is orthogonal basis important? Lets take the $\mathbb{R}^3$ space as example. Any point in the $\mathbb{R}^3$ space can be represented by 3 linearly independent vectors that need not be orthogonal to each other. What is that special quality of orthogonal basis (extending to orthonormal) that we choose them over non-orthogonal basis?

If $\{v_1, v_2, v_3\}$ is a basis for $\mathbb{R}^3$, we can write any $v \in \mathbb{R}^3$ as a linear combination of $v_1, v_2,$ and $v_3$ in a unique way; that is $v = x_1v_2 + x_2v_2+x_3v_3$ where $x_1, x_2, x_3 \in \mathbb{R}$. While we know that $x_1, x_2, x_3$ are unique, we don't have a way of finding them without doing some explicit calculations.

If $\{w_1, w_2, w_3\}$ is an orthonormal basis for $\mathbb{R}^3$, we can write any $v \in \mathbb{R}^3$ as $$v = (v\cdot w_1)w_1 + (v\cdot w_2)w_2 + (v\cdot w_3)w_3.$$ In this case, we have an explicit formula for the unique coefficients in the linear combination.

Furthermore, the above formula is very useful when dealing with projections onto subspaces.

Added Later: Note, if you have an orthogonal basis, you can divide each vector by its length and the basis becomes orthonormal. If you have a basis, and you want to turn it into an orthonormal basis, you need to use the Gram-Schmidt process (which follows from the above formula).

By the way, none of this is restricted to $\mathbb{R}^3$, it works for any $\mathbb{R}^n$, you just need to have $n$ vectors in a basis. More generally still, it applies to any inner product space.

The important thing about orthogonal vectors is that a set of orthogonal vectors of cardinality(number of elements of a set) equal to dimension of space is guaranteed to span the space and be linearly independent. If you have not covered this fact in class, you soon will.

As far as your second question goes, there are no prerequisites for linear algebra, apart from elementary mathematics you learn in high school.

Added later:- The main thing is that orthogonality guarantees linear independence, which is rather convenient.


prove the limit of $f(x,y)=\frac{x^3−y^3}{x^2−y^2}$ exists with an $\epsilon - \delta$ prove Given the function $f(x,y)=\frac{x^3−y^3}{x^2−y^2}$. Defined by $x^2\neq y^2$ and $(x,y) \in \Bbb R$.
Proof that $\lim\limits_{(x,y) \rightarrow (0,0)} f(x,y)$ exists.

I found that the limit went to $0$. And I was trieing to prove this with an $\epsilon$ $ \delta$ prove, but I'm stuck.

I got:

Let $\epsilon > 0$ be given, take $\delta= ....$.
Than $|(f(x,y)-L|$=$|\frac{x^3-y^3}{x^2-y^2}|$= $|\frac{x(x+y)+y^2}{x+y}|$=$|x+\frac{y^2}{x+y}|$

I don't know how to go on.

You can't prove a limit exists when it doesn't take for example $y\to - x$


How to solve this Diophantine Equation. Step by Step What are all the solutions to the diophantine equation:
6x-6y-xy=0.


Is there a shape with infinite area but finite perimeter? Is this really possible? Is there any other example of this other than the Koch Snowflake? If so can you prove that example to be true?

One can have a bounded region in the plane with finite area and infinite perimeter, and this (and not the reverse) is true for (the inside of) the Koch Snowflake.

On the other hand, the Isoperimetric Inequality says that if a bounded region has area $A$ and perimeter $L$, then
$$4 \pi A \leq L^2,$$
and in particular, finite perimeter implies finite area. In fact, equality holds here if and only if the region is a disk (that is, if its boundary is a circle). See these notes (pdf) for much more about this inequality, including a few proofs.

(As Peter LeFanu Lumsdaine observes in the comments below, proving this inequality in its full generality is technically demanding, but to answer the question of whether there's a bounded region with infinite area but finite perimeter, it's enough to know that there is some positive constant $\lambda$ for which
$$A \leq \lambda L^2,$$
and it's easy to see this intuitively: Any closed, simple curve of length $L$ must be contained in the disc of radius $\frac{L}{2}$ centered at any point on the curve, and so the area of the region the curve encloses is smaller than the area of the disk, that is,
$$A \leq \frac{\pi}{4} L^2.)$$

NB that the Isoperimetric Inequality is not true, however, if one allows general surfaces (roughly, 2-dimensional shapes not contained in the plane. For example, if one starts with a disk and "pushes the inside out" without changing the circular boundary of the disk, then one can make a region with a given perimeter (the circumference of the boundary circle) but (finite) surface area as large as one likes.

It depends on the geometry.
In Euclidean: definetely no.
On a fractal surface: sure.


Economics maximization problem linear activity Consider the vectors:

$a_1 = \begin{pmatrix} 0 \\ -1 \\ 1 \\0 \end{pmatrix}, a_2 = \begin{pmatrix} 0 \\ 0 \\ -1 \\1 \end{pmatrix}, a_3 = \begin{pmatrix} 2 \\ 0 \\ 0 \\ 1\end{pmatrix}$

Find a single vector $p$ which maximizes $pa_i$ for $i = 1,2,3$.

To put this in context this is an economics profit max problem where p is a price and each component of the above vectors represents the quantity of the good.

I honestly have no idea how to find this $p$ vector. It doesn't even seem possible to me that a single vector can maximize these three vectors.

I think your question is incomplete. First you should determine your goal function, e.g., you can choose the sum of all profits to maximize, max: $pa_{1}+pa_{2}+pa_{3}$. Also you should determine your boundaries. E.g., total number of goods $d_1, d_2, d_3, d_3$.

A paraphrased version of this question is crossposted on Economics.SE, and as a result, we are closing it. Using the information available there and here, I believe I have an answer.

Basically, you can perform any activity any number of times. Suppose you perform $a_i$ $x_i$ times and the price of each good $j$ is $p_j$. Let $A$ denote the matrix defined by $a_1,a_2,a_3$, let $x$ denote the row vector $(x_1, x_2, x_3)$ of activites and  let $p$ denote the column vector $(p_1, p_2, p_3)$ of prices.
Then the problem is
$$
\max_x x \cdot A \cdot p.
$$
Suppose this maximum exists and
$$
c^* = \max_x x \cdot A \cdot p,
$$
$$
x^* = arg\max_x x \cdot A \cdot p.
$$
As there is no constraint on $x$ the activity vector, $2 \cdot x^*$ yields a profit of
$$
(2 \cdot x^*) \cdot A \cdot p = 2 \cdot c^*.
$$
If $c^* > 0$, then $2 \cdot c^* > c^*$ so $c^*$ was not maximal after all. A similar argument exists for negative profits. Therefore if a profit maximum exists in case of linear activities without resource constraints it is zero. Hence you need to find a vector $p$ for which
\begin{eqnarray*}
a_1 \cdot p & = & 0 \\
\\
a_2 \cdot p & = & 0 \\
\\
a_3 \cdot p & = & 0.
\end{eqnarray*}
From here on out it is a straightforward exercise in linear algebra.


Give an example of a simply ordered set without the least upper bound property. In Theorem 27.1 in Topology by Munkres, he states "Let $X$ be a simply ordered set having the least upper bound property.  In the order topology, each closed interval in $X$ is compact."

(The LUB property is if a subset is bounded above, then it has a LUB.)

I don't understand how you could have a simply ordered set (a chain) WITHOUT the LUB property.  If a subset is bounded and it is a chain, then how can it not have a LUB?
Can someone give an example?

Thanks!

$\Bbb Q$

Added: $\{q\in\Bbb Q:q<\sqrt2\}$ is a chain, it’s bounded above by $2$, say, and it has no least upper bound in $\Bbb Q$, because $\sqrt2$ is irrational.

One more answer, not discussed till now.

Consider $A = \{x,y,z\}$. So $\mathbb{P}(A)=  \{\phi, \{x\}, \{y\}, \{z\}, \{x, y\}, \{y,z\}, \{x, z\}, A\}$.

Consider partial order relation $<$ on $\mathbb{P}(A)$ as $a<b$ iff $a\subset b$.

Consider the element $\phi \in \mathbb{P}(A)$. See nay element of $\mathbb{P}(A)$ is an upper bound of the set $\phi$. What is the least upper bound ? Not exists.


Show that the function $f(x) = x^5 - 10x^3 + 50x - 21$ is an increasing function for all x values. I have worked out that $dy/dx = 5x^4 - 30x^2 + 50$ and this must be greater than $0$ if it is an increasing function.

I simplified this to $x^4 - 6x^2 + 10 > 0$. I know that I must prove this inequality true but I cannot work out how.

I am able to input various values of x and show them to be larger than 0 but am unsure how to do it algebraically.

I know for a quadratic, I can put it into a 'complete the square' form and then you can prove it to be larger than 0 as squaring something = positive and then adding a positive will remain positive. But am unsure how to do it where power of x > 2.

Let rewrite that function as
$$f(x)=(x^5+50x)-(10x^3+21)$$
Obviously,
$$(10x^3+21)<(x^5+50x)$$
Hence, function is increasing :D


How to calculate $\liminf$ of the function $f(x)=1-x^2$ Let $f:\mathbb{R} \longrightarrow \mathbb{R}$ given by
$$f(x)= 1-x^2, \; \forall \; x \in \mathbb{R}.$$
Question. How to evaluate
$$\liminf_{|x|\rightarrow \infty} f(x)?$$
What I got to think about is
$$\liminf_{|x|\rightarrow \infty} f(x)=\min\{c \in \mathbb{R} \; ; \; x_n \rightarrow \infty \: \text{and} \: f(x_n)=c\}.$$
It's true? Or is my reasoning wrong?

The function $f(x)=1-x^2$ is not bounded at infinity, that is, for all $A>0$ and $k>0$  there exists $x_0 \in \mathbb{R}$ such that $x_0>A$ we have $|f(x_0)|>k$ $(\lim_{x \to \infty} 1-x^2=-\infty)$.
Since $$\liminf_{x \rightarrow \infty} f(x)=\sup_{N>0}\inf_{x \geq N} f(x)$$ and for all $N>0$ we have $$\inf_{x \geq N} f(x)=-\infty$$ then $$\liminf_{x \rightarrow \infty}(1-x^2)=-\infty.$$
Similarly, we can handle the other case.

$$\lim_{|x| \rightarrow +\infty} f(x) = -\infty$$
so $$\liminf_{|x| \rightarrow +\infty} f(x) = \limsup_{|x| \rightarrow +\infty} f(x) =-\infty$$


How to calculate probability of these two pair-of-sums ($S_{n}$ and $T_{n}$) and ($SEvenF_{n}$ and $SOddF_{n}$) being the same? Say we have a sequence of $n$ positive integers, we can assume they're randomly chosen, let's call it $U_{n}$.

Let $S_{n}$ = sum of $U_{n}$ from $1$ to $n$.

Let $T_{n}$ = sum of $n$ from $1$ to $n$. In other words, $T_{n}$ is the triangle number $T_{n} = \frac{1}{2}n(n+1)$.

Define $F_{n} = U_{n} + n$.

(note that $SF_{n}$, sum of $F_{n}$, is equal to $S_{n} + T_{n}$)

Define two sets $EvenSet$ and $OddSet$ containing $F_{n}$ when it is even and odd respectively.

Let $SEvenF_{n}$ be the sum of numbers in $EvenSet$ and $SOddF_{n}$ be the sum of numbers in $OddSet$.

(note that $SEvenF_{n} + SOddF_{n} = SF_{n} = S_{n} + T_{n}$)

What is the probability of the two pairs ($S_{n}$ and $T_{n}$) and ($SEvenF_{n}$ and $SOddF_{n}$) are the same pairs?

In other words, either "$S_{n} = SEvenF_{n}$ thus $T_{n} = SOddF_{n}$" or "$S_{n} = SOddF_{n}$ thus $T_{n} = SEvenF_{n}$".

How to calculate this probability?

PS: I have a sample data with $n = 114$ in here
http://goo.gl/k96FZ

PS2: Is there a way or an algorithm to generate such sequence?

PS3: @ZefChonoles has correctly pointed a concern about Probability Measure. Originally I intended this question to have no restriction on the positive integer set, that is to work on the infinite set of $\mathbb Z^{+}$.

But I believe you are free to impose certain restriction to this problem, as long as you can share insights on approaches in solving this problem.

For example, you can restrict $U_{n}$ and $F_{n}$ to be within $[1, 400]$ for the $n=114$ case. Or you can restrict them to be within $[1, Cn]$ with some constant $C$ for the general case. Or within $[1, Tn]$. Anything that helps, basically. Thanks!

The integer wont exceed 3300
If Sn = S even Fn;
Sn = S even n + S odd n
S even F n = S even n + T even n

Then
S even n + S odd n = S even n + T even n
S odd n = T even n
T even n is 3306
S odd n = 3306
The sum of 57 random positive integer is 3306.
Is it safe to use combination of 3306 and 57?
As for the other one:
Tn = S even Fn
Tn = T even n + T odd n
S even Fn = S even n + T even n
Thus
T even n + T odd n = S even n + T even n
T odd n = S even n
T even n is 3249
S even n = 3249

Combination of 3249 and 57

What I still wonder is, how to determine all possibility used in this to divide them


Calculate limit of large function How can I calculate this:
$$\lim_{x\to 0}{\frac{(\sin x)^{13} - \ln(1 + (\sin x) ^{13} )} {(\tan x) ^{26} }} $$
I tried taking $\frac{1}{{\tan^{13} x}} $ common and then evaluating the limit and I got answer zero but the answer is $\frac{1}{2}$. Any suggestions?

Let $\sin^{13}x=u$

$$\lim_{u\to0}\dfrac{u-\ln(1+u)}{u^2}\cdot\lim_{x\to0}\cos^{26}x =\lim_{u\to0}\dfrac{u-\ln(1+u)}{u^2}$$

Now use Series Expansion or L'Hospital's Rule

I think you need to use approximations ie $$\tan(x)=\sin(x)\approx x $$ for small x and taylor expansion for $ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}...$


How do you prove that the dunce cap is not a surface? The dunce cap results from a triangle with edge word $aaa^{-1}$. At the edge, a small neighborhood is homeomorphic to three half-disks glued together along their diameters. How do you prove this is not homeomorphic to a single disk?

If you have two homotopic maps $f,g: S^1 \to X$, then $X \cup_f D^2$ is homotopy equivalent to $X \cup_g D^2$.

You can use this to show that the dunce cap is homotopy equivalent to $D^2$, and thus contractible. Since no closed surface is contractible (using classification of surfaces), the dunce cap is not a surface.

$D^2$ is the closed unit disk. By $X \cup_f D^2$, I mean gluing $D^2$ via the map $f: S^1 = \partial D^2 \to X$. This is the quotient space of $X \sqcup D^2$ identifying each point of $\partial D^2$ with its image under $f$ in $X$. So in our specific case, $D^2$ is homeomorphic to $S^1$ glued to $D^2$ under the identity map $S^1 \to S^1$. On the other hand, we have that the dunce cap is constructed by gluing $D^2$ to $S^1$ under the map $g: S^1 \to S^1$ given by
$$
g(e^{i\theta}) =
\begin{cases}
exp(4 i \theta) & 0 \leq \theta \leq \pi/2\\
exp(4 i (2 \theta - \pi)) & \pi/2 \leq \theta \leq 3\pi/2\\
exp(8 i(\pi - \theta)) & 3\pi/2 \leq \theta \leq 2\pi
\end{cases}$$

It is not hard to show that $g$ is homotopic to to the identity map, and so (using the result I mentioned above), $D^2$ is homotopy equivalent to the dunce cap. So the dunce cap must be contractible.

Edit: I have now realized that the above answers the question in the title, which is not the question posed by the OP. To see that the dunce cap is not homeomorphic to $D^2$, you can simply note that the dunce cap is a disk glued along its boundary (albeit in a strange way), and thus has no 2-dimensional boundary, while $D^2$ does.

The question is essentially answered in the way that it is posed.  No point of a disc has a neighbourhood homeomorphic to three half discs glued along their diameters.

The question has been changed since I wrote this reply.

In the talk about the topological dunce hat, I have described a simple method of contracting it.


Does $g$ behave like $t^k$ near the origin? I asked a similar question early ago, however, my function satisfies more hypothesis, so I am gonna ask it again, please be patient with me. Suppose $G:[0,\infty]\to [0,\infty]$ defined by $G(t)=\int_0^t g(s)ds$ is a (Young or N) function with $g:[0,\infty]\to [0,\infty]$ and $g\in C^1(0,\infty)$ (for na definition of Young function see here page 13). Assume that there exist constants $b_1,b_2>1$, $b_1<b_2$ such that (condition $\Delta_2$) $$b_1\leq \frac{tg(t)}{G(t)}\leq b_2$$

Then, we can conclude that

$$G(1) t^{b_2}\leq G(t)\leq G(1) t^{b_1},\ \forall\ t\in [0,1]$$

Is it possible to find constants $c_1,c_2,k>0$ such that $$c_1 t^k\leq G(t)\leq c_2 t^k$$

for small $t$?

Update 1: I added some hypothesis that fits better in what I want. Also I would like to point out, that this problem come from the paper of Montenegro:  Uniqueness of nonnegative solutions of quasilinear elliptic equations. Dynam. Systems Appl. 6 (1997), no. 1, 125–137. The statement is in inequality $(22)$ of the paper.

Update 2: I have asked this question. I think that in our case here, $k=k_1=k_2$.

Thanks

I assume you are asking "whether $G(t)$ always behaves like $t^k$" (as in your question) instead of "whether $g(t)$ always behaves like $t^k$" (as in the title) for small $t$.

The answer is No.

The question is equivalent to given a monotonic increasing $C^2$ function $G : [0,\infty] \to [0,\infty]$ satisfying:
$$G(0) = 0 \quad\quad\text{and}\quad\quad 1 < b_1 \le \frac{d \log G(t)}{d \log t} \le b_2 < \infty\text{ for }t \in (0,\infty)$$

Is it always true that there exists some $k_G > 0$ such that:

$$k_G = \lim_{t\to 0+} \frac{\log G(t)}{\log t} \quad\text{ and }\quad \limsup_{t\to 0+} | \log G(t) - k_G \log t| < \infty$$

Define $G(t)$ by:
$$G(t) = \begin{cases}t^3 e^{f(\log t)}, & t > 0\\0, & t = 0\end{cases}$$
where $f(s) = \sqrt[4]{s^2+1}$. It is not hard show for any $s \in (-\infty,\infty)$,

$$|f'(s)| = |\frac{s}{2(s^2+1)^{\frac34}}| \le \frac{1}{\sqrt[4]{108}} \sim 0.3102 $$

This implies for $t \in (0,\infty)$,

$$\frac{d \log G(t)}{d \log t} = 3 + f'(\log t) \implies 2.6 < \frac{d \log G(t)}{d \log t} < 3.4 $$
Furthermore, the $t^3$ factor in $G(t)$ make it falls to zero fast enough as $t$ approaches $0$ and turn $G(t)$ into a $C^2$ function over $[0,\infty)$. More precisely, this mean:

$$\begin{align}
\lim_{t\to0+} G'(t)  &= 0 = G'(0)  \stackrel{def}{=} \lim_{h\to 0+}\frac{G(h)}{h}\\
\lim_{t\to0+} G''(t) &= 0 = G''(0) \stackrel{def}{=} \lim_{h\to 0+}\frac{G'(h)}{h}
\end{align}$$
Notice

$$\lim_{t\to0+}\frac{\log G(t)}{\log t} = \lim_{s\to-\infty}\frac{3s+f(s)}{s} = 3$$

$k_G$ for this particular $G$ is $3$. However,

$$\limsup_{t\to0+}|\log G(t) - k_G \log t| = \limsup_{s\to-\infty}|f(s)| \ge \limsup_{s\to-\infty}\sqrt{|s|} = \infty$$

This means in general, $G(t)$ need not behave like any $t^k$ for small $t$.

You can think of your statement as find the constants $c_{1}, c{2}, k>0$ such that $c_{1}t^k \leq a_{1}t^{b_{2}} \leq a_{2}t^{b_{1}} \leq c_{2}t^k$ $\forall t\in[0,1]$
$c_{1} \leq a_{1}t^{b_{2}-k} \leq a_{2}t^{b_{1}-k} \leq c_{2}$  $\forall t\in[0,1]$
Now if $k < b_{2} \Rightarrow \lim_{t\to0}a_{1}t^{b_{2}-k}=0 \Rightarrow c_{1}=0$ so impossible.
Then $k \geq b_{2}$.
Now if $k > b_{1} \Rightarrow \lim_{t\to0}a_{2}t^{b_{1}-k}=\infty \Rightarrow c_{2}=\infty$ so again impossible.
Then $k \leq b_{1}$.
Since $b_{2} > b_{1}$ no such $k$ exists.
And to give you a hint how to imagine a counterexample, think if $g(t)$ oscilates between the 2 boundry functions without braking its convexity.


How to compute the CDF of $X\cdot Y$ if $X,Y$ are independent and uniformly distributed over $(-1,1)$? Just like in the title:

  How to compute the CDF of $X\cdot Y$ if $X,Y$ are independent random variables, uniformly distributed over $(-1,1)$?

I tried using the next formula:
the density of $X\cdot Y$ is the integral  of
$f(u/v)\cdot g(v)$ where $f$ is the density of $X$ and $g$ is the density of $Y$ but it has no sense.

Let $Z=X\cdot Y$ and $f_Z,f_Y,f_X$ be the pdfs of $X,Y,Z$.

$f_Z$ is quite trivially an even function, supported on $[-1,1]$.

So, let we compute, for any $p\in(0,1)$:

$$\mathbb{P}[0\leq Z\leq p]=2\int_{0}^{1}\int_{0}^{\min\left(1,\frac{p}{x}\right)}\frac{1}{2}\cdot\frac{1}{2}\,dy\,dx=\frac{1}{2}\int_{0}^{1}\min\left(1,\frac{p}{x}\right)\,dx\tag{1}$$
that gives:
$$\mathbb{P}[0\leq Z\leq p]=\frac{1}{2}\left(\int_{0}^{p}1\,dx+\int_{p}^{1}\frac{p}{x}\,dx\right)=\frac{p-p\log p}{2}\tag{2}$$
hence:
$$ \int_{0}^{p}f_Z(t)\,dt = \frac{p-p\log p}{2}\tag{3} $$
gives, through differentiation, $f_Z(p)=-\frac{\log p}{2}$. At last,
$$ f_Z(z) = -\frac{\log|z|}{2}\cdot\mathbb{1}_{(-1,1)}(z),\\ F_Z(z)=\left[\frac{1}{2}-\frac{(1-|z|)+|z|\log |z|}{2}\text{Sign}(z)\right]\cdot\mathbb{1}_{(-1,1)}(z)\tag{4}$$
are the pdf and cdf of $Z$.

Hint : Try jacobian transformation with $u=xy$ and $v=y$ then integrate on v between(-1,1).


How to show that $f(x) = x^2 \sin(1/x^2)$ with $f(0)=0$ is differentiable and $f'$ is unbounded on $[-1,1]$? If the function $f$ is defined by
$$f(x) = \begin{cases}x^2 \sin(\frac{1}{x^2}) & x \ne 0 \\
0 & x = 0
\end{cases}
$$
then $f$ is differentiable and $f'$ is unbounded on $[-1,1]$.

I know this is true. How should I go about proving this?

The cool point of the problem is that although for any positive value $M$ and positive $\epsilon$ we can find some value of $x$ with $|x|<\epsilon$ such that $|f'(x)| > M$, yet $f$ is differentiable at $x=0$.

In fact, $f'(0) = 0$, and we can see that by applying the definition of derivative:
$$
\left. \frac{df(x)}{dx} \right|_{x=0} = \lim_{h\rightarrow 0} \frac{h^2 \sin\frac{1}{h^2}-f(0)}{h} = \lim_{h\rightarrow 0} {h \sin\frac{1}{h^2}}-0 =0
$$
since $|\sin\frac{1}{h^2}|$ is bounded by 1 and that is being multiplied by $h$.

On the other hand, look at the expression for $f'(x)$ when $x \neq 0$.  At points where $\frac{1}{x^2} = (n+\frac{1}{2})\pi$,
$$
|f'(x)| > \frac{2}{x} - 2x
$$
and this is unbounded as $x$ approaches zero.

Using the product and chain rule, one gets that
$$f'(x) = 2x\sin(1/x^2)-2\frac{\cos(1/x^2)}{x} $$
Notice that this function is unbounded on the given interval.


If $f:\mathbb R\to\mathbb R$ measurable and negative, there is a sequence $(\varphi _n)$ of simple function s.t. $\varphi _n\nearrow f$ Theorem: I know that if $f:\mathbb R\to\mathbb R$ is measurable and positive, there is a sequence of simple function $(\varphi _n)$ s.t. $\varphi _n(x)\nearrow f(x)$. I kniw that this result is not true if $f$ is not positive.

My question :

Let $f:\mathbb R\to\mathbb R$ measurable. Let $f\leq 0$ and $$\varphi _n(x)=\frac{\left\lfloor 2^n f(x)\right\rfloor }{2^n}\vee (-n),$$
where $\lfloor x\rfloor$ is the floor function and $a\vee b$ denote the maximum between $a$ and $b$. We have that $$\varphi _n(x)\to f(x)$$
and that $\varphi _n$ is increasing even since $$\varphi _n(x)=\frac{\lfloor 2^n f(x)\rfloor }{2^n}\leq \frac{2^{n}f(x)}{2^n}=\frac{2^{n+1}f(x)}{2^{n+1}},$$
and thus $\varphi _n(x)\leq \varphi _{n+1}(x) $.

Why is this theorem is not true if $f<0$ ? The fact that there is a sequence of simple $\varphi _n(x)\nearrow f(x)$ looks true even if $f\leq 0$, no ?

It can't be true if $f$ is not bounded below. Because a simple function is certainly bounded below, and if $\phi_n$ increases to $f$ then $f\ge\phi_1$.

This result is true even for $f \leq 0$. What was probably meant is that you can't use this method to integrate functions that aren't positive almost everywhere.


If $K\subseteqℝ^2$ is a closed triangle spanned by $z^1,z^2,z^3$ and $P:K\toℝ$ is a polynomial of degree at most $1$ with $P(z^i)=0$, then $P\equiv0$ Let

$K\subseteq\mathbb R^2$ be a closed triangle spanned by $z^1,z^2,z^3$ (and assume that $z^i\ne z^j$)
$P:K\to\mathbb R$ be a polynomial of degree at most $1$ with $$P(z^i)=0\;\;\;\text{for all }i\in\left\{1,2,3\right\}\tag1$$

  How can we show that $P\equiv 0$?

My idea is the following:

Let $L_1\subseteq K$ be the edge between $z^2$ and $z^3$
Assume, for the moment, that $z_1^2\ne z_1^3$
Let $$I_1:=[\min(z_1^2,z_1^3),\max(z_1^2,z_1^3)]$$
It's clear that$^1$ there is a linear function $f_1:\mathbb R\to\mathbb R$ with $$f_1(I_1)=L_1\tag2$$
Let $$P_1(x):=P(x,f_1(x))\;\;\;\text{for }x\in I_1$$
It's easy to see that $P_1$ is a polynomial of degree at most $1$
Using the result of my other question, we obtain the existence of some $C\in\mathbb R$ with $$P_1=C_1\left.f_1\right|_{I_1}\tag3$$
Now, we know that $$C_1f_1(z_1^2)=P(z^2)=0=P(z^3)=C_1f_1(z_1^3)\tag4\;,$$ but is this sufficient to conclude?

  So, the question is: How can we conclude and what do we do, if $z_1^2=z_1^3$?

$^1$ Since $z_1^2\ne z_1^3$, we can choose $$f_1(x):=z_2^2+\frac{x-z_1^2}{z_1^3-z_1^2}\left(z_2^3-z_2^2\right)\;\;\;\text{for }x\in\mathbb R\;.$$

Every point $x$ in the triangle $T$ can be written as
$$
x = \sum_{i=1}^3 \lambda_i x_i, \ \sum_{i=1}^3 \lambda_i=1, \ \lambda_i\ge 0\forall i.
$$
Take an affine linear function $f(x)=a^Tx+b$ with $f(x_i)=0$ for all $i$.
Then with the characterization of $x\in T$ above
$$
f(x) = a^Tx + b = (\sum_{i=1}^3 \lambda_i a^Tx_i )+ b
=\sum_{i=1}^3\lambda_i (a^Tx_i+b) = \sum_{i=1}^3\lambda_if(x_i)=0.
$$
Hence $f$ is zero on the triangle.
If the triangle is not degenerate (collapsing to line or point), then the expansion
$$
x = \sum_{i=1}^3 \lambda_i x_i, \ \sum_{i=1}^3 \lambda_i=1
$$
is valid for all $x\in \mathbb R^2$, which gives $f\equiv0$.

The key observation is that any point of $K$ can be written as $\ z\ =\ t\cdot z_r+(1-t)\cdot z_s\ $ for $t\in [0,1]$ and $r,s\in\{1,2,3\}$.

Now write a general polynomial $p$ of degree $1$ as $\ p(x,y)=ax+by+c\ $ (with $a,b,c\in\Bbb R$), then express $p(z)$ by means of $p(z_r)$ and $p(z_s)$.


If $N \lhd G$ and $A \lhd B \le G$ then $AN \lhd BN$? The content of the question is in the title. I'm asking this in relation to a proof that (assuming the above conditions): $$BN/AN \cong B/A(B \cap N)$$
Which I believe follows directly from an application of the second isomorphism theorem for groups.


Find all integers n > 1 so that n, n + 2, n + 4 are all prime Find all integers n > 1 so that n, n + 2, n + 4 are all prime.

Have no idea what to do, anyone know how to solve this one?

Hint : One of the three numbers is divisible by $3$.

The two first must be twin numbers. For the rest I had a lapse. Sorry.

The question is then Find all the triples of consecutive primes such that the two first ones and the two last ones are both twin numbers.

The integer $n$ can be equal to $3m, 3m+1, 3m+2$. Taking these three only possibilities we see clearly that of $n,n+2,n+4$ one of them is necessarily divisible by 3.Hence the only solution $(3,5,7)$. This is a very easy problem indeed.


Proving that the axioms for addition hold in $R$ -- Associativity. Principles of Mathematical Analysis by Walter Rudin. I'm studying the proof for theorem 1.19 from Principles of Mathematical Analysis by Walter Rudin (public online copy here).

There exists an ordered field $R$ which has the least-upper-bound property.
Moreover, $R$ contains $Q$ as a subfield.

The lengthy proof for this theorem is contained in the appendix to chapter 1 on page 17.
I am stuck on Step 4:

If $\alpha \in R$ and $\beta \in R$ we define $\alpha + \beta$ to be the set of all sums $r + s$, where $r \in \alpha$ and $s \in \beta$.
We define $0^*$ to be the set of all negative rational numbers. It is clear that $0^*$ is a cut. We verify that the axioms for addition (see Definition 1.12) hold in $R$, with $0^*$ playing the role of $0$.

Specifically, the author does not prove associativity (A3):

(A3) As above, this follows from the associative law in $Q$.

I don't really understand what the author is saying here; nor am I sure how to demonstrate this myself (I've tried).
Definition 1.12:

(A3) Addition is associative: $(x + y) + z = x + (y + z)$ for all $x, y, z \in F$.

The entire reason students such as myself buy textbooks is to have these main concepts explained to them, which Rudin has, for whatever reason, decided not to do. I would greatly appreciate it if people could please take the time to fill this gap by proving the associative property.

For $x, y, z \in R$, which are 3 cuts of $Q$, you have to prove that:
$$(x+y)+z = z+(y+z).$$

What is an element of the cut $(x+y)+z$? It is the sum of an element $p + s$ where $p \in x+y$ and $s \in z$. And the element $p$ can be written (by definition of the cut $x+y$) as $q+r$ where $q \in x$ and $r \in y$.

Therefore an element of $(x+y)+z$ is having the form $(q+r) + s$ where $(q,r,s) \in (x,y,z)$. Now you can use the associative property of the addition in $Q$ to say that $(q+r)+s = q+(r+s) \in x+(y+z)$. We have proven that:
$$(x+y)+z \subseteq x+(y+z)$$.

You can prove in a similar way the converse inclusion which brings you to the conclusion.

As discussed in the comments to mathcounterexamples.net's answer, there seem to be errors?

I think a correct proof is as follows:

We want to show that $(x + y) + z = x + (y + z)$.

Let $p \in x + y$. Then $p = r + s$, where $r \in x, s \in y$.

Let $c \in z$.

$\therefore (x + y) + z = (r + s) + c$

$(r + s) + c = r + (s + c) \in x + (y + z)$

$\therefore (x + y) + z \subset x + (y + z)$


Explain branches of geometry for non-mathematician Some background -
I'm an advanced physics undergrad and lately was motivated to self study basic contemporary geometry to get a better grip on general relativity (maybe there is a more appropriate category for what I call here geometry). I've found many recommendations for books and resources, yet, as an "outsider" to the field of mathematics, I understand very little the difference between the branches of geometry. Wikipedia's definitions didn't help much to get a clear distinction between the different fields, and I'd like to know some more before I invest significant amount of time in this.

The question is - can you describe (or refer to simple description) for a mathematical layman like me what kind of problems the following branches deal with -  topology, differential geometry , Riemannian geometry and the differences between them (especially between the latter two).

Topology is really distinct from the other two.  Roughly speaking, topology is the study of those properties of a geometric space that do not depend on notions such as distance or angles.  Within topology, the surface of a sphere is considered to be the same as the surface of a cube, though these can be distinguished from the surface of a doughnut (i.e. a torus).  Topology is also concerned with a much broader class of geometric spaces than the other two, including manifolds, complexes, spaces of functions, and more general topological spaces.  The basics of topology are a foundational subject for large portions of mathematics, and most advanced (i.e. graduate-level) treatments of differential geometry assume at least some topology as a prerequisite.

Differential and Riemannian geometry, by contrast, are concerned mainly with manifolds (i.e. curves, surfaces, and their higher-dimensional analogues), and involve notions such as distance, angle, and curvature, as well as the heavy use of differential calculus.

The distinction between differential geometry and Riemannian geometry is subtle.
Riemannian geometry describes a certain approach or point of view towards differential geometry, where the primary objects of study are abstract differentiable manifolds endowed with Riemannian metrics.  This approach has been extraordinarily fruitful, and at this point the vast majority of differential geometry clearly falls under the Riemannian framework.  Indeed, "differential geometry" and "Riemannian geometry" are sometimes described as synonyms.

At the same time, there are a few subjects within differential geometry that don't really fall within Riemannian geometry.  For example, the relatively new fields of symplectic geometry (which is related to Hamiltonian mechanics) and contact geometry study geometric structures that can be placed on a differentiable manifold that are quite distinct from a Riemannian metric, though some would argue that these subjects aren't part of differential geometry either.  But even Lorentzian manifolds (quite important in general relativity) aren't strictly speaking Riemannian manifolds, though they are considered "pseudo-Reimannian" and the study of them is quite similar to Riemannian geometry.  Finally, folks who study arbitrary connections or even synthetic differential geometry have arguably moved on from Riemannian geometry to something more general.

There are also classical aspects of the study of differential geometry that haven't quite been absorbed into the Riemannian framework, such as the study of curves and surfaces in two and three dimensions.  Roughly speaking, Riemannian geometry is concerned with the intrinsic properties of a manifold, so any geometric properties that are extrinsic, i.e. depend on an embedding into an ambient space, don't really fall under Riemannian geometry.  Thus Gaussian curvature is a Riemannian concept, but mean curvature and the study of minimal surfaces (i.e. the shapes made by soap films) is really distinct from Riemannian geometry.

Topology: As a general mind setup, the study of a particular branch of Geometry can be seen as the study of properties of geometrical objects that remain unchanged after a beforehand well-precised set of transformations. For instance, Euclidean Geomety is what you get if you allow only orthogonal transformations of the space: distances are preserved, angles are preserved, lines remain lines and so on. If you allow all affine transformations (an immediate generalizations of linear transformations) instead of just the orthogonal ones, metric notions such as distance or angles lose meaning, but lines remain lines- Topology is what you get when you allow the large set of all bicontinuous invertible transformations of the space. Now also linearity is lost: a line can be transformed into a non straight curve and so on. Yet some basic core properties are preserved: connectedness, compactness and one realizes that some basic properties of continuous functions (e.g. Bolzano's theorem) are toplogical in nature.

Differential Geometry: Object of study of DG are the real and complex varieties (read: geometrical "shapes") that can be described as zero-loci of analytic functions. Here Geometry and Analysis come together, so to speak, allowing the use of analytical tools to study geometrical properties.

Riemannian Geometry: RG is sort of subset of DG: in RG varieties are endowed with an intrinsic notion of distance and curvature. Namely there's little if any reference to an ambient space where these varieties may be embedded. As a (big) example consider Einstein's spacetime of GRT which is locally related to Minkowski's spacetime.

For both DG and RG, the notion of variety (aka, manifold) is fundamental: I suggest to start with a good textbook.


Finding the left and right cosets of H = {(1), (12), (34), (12) ○ (34)} in S4 I have an exercise where I am supposed to find the left and right cosets of H = {(1), (12), (34), (12) ○ (34)} in S4. But how do I generate the cosets? As I have understood it you are supposed to pick a number that is not in the set H and multiply it with every number in H. But this does not exactly give the right answer. I would really appreciate it if someone gave an easy to understand explanation of how to generate the left and right cosets.

Find all distincr right and left costs of the subgroup H={0,2,4} of the group (z6,+6) and obtain the right coset decomposition of z6.


Determine the highest order of an element of a Rubik's Cube group The period of a sequence of moves on a Rubik's Cube is the number of times it must be performed on a solved cube before the cube returns to its solved state. For example, a $90$° clockwise turn on the right face has a period of four; a $180$° clockwise turn on the right face and a $180$° turn on the top face has a period of $12$.

Let's make a $3\times3\times3$ Rubik's Cube group $G$. Each element of $G$ corresponds to each possible scramble of the cube - the result of any sequence of rotations of the cube's faces. Any position of the cube can be represented by detailing the rotations that put a solved cube into that state. With a solved cube as a starting point, each of the elements of $G$ directly align to each of the possible scrambles of the Rubik's Cube.

The cardinality of $G$ is $|G|=43{,}252{,}003{,}274{,}489{,}856{,}000=2^{27}3^{14}5^{3}7^{2}11$ and the largest order/period of any element in $G$ is $1260$. To elaborate, no algorithm needs to be performed on a cube more than $1260$ times to return it to the solved state.

Now let's say we extended $G$ for other sizes of cubes, so $G_3$ is the group of a $3\times3\times3$ and $G_4$ is a the group of a $4\times4\times4$. (If this isn't a valid naming convention, forgive me, I've just begun learning group theory).

Is there a way to find the highest order for any sequence of moves in $G_x$? For example, could I define a function $f$ such that $f(x)$ would give the highest order for any sequence of moves in $G_x$? What would $f$ look like? Would such a function be possible for any size of cube?

Thanks a lot in advance. Once again I apologize for any mistakes I've made; feel free to point them out or correct them.

The maximum orders for a $n\times n\times n$ Rubik's cube group:
\begin{align*}n=2: &\quad 3^2\cdot 5 = 45\\
n=3: &\quad 2^2\cdot 3^2\cdot 5\cdot 7 = 1260\\
n=4: &\quad 3^2\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17 = 765765\\
n=5: &\quad 2^4\cdot 3^2\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 23 = 281801520\\
n\ge 6: &\quad 2^4\cdot 3^2\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23 = 5354228880\end{align*}

In general, there are five types of piece to consider:

For odd $n$, each face has a center. By convention, we treat these as fixed. If we don't, they can be rotated like a solid cube, for a copy of $S_4$. Highest order $4$, least common multiple $2^2\cdot 3=12$. As the largest possible order is divisible by $12$ for all odd $n\ge 3$, whether we include these doesn't affect the answer.
For any $n\ge 2$, there are $8$ corner pieces which can be permuted and rotated, a semidirect product of $S_8$ and $(\mathbb{Z}/3)^8$ - except that that the "sum" of those rotations is zero, dropping one factor of $3$ from the latter part. The order of an element is at worst $3$ times the order of an element of $S_8$. Highest possible order $15\cdot 3=45$, least common multiple $2^3\cdot 3^2\cdot 5\cdot 7$.
For any odd $n\ge 3$, there are $12$ centers of edges, which can be permuted and reflected. The number of reflections must also be even, for a semidirect product of $S_{12}$ and $(\mathbb{Z}/2)^{11}$. Highest possible order $60\cdot 2=120$, least common multiple $2^4\cdot 3^2\cdot 5\cdot 7\cdot 11$.
For any $n\ge 4$, there are $\left\lfloor\frac{n}{2}\right\rfloor-1$ orbits of 24 off-center edges, which can't change how far from the center of their edge they are. The pair of edges on each side are distinguishable due to being reflections of each other, and the group for each set of these pieces is a copy of $S_{24}$. Highest possible order $8\cdot 7\cdot 5\cdot 3 = 840$, least common multiple $2^4\cdot 3^2\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23$.
For any $n\ge 4$, there are $\left\lfloor\frac{(n-2)^2}{4}\right\rfloor$ orbits of 24 off-center face pieces each. These can be rotated in their face, but not reflected;  mirror images across a symmetry line within a face (in size $\ge 6$) are separate orbits. Each piece of the same color in one of these sets is indistinguishable, so that gets us an action of $S_{24}$, with $\frac{24!}{24^6}$ possible states. The possible orders are the same as in the previous case. In terms of what we need to solve the cube, there are three variants based on whether they lie on a symmetry line of a face - but they're all the same for the count we care about here.

So then, the highest possible order in any Rubik group is the least common multiple of all of these, namely $2^4\cdot 3^2\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23$. This is achieved for all $n\ge 6$; we have at least six of the size-24 orbits, which we can use to produce orders of $16\cdot 5$, $9\cdot 11$, $7\cdot 13$, $17$, $19$, and $23$.

Now, all of this assumes that we can modify the pieces in each orbit independently. That isn't quite true - several combinations must be even permutations. First, for odd $n$, the combined permutation of corners and edge centers is even; they're modified by face rotations, each of which is a 4-cycle of edges and a 4-cycle of corners. Second, for $n\ge 4$, the combined permutation of the corners and an orbit of pieces on face diagonals is even. Next, for $n\ge 6$, each mirrored pair of orbits of off-center face pieces must have a combined even permutation; each face rotation and each relevant slice rotation is a 4-cycle in each of the orbits. Finally, for odd $n\ge 5$, the combined permutation of off-center edge pieces, pieces on face diagonals, and pieces on face midlines at a fixed distance from the center is even. These parity restrictions are the only additional restriction; other than that, the orbits can be manipulated independently.
How does this change things? For the $n\ge 6$ case, all we have to do is implement the order $16\cdot 5$ piece as the product of a $16$-cycle, a $2$-cycle, and a $5$-cycle. Then all of the permutations we're working with are even, so there aren't any problems. In the smaller cases, we'll have to check.

For $n=2$, only the corners matter, and the highest order is $45$.
For $n=3$, we have corners and edge centers. With $17$ possible orders for the first and $31$ for the second, we can just scan all of the possible combinations. The highest LCM is $2520$, from order $45$ (3-cycle, 5-cycle, rotations in the 3-cycle don't add to zero) in the corners and order $56$ (4-cycle, 7-cycle, reflections in the 4-cycle don't add to zero) in the edges. This, however, doesn't pass the parity check; the combined permutation is odd, and there's no wiggle room to change that. That leaves us dropping down to the second-highest possibility of $1260$, possible in several ways ($45$ and $28$, $45$ and $84$, $35$ and $36$, $70$ and $36$). Three of those fail the parity check, leaving only the pair of order $45$ in the corners and order $28$ (Two 2-cycles, a 7-cycle, reflections in at least one 2-cycle don't add to zero) in the edges. Highest order $1260$, as previously noted.
For $n=4$, we have the corners, one orbit of off-center edges, and one orbit of off-center face pieces. With over a hundred possible orders in $S_{24}$, finding the least common multiple isn't the easiest search. Pruning everything with an order that divides another gets that down to a manageable total of $27$, and $5$ for the corners. Running it in the spreadsheet I've been using, the LCM is $765765=3^2\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17$; $45$ in the corners, $11\cdot 13$ in the edges, and $7\cdot 17$ in the faces. Checking parity... all the pieces are odd permutations, so we'll actually have something of that order.
For $n=5$, we have corners, edge centers, three size-24 orbits, and two parity restrictions that cross orbits. The best I've got here is $2^4\cdot 3^2\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 23 = 281801520$; it could use an independent check, but I'm pretty sure it's right. The three size-24 orbits get us orders of $23$, $7\cdot 17$, and $11\cdot 13$ while the corners give us order $45$ and the edges give order $16$ (an 8-cycle and a 2-cycle, with an odd number of reflections in the 8-cycle). They're all odd permutations, so the parity checks pass.
For $n\ge 6$, as previously noted, we reach the maximum possible order $2^4\cdot 3^2\cdot 5\cdot 7\cdot 11\cdot 13\cdot 17\cdot 19\cdot 23 = 5354228880$.

Notations are from https://ruwix.com/the-rubiks-cube/notation/

$RY$ is an element of order 1260.
It's easy to check that $(RY)^{36}$ keeps all the corners intact but permutes the edges in 2 disjoint groups of order 5 and 7, respectively. So, the total order is $36*5*7=1260$.

$RY$ applied 36 times - https://ruwix.com/saved-rubiks-cube/?moves=RYRYRYRYRYRYRYRYRYRYRYRYRYRYRYRYRYRYRYRYRYRYRYRYRYRYRYRYRYRYRYRYRYRYRYRY


Represent recurrent digit in numbers I have try google and this website search engine, didnt find anything.

Let's say that I have a number like this : 111118883458888.

Is there a way to express this number like this (reg exp) : ('1'^5)
 ('8'^3) '345' ('8'^4), but in a mathematical way?

Basically, is there a way to represent recurent digit in numbers?

EDIT -

I'll add why I'm asking the question, basically I wanted a way to represent binary number (like 256 and 512 bits) and you don't feel sick reading them. Thanks for the answers!

An overbar can be used to denote infinite repetition of decimals, say

$$\frac17=0.\overline{142857}$$

and I don't think there is a strong need to denote the finite repetition of digits.

What about indexed overlines (also for groups) ?

$$\overline{1}_5\overline{821}_3345\overline{8}_4$$

and also recursive ?

$$\overline{1}_5\overline{\overline{821}_3345\overline{8}_4}_5$$


Is an infinitely-long integer in the natural numbers? Say you constructed a number by counting out the positive odd numbers, in sequence:

$\{1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41, 43, 45, 47, 49, 51, 53, 55...\}$

$135791113151719212325272931333537394143454749515355575961636567697173757779...$

This number would clearly be infinitely long, as there are an infinite number of odd numbers to choose from. Would this number be considered a natural number?

Follow-up:

If this number is not natural, what specifically disqualifies it? It certainly appears on the number line, just infinitely far from the origin, and it is a whole number. Moreover, which of the basic sets is this number a member of, and why?

No, it is not an integer. Nor a real number, nor anything meaningful, until you give it a meaning.

Answer this: If that is an integer $n$, what is $n+1$? Or what is $10\times n$? Looks like $10\times n=n$.

On any line, we have the concept of "between-ness." If $x,y,z$ are three distinct points on a line, then one of the points is between the other two.

Now, if your thing represents a number, let $x=0,y=123\dots,z=135\dots$.

You might first say, "Obviously, $y$ is between $x$ and $z$." But wait, is $z_1=0135\dots$ the same as $z$? If so, then it looks like $z$ should be between $0$ and $y$! So now, you are stuck trying to give meaning to the above expression, where $z_1=0135\dots$ is a different thing that $z_2=135\dots$.

Again, notation is given meaning by definitions. Just because we can write something down and it sort-of looks like something we've seen before doesn't mean it has meaning.

The natural numbers (positive integers) have a lot of representations. Base $10$ is a notation system for representing natural numbers, and we use it, and other notations (like Roman numerals, or base $2$, etc.) because the most primitive way to write a number $n$ is as a repetition of $n$ symbols, which is impractical. So we might write $10$ as $IIIIIIIIII$. That gets horrible fast, and it is very hard to write out positive integers in this way, so we find other notation to let us represent this, for clarity and practicality reasons.

Roman numerals are instructive for this reason, because $I,II,III$ are $1,2,3$. But then,  somebody realized writing all those $I$s got tedious, and started "shortening" larger numbers in a predictable way.

But every positive integer is necessarily representable by a sequence of one symbol. It is unclear, even if you allowed infinite sequences of of that symbol, how you would distinguish $1357\dots$ from $1111\dots$ as different infinite sequences of dots.

Notation does not exist in a vacuum. Notation has meaning because we define it to have meaning. Escher can draw pictures of things that cannot exist, and you can write symbols that sort-of look somewhat like normal notation but aren't, but which has no useful meaning.

There does exist a type of number that has infinite digits, called $g$-adic numbers, were $g$ is a base. But these numbers start at the right. For example, there is a $10$-adic number:

$$...131197531$$

The $g$-adic numbers are not ordered, so we can't put them on a line. When $g$ is divisible by two or more distinct primes, they even have zero divisors - there are $a,b\neq 0$ with $ab=0$.

But $g$-adic numbers, while containing all the integers, also contain non-integers, and have very foreign behaviors.

When $g$ is prime, though, they become very useful and interesting for solving certain types of problems.

"This is the follow-up:

"If this number is not natural, what specifically disqualifies it?

It's infinite.  All real numbers are finite in value.

"It certainly appears on the number line, just infinitely far from the origin,

Which means it doesn't appear on the number line.  The number line is the collection of all points that are zero or finite distance from the origin.  This number is not so it doesn't appear on then number line.

"and it is a whole number."

No, it isn't.  Now you are just making stuff up.

"Moreover, which of the basic sets is this number a member of, and why?"

Well, it's a member of the set of divergent series.  X = $\sum_{k = 0}^{\infty} b_k*10^k$ which diverges...


Solve $T(n) = T( \lfloor 2n / 10\rfloor ) + \Theta(n^3)$ recurrence I don't know how to solve the following recurrence. Particularly, I don't know how to cope with the $2n$ term:
$$T(n) = T( \lfloor 2n / 10\rfloor ) + \Theta(n^3)$$

Master theorem of recurrences doesn't really explains how to... I was thinking that this would solve it:

$$T(n) = 2T(\lfloor n / 10\rfloor) + \Theta(n^3)$$

But I'm not sure. How do I approach this?

\begin{align}T(n) &= T(\lfloor2n/10 \rfloor) + \Theta(n^3) \\
&= T(\lfloor n/5\rfloor) + \Theta(n^3) &(1)\\
&=T(n/5) + \Theta(n^3) & (2)\\
&= \Theta(n^3) & (3)\end{align}

I have applied:

$(1)$ Fraction simplification: $2n/10 = n/5$.

$(2)$ Removing floors. In general you can remove all floors and ceilings in recurrences that may be solved with the master theorem, that is, recurrences of the form $$T(n) = a T(n/b) + g(n)$$ (this is a proved fact).

$(3)$ Application of the master theorem.


Is there any integral for the Golden Ratio? I was wondering about important/famous mathematical constants, like $e$, $\pi$, $\gamma$, and obviously the golden ratio $\phi$.
The first three ones are really well known, and there are lots of integrals and series whose results are simply those constants. For example:

$$ \pi = 2 e \int\limits_0^{+\infty} \frac{\cos(x)}{x^2+1}\ \text{d}x$$

$$ e = \sum_{k = 0}^{+\infty} \frac{1}{k!}$$

$$ \gamma = -\int\limits_{-\infty}^{+\infty} x\ e^{x - e^{x}}\ \text{d}x$$

Is there an interesting integral* (or some series) whose result is simply $\phi$?

* Interesting integral means that things like

$$\int\limits_0^{+\infty} e^{-\frac{x}{\phi}}\ \text{d}x$$

are not a good answer to my question.

Potentially interesting:

$$\log\varphi=\int_0^{1/2}\frac{dx}{\sqrt{x^2+1}}$$

Perhaps also worthy of consideration:

$$\arctan \frac{1}{\varphi}=\frac{\int_0^2\frac{1}{1+x^2}\, dx}{\int_0^2 dx}=\frac{\int_{-2}^2\frac{1}{1+x^2}\, dx}{\int_{-2}^2 dx}$$

A development of the first integral:

$$\log\varphi=\frac{1}{2n-1}\int_0^{\frac{F_{2n}+F_{2n-2}}{2}}\frac{dx}{\sqrt{x^2+1}}$$

$$\log\varphi=\frac{1}{2n}\int_1^{\frac{F_{2n+1}+F_{2n-1}}{2}}\frac{dx}{\sqrt{x^2-1}}$$

which stem from the relationship $(x-\varphi^m)(x-\bar\varphi^m)=x^2-(F_{m-1}+F_{m+1})x+(-1)^m$, where $\bar\varphi=\frac{-1}{\varphi}=1-\varphi$ and $F_k$ is the $k$th Fibonacci number. I particularly enjoy:

$$\log\varphi=\frac{1}{3}\int_0^{2}\frac{dx}{\sqrt{x^2+1}}$$
$$\log\varphi=\frac{1}{6}\int_1^{9}\frac{dx}{\sqrt{x^2-1}}$$
$$\log\varphi=\frac{1}{9}\int_0^{38}\frac{dx}{\sqrt{x^2+1}}$$
$$\log\varphi=\frac{1}{12}\int_1^{161}\frac{dx}{\sqrt{x^2-1}}$$

This is my attempt to find another integral representation for Golden ratio using some special function as shown in the above image !!!!!!!


Why do mathematicians use single-letter variables? I have much more experience programming than I do with advanced mathematics, so perhaps this is just a comfort thing with me, but I often get frustrated when I try to follow mathematical notation. Specifically, I get frustrated trying to keep track of what each variable signifies.

As a programmer, this would be completely unacceptable no matter how many comments you added explaining it:

float A(float P, float r, float n, float t) {
  return P * pow(1 + r / n, n * t);
}

Yet a mathematician would have no problem with this:

  $A = P\ \left(1+\dfrac{r}{n}\right)^{nt}$

  where
  $A$ = final amount
  $P$ = principal amount (initial investment)
  $r$ = annual nominal interest rate (as a decimal)
  $n$ = number of times the interest is compounded per year
  $t$ = number of years

So why don't I ever see the following?

$\text{final_amount} = \text{principal}\; \left(1+\dfrac{\text{interest_rate}}{\text{periods_per_yr}}\right)^{\text{periods_per_yr}\cdot\text{years}}$

I think one reason is that often one does not want to remember what the variable names really represent.

As an example, when we choose to talk about the matrix $(a_{ij})$ instead of the matrix $(\mathrm{TransitionProbability}_{ij})$, this expresses the important fact that once we have formulated our problem in terms of matrices, it is perfectly safe to forget where the problem came from originally -- in fact, remembering what the matrix "really" describes might only be unnecessary psychological baggage that prevents us from applying all linear-algebraic tools at our disposal.

(As an aside, have you ever seen code written by a mathematician? It very often looks exactly like your first example.)

Yet another aspect...

In fact not all symbols consist of only one letter:
I have to get pretty creative within my publications not to exhaust all symbols within the one publication - including capitalization, other fonts and greek letters. I am not at all able to stay consistent within my entire body of work precisely because there are not enough letters. So, I actually use - at times - symbols with more than one letter. Thanks to that I have the iron rule to always use a dot for multiplication.
However, I try to avoid it, because of many of the mentioned reasons. It also helps to think hard before introducing a new symbol. This leads me to think whether I really need this new symbol or whether there is an underlying principle that allows me to abstract multiple concepts into a single concepts. This makes me a better researcher and actually leads to new publications by itself. (You could say the publications write themselves :-P.)
However, most publications in general do not need that many symbols and even I do not even nearly need as many symbols as I do while programming.


How to compute this sum? I want to sum the following:

$$f(n) = \sum_{i=1}^n (i^3 \cdot (n \mod i))$$

Since the sum can be huge I have to output the sum modulo some given number m.
How can I approach this problem?

Also, n can be upto $10^{10}$.

As Duplicate question of this has been deleted, I am posting the snapshot of answer posted there.


Proofs that every mathematician should know. There are mathematical proofs that have that "wow" factor in being elegant, simplifying one's view of mathematics, lifting one's perception into the light of knowledge, etc.

So I'd like to know what mathematical proofs you've come across that you think other mathematicans should know, and why.

Here is my favourite "wow" proof .

Theorem
There exist two  positive irrational numbers $s,t$  such that $s^t$ is rational.
Proof
If $\sqrt2^\sqrt 2$ is rational, we may take $s=t=\sqrt 2$ .
 If $\sqrt 2^\sqrt 2$ is irrational , we may take $s=\sqrt 2^\sqrt 2$ and  $t=\sqrt 2$ since  $(\sqrt 2^\sqrt 2)^\sqrt 2=(\sqrt 2)^ 2=2$.

It is propably not something which everybody should know, nevertheless it is simply beautiful!

The stable Hurewicz theorem using that the sphere spectrum is a compact generator of the stable homotopy category.
In particular Serre's theorem that the rational stable homotpy groups of spheres are trivial for degrees bigger than 1.


How to tell whether the roots are the only rational roots for a given polynomial Find all the rational roots of the polynomial $p(x)=2x^4-5x^3+7x^2-25x-15$. I only found $x=3, -\frac{1}{2}$. I am not sure whether there is any other rational roots.

Is there a way to tell whether these are the ONLY rational roots other than doing the long polynomial division?

Thanks.

You can use Vieta's relations: the sum of the roots (in $\mathbf C$) is $\frac 52$. Now the sum of the found roots is already $\frac 52$. So if  there were other rational roots, they would be opposite. Now the equation can be written as
$$2x^4+7x^2+15=5x(x^2+5),$$
and having opposite roots would imply they're roots of $x^2+5$, which has no real root.

Develop $(x-3)(x+\frac{1}{2})(ax^2+bx+c)$. Find $a,b,c$ by identification. Then, you can calculate the other roots.


Evaluation of $\sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots+\sqrt{\frac{1}{2^n}}}}}$ I was just playing around with a calculator, and came to the conclusion that:

$$\sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots+\sqrt{\frac{1}{2^n}}}}} \approx 1.29$$

Now I'm curious. Is it possible to evaluate the exact value of the following?

This question is related to at least five others:

Problem 6 - IMO 1985

How to find this limit: $A=\lim_{n\to \infty}\sqrt{1+\sqrt{\frac{1}{2}+\sqrt{\frac{1}{3}+\cdots+\sqrt{\frac{1}{n}}}}}$

Find the limit $L=\lim_{n\to \infty} \sqrt{\frac{1}{2}+\sqrt[3]{\frac{1}{3}+\cdots+\sqrt[n]{\frac{1}{n}}}}$

Evaluating the sum $\lim_{n\to \infty}\sqrt[2]{2+\sqrt[3]{2+\sqrt[4]{2+\cdots+\sqrt[n]{2}}}}$

Limit of $\lim_{x\rightarrow 1}\sqrt{x-\sqrt[3]{x-\sqrt[4]{x-\sqrt[5]{x-…}}}}$

This makes a short answer possible (and desirable).
For a numerical calculation backward recursion is proposed (again):
$$
  a_{n-1} = \sqrt{1/2^n+a_n} \qquad \mbox{with} \quad \lim_{n\to\infty} a_n = 0
$$
Here comes the Pascal program snippet that is supposed to do the job:

program apart;
procedure again(n : integer);
var
  a,two : double;
  k : integer;
begin
  two := 1;
  for k := n downto 2 do
    two := two/2;
  a := 0;
  for k := n downto 2 do
  begin
    a := sqrt(two+a);
    two := two*2;
  end;
  Writeln(a);
end;
begin
  again(52);
end.

Note that an error analysis is not implemented in the program. This has not much sense
because the accuracy is determined by the smallest $1/2^n$ that can be represented
with some significance; that is for $n\approx 52$ in double precision Pascal.
The outcome is, of course, in concordance with the value already found by
Lucian:

 1.28573676335699E+0000

Disclaimer. I certainly would have tried the closed form
- whatever that means in modern times - if I only could believe that such
a thing does indeed exist here.

$$y=\sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots+\sqrt{\frac{1}{2^n}}}}}\equiv\sqrt{\frac12+\frac{1}{\sqrt{2}}\left(\sqrt{\frac12+\sqrt{\frac12+\cdots+\sqrt{\frac{1}{2}}}}\right)}$$

$$y=\sqrt{\frac12+\frac{1}{\sqrt{2}}\left(\sqrt{\frac12+\sqrt{\frac12+\cdots+\sqrt{\frac{1}{2}}}}\right)}$$ But the Let the term in bracket be x
therefore:$$x=\sqrt{\frac12+\sqrt{\frac12+\cdots+\sqrt{\frac{1}{2}}}}\equiv\sqrt{\frac12+x}$$
squaring both sides
$$x^2=\frac12+x$$
$$x^2-x-\frac12=0$$solving the equation gives:
$$x=\frac{1+\sqrt{3}}{2}$$
but we have $$y=\sqrt{\frac12+\frac{1}{\sqrt{2}}\left(\sqrt{\frac12+\sqrt{\frac12+\cdots+\sqrt{\frac{1}{2}}}}\right)}=\sqrt{\frac12+\frac{1}{\sqrt{2}}x}$$
Putting the value of x into y gives$$y=\sqrt{\frac12+\frac{1}{\sqrt{2}}\frac{1+\sqrt{3}}{2}}=1.211$$
Therefore $$\sqrt{\frac12+\sqrt{\frac14+\sqrt{\frac18+\cdots+\sqrt{\frac{1}{2^n}}}}}=1.211$$
This is an exact value.


Expressing an integral in closed form Is there a closed-form expression for this integral?

$$\int \frac{\sin(Ax/2)}{A\sin(x/2)}\mathrm{d}x$$


Constructing a bounded set of real numbers with exactly three limit points Credit for the problem goes to Baby Rudin, Chapter 2, Exercise 5.

We are to construct a bounded set of real numbers with exactly three limit points. Seeing as there are few "computation"-tasks in the book, I haven't really had a chance to build up intuition the way I am used to, however, loeoking at the text in Topology, it seems like I have to get used to this.

I let $X$ be the metric space $\mathbf{R}$. I let $E$ be its subset $\{(1-\epsilon,1+\epsilon),(2-\epsilon,2+\epsilon),(3-\epsilon,3+\epsilon)\}$ for small $\epsilon > 0$. Clearly, $1, 2$ and $3$ are limit points of $E$. However, if I have understood the concept of a limit point (and a neighborhood) correcly, so is $1+\frac{\epsilon}{2}$. I'm having a hard time seeing how I am to limit the number of limit points (no pun intended.)

Take the set
$$\left\{\sin{2\pi n\over3}+{1\over n}\ \biggm|\>n\in{\mathbb N}_{\geq1}\right\}\ .$$


Need a generalized way to solve $127=1+2+2^2+...$ How can I solve this equation generally? I can solve it by checking in my calculator. But I don't know any generalized way.

$$127=1+2+2^2+2^3+....+ 2^{x-1}$$

Hint: $127=a\cdot \dfrac{r^{\text{number of terms}}-1}{r-1}$, where $a=$starting term, $r=$common ratio


Can I show $\frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \leq 1$ by induction? It is known that $a_{n+1} > a_n$. I tried to prove
$$\frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \leq 1$$
via induction.
Base case: $a_0 = \frac{1}{0+1} = 1 \leq 1$
Inductive step: assume $a_n \leq 1$, prove $a_{n+1} \leq 1$.
$a_n = \frac{1}{n+1} + \frac{1}{n+2} + \cdots + \frac{1}{2n} \leq 1$
$\frac{1}{n+2} + \frac{1}{n+3}+ \cdots + \frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2} \leq 1-\frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+2}$
But $-\frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+2}$ is positive, so I'm not sure where to go from here.
Normally, the inductive step would involve a change on the right hand side, but in this case, it remains the same. Makes me wonder if induction is even the correct approach?

The result is "obvious." There are $n$ terms in the sum, each $\le \frac{1}{n+1}$. So the sum is $\le \frac{n}{n+1}$, which is $\lt 1$.

If we, unreasonably, want to prove the result by induction, we run into the problem mentioned in the OP.  The solution is to strengthen the induction hypothesis. We prove by induction that our sum is $\le \frac{n}{n+1}$.

So suppose we know that for a particular $k$ we have
$$\frac{1}{k+1}+\frac{1}{k+2}+\cdots +\frac{1}{2k}\le \frac{k}{k+1}.\tag{1}$$
We want to show that
$$\frac{1}{k+2}+\frac{1}{k+3}+\cdots +\frac{1}{2k}+\frac{1}{2k+1}+\frac{1}{2k+2}\le \frac{k+1}{k+2}.\tag{2}$$
Now if we use the induction hypothesis (1) and play a bit with inequalities, we can prove (2).

According to Wolfram Alpha, this series is equivalent to $ \psi^{(0)}(2n+1)-\psi^{(0)}(n+1)$ where $ \psi^{(0)}(x)$ is the $n^{th}$ derivative of the digamma function. Using this, you could show the continuity for the function for values of $n > 0$, show that the limit as n approaches infinity is $log(2)$ (easily proven), and show that the function is continuously increases (by showing derivative is positive for $n>0$?)

Note that this actually shows that the series is always $< log(2)$, a stronger statement than that given by the induction argument.


If $(1 + 2i)$ and $(3 - 2i)$ are two roots of $x^5 + ax^4 + bx^3 + cx^2 + dx + 4$, then $a$ =? Consider the polynomial $x^5 + ax^4 + bx^3 + cx^2 + dx + 4$ where $a, b, c, d$ are real
numbers. If $(1 + 2i)$ and $(3 - 2i)$ are two roots of this polynomial then what is the
value of a?

Well, I know only the 4 roots, which are obvious from what are given (the conjugates serve as the other two), but what next?

Hints.

The polynomial has real coefficients.  By looking at the two roots you are given, you can write down two more roots without calculation.
What do you know about the product of the roots of a polynomial?
What do you know about the sum of the roots of a polynomial?

i think the solution will come from vieta's theorem .
after this multiple of all the roots will give a 4 . let the fifth root be r then ,

Acc to vieta theorem: product of roots = (-1)^n (a0/an) = -4
-4 = (1+ 2i)(1-2i)(3-2i) (3+2i) r
find the value of r from this ..
r= -65/4
 now after this again from vietas theorem

a = (-1)((1+2i) + (1-2i) + (3+2i) +(3-2i) + r ( r is the value u got above )
a= 33/4


Density of outputs of divisor function We define the function:
$$\sigma_k{(n)}=\sum_{d \mid n} d^k$$
as the sum of divisors function of the $k$th power, where $k \in \mathbb{N_0}$, for $n \in \mathbb{N}$. Now, we define:
$$S_k=\{a \space | \space \exists \space m \in \mathbb{N} \space ; \sigma_k{(m)}=a\}$$
It is clear that $S_0$ is the same as $\mathbb{N}$ since for any $n \in \mathbb{N}$, we have $\sigma_0{(2^{n-1})}=n$. However, it is clear than this is not true for $S_k$ where $k>0$.

  What is the density of $S_k$ in $\mathbb{N}$ for $k \in \mathbb{N}$?

Edit: As pointed out by @ThomasAndrews in the comments below, $k>1$ has zero density by bounding. Now, the problem is solved if anybody is able to find the density of the sum of divisors function $\sigma(x)$ ($1$st power).

The point is that $\sigma(n)$, as $n$ ranges, is usually very even (i.e. many powers of $2$ divide it).

Fix $M \ge 1$ large. Since $\sigma(n) \ge n$ for each $n$, $$\#\{1 \le m \le M : m = \sigma(n) \text{ for some } n\} \le \#\{\sigma(n) : 1 \le n \le M\} \cap [M].$$ Up to a set of density 0, every positive integer $n$ has at least $\frac{1}{2}\log\log n$ prime divisors. Therefore, we expect there to be at least $\frac{1}{4}\log\log n$ prime divisors of $n$ with highest exponent dividing $n$ being odd. The point is that for such prime divisors, since $\sigma$ is multiplicative, they contribute a power of $2$ to $\sigma(n)$, so we expect $\frac{1}{2}\log\log n$ powers of $2$ dividing $\sigma(n)$, so $$\#\{\sigma(n) : 1 \le n \le M\} \cap [M] \le \#\{m \le M : 2^{\frac{1}{4}\log\log M} \mid m\}$$ has density $0$. I don't think it should be hard to make this more rigorous, but I'm watching the finals.

Below shows that $\sigma_k(n) > n^k$ for $n>1$ and $k>0$. $\sigma_1(n) = \sum_{d|n}d = \sigma(n)$ is the very well known divisor function. Basic properties of $\sigma_k(n)$ and fundamental theorem of arithmetic will lead us to the result.

$\textbf{Property 1:}$ If $p$ is a prime number and $q \in \mathbb{N}$, then

$\sigma_k(p^q) = 1+p^k+p^{2k}+...+p^{qk}$

Proof of above property is straight forward as any integer of the form $p^i$ for $i \in \mathbb{N}$ and $i \leq q $ will divide $p^q$ with a quotient of $p^{q-i}$

$\textbf{Property 2:}$ If $gcd(a,b)=1$ then $\sigma_k(ab) = \sigma_k(a)\sigma_k(b)$

Proof: Let $d_a,d_b$ be such that $d_a|a$ and $d_b|b$. As $gcd(a,b) =1$ we also have $gcd(d_a,d_b) =1$ and thus there's a bijection between the sets $(d_a,d_b)\rightarrow d_ad_b$. Denote $d_{ab} = d_ad_b$ and we also have $d_{ab}|ab$.

Now $$\sigma_k(a)\sigma_k(b) = (\sum_{d_a | a} d_a^k)(\sum_{d_b | b} d_b^k) = \sum_{d_a | a, \ d' | b} (d_ad_b)^k = \sum_{d_{ab} | ab} d_{ab}^k=\sigma_k(ab)$$

Look at a different approach for $k=1$ here: The sigma function (sum of divisors ) multiplicative proof

Let me know in the comments if the bijection is not obvious, I write some more on that if needed.

Now using the fundamental theorem of arithmetic, for any $n\in \mathbb{N}$, there exists $p_1,p_2,...,p_q,a_1,a_2,...,a_q$ where $p_1,p_2,...,p_q$ are prime numbers and $a_1,a_2,...,a_q$ are positive integers such that,

$$n = p_1^{a_1}p_2^{a_2}...p_q^{a_q}$$

Then, $\sigma_k(n) = \sigma_k(p_1^{a_1}p_2^{a_2}...p_q^{a_q}) = \sigma_k(p_1^{a_1})\sigma_k(p_2^{a_2})...\sigma_k(p_q^{a_q})$

$=(1+p_1^k+p_1^{2k}+...+p_1^{a_1k})(1+p_2^k+p_2^{2k}+...+p_2^{a_2k})...(1+p_q^k+p_q^{2k}+...+p_q^{qk}) > p_1^{a_1k}p_2^{a_2k}...p_q^{a_qk}$

$= (p_1^{a_1}p_2^{a_2}...p_q^{a_q})^k = n^k$

Note that the inequality holds true only if $k \neq 0$ and $n \neq 1$.

Thus the Natural Density (https://en.wikipedia.org/wiki/Natural_density) of $\sigma_k(n)$ is $0$ for $k>0$


$x e ^{-\frac{x^2}{2}}$ at $\infty$ to $-\infty$ I want to know how to explain $\left. \left(x e ^{-\frac{x^2}{2}} \right) \right|_{- \infty} ^{\infty}$ is zero?

Is it because the speed of exponentiation is greater than that of linear? How to prove? If using Bernoulli's rule, is there any other way to give an intuitive explanation?

Thanks!

$\exp -\frac{x^2}{2}$ is even.

$x$ is odd.

An odd function times an even function is odd.

The integral of an odd function over the reals is zero.


Find probability of exactly one $6$ in first ten rolls of die, given two $6$s in twenty rolls I am trying to calculate the probability that, when rolling a fair die twenty times, I roll exactly one $6$ in the first ten rolls, given that I roll two $6$s in the twenty rolls.

My thoughts

Let $A = \{\text {Exactly one 6 in first ten rolls of a die} \}$ and $B = \{\text {Exactly two 6s in twenty rolls of a die} \}.$
Then I want to find
$$P[A\mid B] = \frac{P[A \cap B]}{P[B]}.$$

By the binomial distribution formula, we get that
$$P[B] = {20 \choose 2} \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^{18}.$$

Furthermore I think that $P[A \cap B]$ is equal to the probability of rolling exactly one $6$ in ten rolls and then rolling exactly one $6$ in another set of ten rolls.  That is,
$$P[A \cap B] = \left[{10 \choose 1} \cdot \left(\frac{1}{6}\right)^1 \cdot \left(\frac{5}{6}\right)^9\right]^2.$$

Am I correct in thinking this?
If so, then it follows that the required probability is
$$P[A \mid B] = \frac{\left[{10 \choose 1} \cdot \left(\frac{1}{6}\right)^1 \cdot \left(\frac{5}{6}\right)^9\right]^2}{{20 \choose 2} \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^{18}},$$
which, I know, can be simplified further!

I think you are over thinking this.  We know we get exatly two sixes in twenty rolls how many ways can that happen? Consider a roll to be 6 or not 6 we don't
care what number it is otherwise.

One of the sixes arived in any of the twenty rolls and the other in an of the nineteen remaining rolls and since a 6 is a 6 we divide by two because the order does not matter.

There are thus $\dfrac{20 \cdot 19}{2} = 190$ ways we can get exactly 2 6's in 20 rolls.

In how many ways can we have exactly one 6 in 10 rolls? Well it can be any of the 10 rolls, and it must be not 6 in the other 9 so there are 10 ways to get exactly one 6 in ten rolls, and 10 ways to get the second 6 in the last 10 rolls making

The answer is simply $\dfrac{10 \cdot 10}{190} = \dfrac{100}{190} = \dfrac{10}{19}$.


Only 3 Basis Vectors for $\mathbb{R}^4$. I've heard that for $\mathbb{R}^n$, there can only be $n$ number of basis vectors for it. Is that really the case though? Or am I having a misconception somewhere.

Let's take the example of $\mathbb{R}^4$. It seems that I can pick only 3 basis vectors $(1,1,0, 0)$, $(0,1,1,0)$ and $(0,0,1,1)$.

Firstly, these basis vectors are linearly independent. For example, no linear combination of $(0,1,1,0)$ and $(0,0,1,1)$ yields $(1,1,0,0)$. This is because these linear combinations will still have a zero as the first number in the vector.

Secondly, it (seems) that these 3 vectors span $\mathbb{R}^4$.

Can $(0,0,0,1)$ be written as a linear combination of these vectors?

The first statement of your question is absolutely correct that's what is referred to as the dimension of the vector space.

But coming to the example, we can see that your assumed basis vector (1,1,0,0), (0,1,1,0), and (0,0,1,1) does not span the entire vector space. If you add a vector (1,0,0,0), this completes your basis vector which spans the entire 4 dimensional vector space.


When we speak of a group, must we explicitly specify a certain binary operation? I was wondering about the ways we say something is a group. For example, the set $$G = \{z \in \mathbb{C} \mid z^n = 1 \text{ for some }n \in \mathbb{Z}^+\}$$ is a group under multiplication. But if I were to just say "Assume that $G = \{1,a,b,c\}$ is a group of order $4$," how can we not specify the binary operation? Is this just saying there exists a binary operation for which it is a group?

Edit: Here is a definition of a group that seems to show what I am saying:

In general, you are right. A group always has to be specified along with a binary operation, e.g. $(G,\times)$. But in many situations, the binary operations are understood. I'll give a few examples:

$\mathbb{C}\backslash \{0\}$ would be a group with multiplication as its binary operation, because if it were addition, $0$ would be in the group.
Group of 2x2 invertible matrices. In this case, we have either matrix multiplication or matrix addition as a choice for binary operation. But matrix addition would mean that the identity element is the zero matrix, which is not invertible.

Some others are just by convention, such as $\mathbb{Z}/n\mathbb{Z}$, the group of integer modulo n, which has the binary operation $+$. As compared to the multiplicative group $(\mathbb{Z}/n\mathbb{Z})^*$, which has the binary operation $\times$.

That said, if there is any ambiguity at all, it is still necessary to state the binary operation.

"What is a group? Algebraists teach that this is supposedly a set with two operations that satisfy a load of easily-forgettable axioms. This definition provokes a natural protest: why would any sensible person need such pairs of operations? "Oh, curse this maths" - concludes the student (who, possibly, becomes the Minister for Science in the future).

We get a totally different situation if we start off not with the group but with the concept of a transformation (a one-to-one mapping of a set onto itself) as it was historically. A collection of transformations of a set is called a group if along with any two transformations it contains the result of their consecutive application and an inverse transformation along with every transformation.

This is all the definition there is. The so-called "axioms" are in fact just (obvious) properties of groups of transformations."

From V. Arnol'd's article "On teaching mathematics", http://pauli.uni-muenster.de/~munsteg/arnold.html


Directional derivative of the determinant Please help me find the mistake in my derivation:

Let $f:M_{n,n}(\mathbb{R}) \to \mathbb{R}$ be the determinant function, $f(A)=det(A)$. Let $p_A(x)$ denote the charecteristic polynomial of $A$. Computing the directional derivative we get:

$$\nabla_If(A)=lim_{x\to0}\frac{det(A+Ix)-det(A)}{x}=lim_{x\to0}\frac{-p_A(-x)-(-p_A(0))}{x}=p_A'(0)$$

According to notes of a course i'm taking: $\nabla_If(A)=tr(A)$.
 Where is my mistake?

EDIT: I just looked again and i got it backwards! It says that the derivative of the determinant at $I$ along any matrix is the trace of the matrix.
$$Df_I(A)=lim_{x\to0}\frac{det(I+Ax)-1}{x}=det(A)lim_{x\to0}\frac{det(A^{-1}+xI)-det(A^{-1})}{x}=det(A)p'_{A^{-1}}(0)$$

The last expression is the trace since $p_{A^{-1}}(x)=\frac{(-x)^n}{det(A)}p_A(\frac{1}{x})$

I would delete if I could...

Ok, just try it with a diagonal matrix:
$$ A+Ix = diag(a_{11}+x,\ldots, a_{nn}+x)$$
then
$$ \det(A+Ix)=\prod_{i=1}^n (a_{ii}+x)$$
Then the derivative is for example if $n=2$
$$ a_{22}+a_{11}$$

Then you can generalize for other matrices.


Integrating the Geometric Mean of the Differentials I'm curious if anyone has seen any literature on the following.  I was playing around with differentials and integration and couldn't help but notice the interesting property of the following integral:

$$
\int{\sqrt{dxdy}}=\int{\sqrt{\frac{dy}{dx}}dx}
$$

A similar integral for functions of n-variables integrating the geometric mean of all the differentials can be written as well.

So the integral of the geometric mean of the two differentials is the integral over $x$ of the square root of the derivative of $y$ with respect to $x$ (or you can also flip $x$ and $y$ around.

In the general case,

$$
\int{(dx_1dx_2\cdots dx_n)^{\frac{1}{n}}}=\int{(\frac{dx_2}{dx_1}\cdots \frac{dx_n}{dx_1})^{\frac{1}{n}}dx_1}
$$

It just seems too elegant to not have an application. Any thoughts greatly appreciated!

I highly suspect this is just misleading notation. the right hand side of your first equation assumes there is a function y(x), such that you can compute it's square root, and proceed Riemann integration. I don't know any definition for the left hand side, so this would need to be done with some rigor before the equation you wrote could be proved, and any application found. You might have implicitly considered x=x(t) and y=y(t), such that:

$$
\sqrt{dxdy}=\sqrt{x'(t) dt y'(t) dt}=\sqrt{x'(t)y'(t)}\sqrt{dt dt}
$$

But even if it seems trivial, I wouldn't even claim:
$$
\sqrt{dt dt} = dt
$$
Because at the very least:
$$
\sqrt{dt dt} = |dt|
$$

But note that $dt$ is not a number, so the expression itself lacks rigor.


Calculator model with mod function? I'm wondering does anyone know of a scientific calculator with a mod function?

In C# this is shown as follows (just in case there are any other mods that a mathematical non-savant such as myself wouldn't be aware of)

5 % 4 = 1

I've looked through the most common ones on the web but the function is either not in the list or not present.

I have a little bit of a bias towards Casio calculators as that's what I'm familiar with but will accept any suggestions that can be currently purchased.

Derek

HP : buy a hewlett packard calculator. the model HP-35S contains hundreds of functions.
there is a description here ; http://www.shopping.hp.com/product/calculator/Scientific/1/storefronts/F2215AA%2523ABA;HHOJSID=FW7TNQVJMqHh075qvYbPbRvjvx2K1ybHFyYXGDY9Vr8kpqFlknwg!492383249

The 32Sii used to have a button just for that, on the other hand the
HP35s is programmable and you can most probably do it easily.

The TI-36x pro has a mod button under the Math menu.


If $f$ is a homeomorphism from an open subset $U$ of $R^n$ to a subset $V$ of $R^n$ If $f$ is a homeomorphism from an open subset $U$ of $R^n$ to a subset $V$ of $R^n$, is $V$ also open in $R^n$?

Yes, this is Brouwer's Invariance of Domain Theorem.

Edited to add (thanks commenters): if the domain of $f$ is only $U$ then you do in fact need the fancy theorem. I read the question too fast and assumed $f$ was a homeomorphism from $\Bbb{R}^2$ to itself. I will leave this answer because the comments on it are valuable.

It needs nothing as fancy as a big named theorem. A homeomorphism is a continuous bijection with continuous inverse. A function $f$ is continuous if and only if the preimage of an open set is always open. So you have these facts:

$f^{-1}$ is continuous, and

$V$ is the preimage of an open set $U$ under $f^{-1}$.

Thus $V$ is open.