Problems that differential geometry solves Recently, I've been studying a course in differential geometry.

Some keywords include (differentiable) manifold, atlas, (co)tangent space, vector field, integral curve, lie derivative, lie bracket, connections, riemannian geometry, symplectic geometry.

However, I've been wondering what problems in pure mathematics that are obvious and interesting can be solved with tools from differential geometry. In other words what questions could one ask that will motivate the study of differential geometry for someone who's interested in pure mathematics mainly.

Please don't limit yourself to merely stating a problem, but include a hint, full solution, or reference on how exactly differential geometry becomes useful.

Here are some possible answers with my own comment that may inspire you:

It's a great language to formulate problems in physics. That may be true, but unfortunately I don't care about pysics (enough).
It's a language to talk about differential equations, which are "naturally interesting". But honestly, I don't think I care about applications to differential equations, knowing that what it comes down to in the end is that the equations are rammed into a computer and brutally 'solved' by numerical methods anyway, no matter how fancy you formulate them.
Perelman's solution of the Poincaré conjecture, which may be considered a topic in pure mathematics, uses differential geometry. Apparently it does, but isn't that a bit like using Wiles' solution to FLT as a motivation for a course in commutative algebra?
It provides deeper insights in the whole of mathematics. Well, I just made that up. But maybe it provides a wealth of examples in, for instance, topology, or maybe techniques can be borrowed from differential geometry and used in other branches. But then again, doesn't it make more sense to study them where they become interesting?

As a final example, a simple question that I consider motivating for exploring groups beyond their definition would be: "how many groups of order 35 are there?": it's an easy question, only refering to one easy definition with a somwhat surprising answer where the surprise vanishes once you've developed a sufficient amount of theory.

ps - Since there is no best answer to my question maybe it should be community wiki. I'm sure some moderator will do what's appropriate.

pps - In reply to Thomas Rot's answer I must apologise for the tone when I'm talking about differential equations. Actually I'm a person who obtained a degree in applied physics before turning over to "pure" (in a sense that I don't mind if it's hard to find applications in real life situations) math. I've seen how these people solve differential equations -- I've seen how I used to do it Myself, actually. No cute mathematical theory, just discretize everything and put it into a computer. If it doesn't work out, let's try a finer grid, let's leave out a term or two, until it works.
Surprisingly they don't use cotangent spaces to even state the problem, still it appears sufficient to calculate the heat distribution in nuclear reactors, or calculate the electron density around a deuteron.
Because I've seen all this and didn't think it is pretty, I've turned away from it. But feel free to change my mind on the subject.

If you find the question: "How many groups are there of order 35?" motivating, why don't you find the question: "How many differentiable manifolds of dimension 2 are there?", motivating as well?

There are millions of applications of manifolds in pure mathematics. Lie groups (continuous symmetries) are a beautiful example.

As an aside, your question got downvoted (not by me), because the tone is somewhat arrogant. For example your statement:

"It's a language to talk about differential equations, which are "naturally interesting". But honestly, I don't think I care about applications to differential equations, knowing that what it comes down to in the end is that the equations are rammed into a computer and brutally 'solved' by numerical methods anyway, no matter how fancy you formulate them."

is an incredibly ill informed view of the subject. The theory of differential equations is extremely rich (both from a pure and applied viewpoint).

Differential geometry is a basic tool in mathematical physics, in particular in mechanics. I strongly recommend the lecture of the book of Arnold "Mathematical Methods of the Classical mechanics", and the book of Penrose "The Road to Reality: A Complete Guide to the Laws of the Universe". Enstein theory of relativity (what is space-time) cannot even be formulated without the basic concepts of diff geometry ; symplectic geometry is now a standard tool in Hamiltonian mechanics etc etc...


Proof for SSS Congruence? I'm hoping that someone can provide a method for deducing the commonly known SSS congruence postulate? The postulate states

  If the three sides of one triangle are pair-wise congruent to the three sides of another triangle, then the two triangles must be congruent.

I am a middle school math teacher (teaching a HS Geometry course) and would like to be able to explain/justify the triangle congruence theorems that I expect students to apply with more clarity. I understand that the Law of Cosines could be used to justify the SSS triangle congruence theorem but I wonder if a proof can use more basic properties. My students are currently learning the congruence theorems and using constructions to justify their validity. I know two ways to demonstrate the validity of SSS congruence: constructions and the Law of Cosines. Is there another way?

I've researched this and have not seen another answer so far. The Law of Cosines is the accepted answer to Proof of ASA , SAS , RHS , SSS congruency theorem. I've also seen a few other questions hinting at this without a clear answer, see: Why is SSS criterion for congruence of triangles referred to as "SSS postulate" in textbooks?. Another question on math.SE asks what I'm asking (how can we derive the triangle congruence postulates from basic axioms?) Need for triangle congruency axioms and the accepted answer suggests only a quick visit to http://en.wikipedia.org/wiki/Taxicab_geometry but does not discuss whether or not the triangle congruence axioms follow from the basic axioms or are independent of them.

Cut-the-Knot's SSS proof page has a number of solutions, including Euclid's. As the author indicates, however, only Hadamard's proof "goes through without a hitch", with the important aside: "assuming of course that isosceles triangles have been fairly treated previously".

I'll give a full development of Hadamard's argument, including the necessary bits needed about isosceles triangles. I include a couple of "obvious" sub-proofs just to make clear which axioms are in play.

Preliminaries:

SAS triangle congruence is an axiom.
(1) implies one direction of the Isosceles Triangle Theorem, namely: If two sides of a triangle are congruent, then the angles opposite those sides are congruent. $[\star]$
(2) implies that A point equidistant from distinct points $P$ and $Q$ lies on the perpendicular bisector of the $\overline{PQ}$. $[\star\star]$

Now, we'll cut to the part of Hadamard's argument where we have $\triangle ABC$ and $\triangle A^\prime BC$ with corresponding edges congruent, constructed with $A$ and $A^\prime$ on the same side of $\overleftrightarrow{BC}$. (That last part is key.) We need only show that $A$ and $A^\prime$ coincide to prove SSS. Hadamard makes this argument by contradiction ...

Assume $A \neq A^\prime$.
$B$ is equidistant from $A$ and $A^\prime$, and therefore, $B$ lies on the perpendicular bisector of $\overline{AA^\prime}$ by (3) above. The same is true of $C$.
Therefore, $\overleftrightarrow{BC}$ is the perpendicular bisector of $\overline{AA^\prime}$.
Therefore, $\overleftrightarrow{BC}$ separates $A$ and $A^\prime$.
This contradicts the fact that we constructed $A$ and $A^\prime$ to be on the same side of $\overleftrightarrow{BC^\prime}$. $[\star\star\star]$
Therefore, the assumption that $A \neq A^\prime$ must be invalid, so that triangles $\triangle ABC$ and $\triangle A^\prime BC$ coincide; this gives us SSS. $\square$

$[\star]$ Proof: Isosceles $\triangle ABC$ with base $\overline{BC}$ is congruent to $\triangle ACB$ by SAS, and those opposite angles are corresponding angles of the congruent triangles.

$[\star\star]$ The midpoint, $M$, of $\overline{PQ}$ is certainly on the perpendicular bisector. A point $R \neq M$ equidistant from $P$ and $Q$ creates isosceles $\triangle RPQ$ with congruent angles at $P$ and $Q$ by (2). Thus, $\triangle RPM \cong \triangle RQM$ by SAS. Corresponding angles $\angle RMP$ and $\angle RMQ$ must be congruent; as supplements, they must be right. $\overleftrightarrow{MR}$, then, is the perpendicular bisector of $\overline{PQ}$. (I'll note that an easier proof could assert that, from the get-go, $\triangle RPM \cong RQM$ by SSS ... but we can't use that argument in a proof of SSS itself.)

$[\star\star\star]$ The contradiction here is one of Euclid's oversights later made explicit by Hilbert as the Plane Separation Axiom. The PSA has slightly-different formulations in different treatments of geometric foundations, but effectively it's the "Because I said so!" assertion that lines break planes into two disjoint "sides" and gives us our contradiction.

Proof of SSS is flawed. Once construct is drawn, the implicit assumption is that A and A' are at the same position, i.e, (A = A'). This is essentially assuming what you want to prove. (A and A' could be on the same side of line BC but their positions could be different).

Using the law of cosines may be possible but it is a bit ridiculous since first one needs to prove the Pythagorean theorem.


Prove: $\sin(\theta) - \sin(\theta)\cos^2(\theta) = \sin^3(\theta)$. Help please? I'm lost. Prove:  $\sin(\theta) - \sin(\theta)\cos^2(\theta) = \sin^3(\theta)$.

Can someone show me how to prove this? Do I use the Pythagorean Identity to prove it?

I will be forever grateful. Thanks!

Do I factor $\sin(\theta)$ out of both terms on the left side of the equal sign
and then divide both sides by $\sin(\theta)$?

Rewrite
$$\sin\theta-\sin\theta\cos^2\theta=\sin\theta(1-\cos^2\theta)$$
Remember that $1-\cos^2\theta=\sin^2\theta$ and then plug it in:
$$
\sin\theta-\sin\theta\cos^2\theta=\sin\theta(1-\cos^2\theta)=
\sin\theta\cdot\sin^2\theta=\sin^3\theta.
$$

Note: there's no need to divide by $\sin\theta$ which is a very dangerous thing to do, because you don't know whether it's zero or not. In this particular case it makes no real difference, because if $\sin\theta=0$ the identity clearly holds. It could be in different situations.

sin(θ)−sin(θ)cos2(θ)=sin3(θ)

First, subtract sin(θ) from both sides.

-sin(θ)*cos2(θ)=sin3(θ)-sin(θ)

Now, divide both sides by -sin(θ). Mind the - sign.

Cos2(θ)= [sin3(θ)-sin(θ)]/-sin(θ)

Factor the top by grouping. Pull out a -sin(θ) from both sides. Mind the signs.

Cos2(θ)=-sin(θ)(-sin2(θ)+1)-sin(θ)

Cancel out -sin(θ) from the top and bottom

cos2(θ)=-sin2(θ)+1

Move the 1

cos2(θ)=1-sin2(θ)

Take the square root of both sides

Cos(θ)= +-√1-sin2(θ)

Bam. Trig identity!


Have historians responded to Raju's critique? C. K. Raju has made some outrageous criticisms of the traditional take on Euclid in particular and Western history in general. Yet he has a book published on the subject with an apparently respectable publisher in India. Have modern historians of the classical period responded to his critique?

Note. One of the responders mentioned a helpful review by Ferreiros here.

We can see:

José Ferreirós' review of C.K. Raju, Cultural Foundations of Mathematics: The Nature of Mathematical Proof and the Transmission of the Calculus from India to Europe in the 16th c. CE, in Philosophia Mathematica Volume 17, Issue 3:

  In his interest to revise traditional historiography and oppose proofcentred mathematics, Raju devotes a lot of effort to questioning the existence
  of Euclid and insisting that the text of the Elements originates at the
  earliest in 370 CE (with Theon) or perhaps even in the tenth century. In
  my opinion, this is useless and does not help advance the author’s main theses. For historical purposes, what is relevant is that Elements represents
  a systematisation of a large portion of geometrical knowledge in the Greekspeaking world before the common era. (‘Euclid’ is simply the name of
  its otherwise unknown author, whose dates—it is true—are dubious; incidentally,
  an interesting question would be whether philologists find reason
  to think that the text of Elements was written by different authors.) Raju
  insists on the idea that Proclus’s views represent the original philosophy
  of mathematics in the Elements (p. 25), and he overemphasizes the connections
  between geometrical proof, Platonism, and Christian religion.

Regarding specifically the possibility that Euclid "the man" was a forgery is totally irrelevant: we know quite nothing about him, but we have "the book": Elements, and its value does not change if it was written by someone else (but in any case by some ancient Greek: we have testimonies...).

If the Elements has been "falsified", this fact does not change of a iota the meaning of its content: the theorems. It is not the same thing as saying that the Holy Bible has not been written by some ancient prophets and in reality is a late Hellenistic forgery.

Regarding in general the approach to history (and history of science as well), my humble opinion is that history is like any other science: conjectures and hypotheses are fundamental, and new insight and a fresh point of view can be useful and necessary, but there is a golden rule: all must be checked with facts.

Euclid is no myth.Applonius of Perga directly refers to him in one of his books on conic sections ( In book 3, an original version of which has survived,in Greek).
I have written an article titled'C.K.Rajus mistake' which is on the net exposing all his false arguments.You can read it.I have introduced a lot of new arguments favouring , an Indian origin to Mathematics, and in the course have introduced some Indian philosophical thoughts,which have influenced greek mathematicians starting from Pythogoras.Independent of whether one agrees to my theory,nevertheless, I have busted Raju's false arguments of Euclid being a myth and many of his arguments against Mathematics(which he calls as formal Mathematics).
My paper was published in the 'Mathematics Teacher', the official publication of the society of Mathematics Teachers of India,located in Chennai,India.
The name of the article is 'Stop the Tirade against Euclid.Why Euclid should be more emphasized in Schools than is currently done'I changed its title to'C.K.Rajus mistake' while posting it on the net,to catch the attention of people who may be following him.
Here is the link to the article.
http://www.vixra.org/abs/1504.0198.
Regards.


Linear Algebra Subspace question Here is the following question and answer for the question.
I don't seem to quite grasp the answers or how the answers are what they are.

Requirements for subspace:

the zero vector is in the subset
if you add 2 elements of the subset they remain in the subset
if you multiply any element in the subset by a constant than this is in the subset

a) Since b1 = b2, doesn't that mean that there are effectively only two vectors that define the subspace -> meaning it creates a plane in R3. Since there are no limitations all three conditions are met. Anything I missed?

b) The plane of vectors? What does that even mean. Since there is only one vector does this not only span a line?

c) All three vectors, b1, b2 and b3 must be 0 if they are to result in 0. That fulfills the zero vector requirement. b1+b2 = 0 and b1b3 = 0 (zero vector), is this not in R3?

d) Scalar multiplier of 0 fulfills the zero vector requirement.  Then I'm lost.

e) Isn't this an example of linearly independent set of vectors?

f) Lost completely.

Anyone give a light note of what they all mean! I'm lost in what they mean basically.

I show the steps for the (a).
Given the plane $(b_1,b_2,b_3)$ with $b_1=b_2$

First, zero vector is contained in the set since $b_1=b_2=0$

Next, suppose there are two vectors fulfilling the equation of plane given by
$$u=(x_1,x_2,x_3) \text{ and } v=(y_1,y_2,y_3) $$
such that $x_1=x_2$ and $y_1=y_2$

Now,
$$u+v=(x_1+y_1,x_2+y_2,x_3+y_3)$$
$u+v$ is contained in the set since $x_1+y_1=x_2+y_2$

Next,
suppose $r\in F$
$$ru=(rx_1,rx_2,rx_3)$$
$ru$ is contained in the set since $rx_1=rx_2$

Since these three conditions is satisfied, so the plane is subspace.

(a) Consider the plane "z = 1" in $R^3$. It's not a subspace, but your argument ("there are no limitations") applies to it just as well as the one in the problem. So your argument cannot be sufficient.

(b) The set of all vectors of the form $(1, p, q)$, where $p$ and $q$ are real numbers, is a plane in 3-space. (You might call it the "x = 1" plane). They're asking if it's a subspace of $R^3$. It's not, because the vector $(0,0,0)$ is not in it.

(c) Once again your reasoning is wrong: any ONE of the three coordinates can be zero and still have $b_1 b_2 b_3 = 0$.

I worry that you may be in the wrong course, since the problems you're having are all with understanding definitions clearly, and with elementary algebra, not with linear algebra.


improper integral and dominated convergence Suppose $f:(a,b] \to \mathbb R$ is a continuous function and $\lim_{x\to a+} f(x)=+\infty$

Suppose $f(x)\leq g(x)$ for all $x\in(a,b]$, for some $g：[a,b]\to \mathbb R$ satisfying  $\int_a^b g(x)dx<+\infty$

Can we conclude that

$\lim_{t\to a^+}\int_t^b f(x)dx \leq \int_a^b g(x)dx$  ?

Thanks.

Yes, we can. First, for sufficiently small $h>0$, we have
$$\int_{a+h}^bf(x)\,dx\leq\int_{a+h}^bg(x)\,dx$$
and since the r.h.s tends to a limit as $h\to 0$, we conclude that the l.h.s is
bounded. Moreover, since $f(x)\to\infty$ as $x\to a^+$, we can choose $h>0$
so that $f(x)$ is positive in $(a,a+h)$, and in particular, the integral $\int_{a+h}^bf(x)\,dx$ increases as $h$ decreases to zero. Thus, the function
$h\to\int_{a+h}^bf(x)\,dx$ is monotonically increasing and bounded as $h\to 0^{+}$, hence the limit exists, and obviously is not larger than the integral of $g$.


Proof convergence of random variables (almost sure) Let $\left\{ X_{n}\right\} _{n\in\mathbb{N}}$ be a sequence of independent random variables. Prove that $X_{n}\overset{a.e.}{\rightarrow}0$ if and only if, for all $\epsilon>0$, $\sum_{n=1}^{\infty}\textrm{P}\left(\left|X_{n}\right|>\epsilon\right)<\infty$.

I guess I have to use the Borel Cantelli Lemma, right? Thanks.

Important inequalities

Williams - Probability with Martingales

Deduced similarly:

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

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

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

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

'if'

Suppose $\forall \epsilon>0$,

$$\sum_{n=1}^{\infty}\textrm{P}\left(\left|X_{n}\right|>\epsilon\right)<\infty$$

By BCL1, we have

$$\textrm{P}\left(\limsup [\left|X_{n}\right|>\epsilon\right]) = 0$$

$$\to \textrm{P}\left(\liminf [\left|X_{n}\right|\le\epsilon\right]) = 1$$

$$\to \textrm{P}\left([\liminf \left|X_{n}\right|]\le\epsilon\right) = 1 \ \text{and} \ \textrm{P}\left([\limsup \left|X_{n}\right|]\le\epsilon\right) = 1$$

$$\to \textrm{P}\left([\liminf \left|X_{n}\right|]=0\right) = 1 \ \text{and} \ \textrm{P}\left([\limsup \left|X_{n}\right|]=0\right) = 1$$

$$\to \textrm{P}\left([\lim \left|X_{n}\right|]=0\right) = 1$$

$$\to \textrm{P}\left([\left|\lim  X_{n}\right|]=0\right) = 1$$

$$\to \textrm{P}\left([\lim X_{n}]=0\right) = 1$$

'only if'

Suppose $\exists \epsilon_0 > 0$ s.t.

$$\sum_{n=1}^{\infty}\textrm{P}\left(\left|X_{n}\right|>\epsilon_0\right)=\infty$$

By BCL2, we have

$$\textrm{P}\left(\limsup [\left|X_{n}\right|>\epsilon_0\right]) = 1$$

$$\to \textrm{P}\left([\limsup \left|X_{n}\right|]>\epsilon_0\right) = 1$$

$$\to \textrm{P}\left([\limsup \left|X_{n}\right|]\le\epsilon_0\right) = 0$$

$$\to \textrm{P}\left(\liminf \left[|X_{n}\right|\le\epsilon_0]\right) = 0$$

However since

$$\textrm{P}\left([\lim X_{n}]=0\right) = 1,$$

we have

$$\textrm{P}\left([\limsup \left|X_{n}\right|]=0\right) = 1$$

$$\to \forall \epsilon > 0, \textrm{P}\left([\limsup \left|X_{n}\right|]<\epsilon\right) = 1$$

$$\to \forall \epsilon > 0, \textrm{P}\left(\liminf \left[|X_{n}\right|< \epsilon]\right) = 1$$

Choose $\epsilon = \epsilon_0$ to get:

$$\textrm{P}\left(\liminf \left|[X_{n}\right|< \epsilon_0]\right) = 1 ↯$$

$\therefore, \forall \epsilon>0, \sum_{n=1}^{\infty}\textrm{P}\left(\left|X_{n}\right|>\epsilon\right)<\infty$ QED


When can we "multiply through" by the kernel of a homomorphism? Let $G/H=K$. I think I am right in saying that if $K\cong K'\trianglelefteq G$ and is such that $K'H=G$ and $K'\cap H=1$ then we have $H\times K'=G$. If we relax the normality requirement of $K'$ to simply $K'\le G$ we have $H\rtimes K'=G$. I think there is no need for Zappa-Szep products here.

Now there will be situations where I don't know whether the image is isomorphic to a subgroup or a normal subgroup, or if $K'\cap H=1$, or if $K'H=G$.

It has been pointed out in the comments that there is no general result for finding out the information about the image and kernel required to make a direct or semidirect product. What are some special cases? Finite abelian groups have been mentioned.

In general $G/H$ is not isomorphic to a subgroup of $G$.

For example, $G = (\mathbb Z,+)$ the group of integers, $H = (2\mathbb Z,+)$ the subgroup of even integers.  The quotient  $G/H$ has order $2$, so it cannot be a subroup of $G$.


Induction Proof: Round Robin In a round-robin tournament, each team plays every other team exactly once. Show that if no games end in ties, then no matter what the outcomes of the games, there will be some way to number the teams so that team 1 beat team 2, and team 2 beat team 3, and team 3 beat team 4,
and so on.

I have the base case of two teams, call them Team A and Team B. Whichever team wins, will be renamed Team 1 and whichever team loses will be renamed Team 2 so that Team 1 beats Team 2. I know that I need to make an inductive hypothesis about having $n$ teams, so that I can prove it for $n+1$ teams, but I have no idea how to go about it.

Consider a round-robin tournament with $n + 1$ teams. Arbitrarily pick a particular team, say team $x$. Then by the definition of a round-robin tournament with no ties, observe that we can partition the remaining $n$ teams into two sets $W$ and $L$ of size $a$ and $b$ respectively, where $W$ is the set of all winners who beat team $x$ and $L$ is the set of all losers that team $x$ managed to beat (so that $a + b = n$). Then since $a,b \leq n$, it follows by the inductive hypothesis that we can relabel the teams so that $W = \{w_1, w_2, \ldots, w_a\}$ and $L = \{\ell_1, \ell_2, \ldots, \ell_b\}$, where:
$$
w_1 \xrightarrow{\text{beats}} w_2 \xrightarrow{\text{beats}} \cdots \xrightarrow{\text{beats}} w_a
$$
$$
\ell_1 \xrightarrow{\text{beats}} \ell_2 \xrightarrow{\text{beats}} \cdots \xrightarrow{\text{beats}} \ell_b
$$
But then we can order all $n+1$ teams as follows:
$$
w_1 \xrightarrow{\text{beats}} w_2 \xrightarrow{\text{beats}} \cdots \xrightarrow{\text{beats}} w_a \xrightarrow{\text{beats}} x \xrightarrow{\text{beats}}  \ell_1 \xrightarrow{\text{beats}} \ell_2 \xrightarrow{\text{beats}} \cdots \xrightarrow{\text{beats}} \ell_b
$$

A simple inductive proof. Easy for 2 teams. Assume you can do it for n team and consider n+1 team. Set aside Team "X". Arrange the n teams so that team k--> k+1. If X beats 1 then X --> 1 -->2...-->n. If not look for the first team j that X beats. Then:
1-->2-->...j-1 -->X -->j -->...n is your "inductive answer."


Use mathematical induction to prove that for all integers $n \ge 2$, $2^{3n}-1$ is not prime I had a homework due yesterday with this problem.

The TA did the problem last week in discussion but I didn't understand it.

She pulled out a $7k$ almost immediately, and I have no idea from where.

It was like, it wasn't prime if $2^{3n}-1$ was 7 times some constant. I understand that that would make $2^{3n}-1$ not prime, but I don't understand how she just used "7".

Where did she get that from? I was thinking, maybe it was like $2^3 \cdot 2^n-1$... which is $8 \cdot 2^n-1$... but you can't just do $8-1$. How?

I’m sure that she used the following factorization

$$\begin{align*}
2^{3n}-1&=(2^3)^n-1\\
&=8^n-1^n\\
&=(8-1)(8^{n-1}\cdot 1^0+8^{n-2}\cdot 1^1+\dots+8^2\cdot 1^{n-3}+8^1\cdot 1^{n-2}+8^0\cdot 1^{n-1}\\
&=7(8^{n-1}+8^{n-2}+\dots+8^2+8+1)\\
&=7\sum_{k=0}^{n-1}8^k\;,
\end{align*}$$

which as Dylan Moreland pointed out is a special case of $$\begin{align*}a^n-b^n&=(a-b)(a^{n-1}b^0+a^{n-1}b^1+\dots+a^1b^{n-2}+a^0b^{n-1})\\
&=(a-b)\sum_{k=0}^{n-1}a^kb^{n-1-k}\end{align*}\;.$$

This is really just the formula for the sum of a finite geometric series in disguise:

$$\begin{align*}
\sum_{k=0}^{n-1}a^kb^{n-1-k}&=b^{n-1}\sum_{k=0}^{n-1}\left(\frac{a}b\right)^k\\
&=b^{n-1}\frac{\left(\frac{a}b\right)^n-1}{\frac{a}b-1}\\
&=\frac{\frac{a^n}b-b^{n-1}}{\frac{a-b}b}\\
&=\frac{a^n-b^n}{a-b}\;,
\end{align*}$$

and multiplying through by $a-b$ yields the factorization formula.


Proof of simple interest formula Can someone please prove to me that $I = PRT$, where $P$ is the principal, $R$ is the interest rate, and $T$ is the number of years/time.  I have seen $I = P(1+TR) = P+PTR$ which does not equal $PRT$, so I am slightly confused.  Any help is appreciated,
Thanks!

You have for simple interest at a fixed interest rate per time period $R$:

$$I=\sum_{i=1}^{T}PR=PRT$$

Where $I$ is the total interest after $T$ time periods. Therefore your other formula should read:

$$P(T)=P(0)(1+RT)$$

Where $P(T)$ is the principle after $T$ time periods.

For 100 we get $r$,then for $p$ we get $(pr/100)$,first time, same  thing we will get after 2nd year similarly 3rd, therefore

s.i after $t$ time=$(pr/100)+(pr/100)+......................(pr/100)=(prt/100)$


Which is larger? $20!$ or $2^{40}$? Someone asked me this question, and it bothers the hell out of me that I can't prove either way.

I've sort of come to the conclusion that 20! must be larger, because it has 36 prime factors, some of which are significantly larger than 2, whereas $2^{40}$ has only factors of 2.

Is there a way for me to formulate a proper, definitive answer from this?

Thanks in advance for any tips. I'm not really a huge proof-monster.

It is probably easier to note that $2^{40} = 4^{20}$. The only ones of the 20 factors in $20!$ that are smaller than $4$ are $1$, $2$ and $3$. But, on the other hand, $18$, $19$ and $20$ are all larger than $4^2$, so we can see
$$ 20! = 1\cdot 2\cdot3\cdots 18\cdot19\cdot 20 > \underbrace{1\cdot1\cdot1}_{3\text{ ones}}\cdot\underbrace{4\cdot4\cdots 4\cdot 4}_{14\text{ fours}}\cdot\underbrace{16\cdot16\cdot 16}_{3\text{ sixteens}} = 2^{40} $$
by comparing factor by factor.

With Stirling's formula:

$$ 4^n \lt \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n \le n! \le e \sqrt{n} \left(\frac{n}{e}\right)^n $$

Both $\le$ are valid for $n\ge1$. Because $e\le 3$ the left inequality is valid a least for all $n \gt 12$.


Prove an inner product space Prove $(L^2[0,1],\|·\|_g)$ is an inner product space, where the norm of $\|·\|_g$ is given by $\|f\|_g=(\int_0^1 g(x)|f(x)|^2dx)^{\frac{1}{2}}$. I know that I need to prove this through proving it satisfies the four properties of inner product space, I just don't know how to express the inner product from the norm. (Consider it in the real space.)

Is $\langle f,g \rangle=\int_0^1 g(x)|f(x)||g(x)|dx$ the corresponding inner product?

The inner product is determined by the norm via the polarisation identity.

$\langle f, h \rangle = {1 \over 4} ( \|f+h\|^2 - \|f-h\|^2)$.

Hence
$\langle f, h \rangle = {1 \over 4} ( \int g \cdot (f+h)^2 -   \int g \cdot (f-h)^2 ) = \int g \cdot f \cdot h $.

The one you have defined is not an inner product because of the absolute values for $f$ and $g$. The absolute value sign destroys linearity.

You have to assume that $g \geq 0$ a.e. for this to be a norm. The inner product is given by $ \langle f, \phi \rangle =\int_0^{1} g(x)  f(x) \phi (x) dx$.

[If you  are considering complex valued functions change $\phi (x)$ to $\overline {\phi} (x)$].


Why is the construction of the real numbers important? There are a lot of books, specially in Real Analysis and set theory, which define the real numbers by Cauchy sequences or Dedekind cuts. So my question is why don't we simply define the Real numbers as a complete ordered field?

What's the importance of studying the construction of the Real numbers? Is it just for historical reasons?

First of all, mathematics is based on intuition and on concrete (imaginary, but concrete) objects, often inspired by reality. Let's stop the formalities for a moment and speak freely: we don't think of $3/4+1/2$ as an operation involving equivalence classes of ordered pairs of Von Neumann integers, we think of it as pouring $3/4$ liters of water and $1/2$ liters of water into a bowl.

With that said, if you're going to be doing reasoning on the basis of intuitive objects like fractions and integers, you need somewhere to start. I'm not even really talking about axioms, I just mean that you need to accept that certain things are reasonable enough to just accept that you understand how they work without having to analyze any further. I accept whole numbers as a basic object of mathematics, I don't need to ask what a whole number is or what it means to add whole numbers. I suppose all I'm saying is that in mathematics, we need undefined objects.

Now, I'm sure you're willing to accept integers, integer addition and whatnot as undefined objects. Probably you accept rational numbers as well - you know what I mean when I talk about "chopping $3$ things into $4$ equal pieces", and what it would mean to "add" two such quantities.

So, so far we're agreed that accepting rational numbers as basic objects without further analysis is philosophically tolerable. Well, I'm sure you've seen the proof that the square root of $2$ is irrational. But check it again - that's not what it proves. What it proves is that there is no rational number which squares to $2$. It doesn't prove that there is such a thing as irrational numbers or that there's any object in the universe worth calling a square root of $2$. In fact, if so far we've accepted rational numbers into our menagerie of philosophically coherent objects, then there's no reason at all why we shouldn't simply stop here and say "well, clearly there's no square root of $2$". Let's be honest, the reason why most students feel so strongly otherwise is because they've never gotten ERR when they typed SQRT(2) into a calculator - it's accepted on the basis that an authority figure told them so, and the philosophy of it all goes unquestioned. But there really is no reason to panic and postulate a bigger number system just because there's no number whose square is $2$, there's also no number whose square is negative, but that didn't bother anyone until they started doing relatively advanced algebra.

But hold on, there is a way to rescue poor $\sqrt 2$. Let's say that you needed a number whose square was $2$ - maybe you needed to draw a square whose area was $2$. Well then you wouldn't be bothered by my ridiculous metaphysical musings, you'd just pick a rational number whose square was close to $2$, say $1$cm and $4$mm, and draw the square that way. Now, if we want better and better approximations - number whose squares are closer an closer to $2$ - we would find that these numbers "converge" to a sort of "ideal point". For example, given any $\epsilon$, I can find an interval of width $\epsilon$ in which all approximations beyond a certain precision must lie. What's more, these intervals will nest in one another as $\epsilon$ decreases, so they really are "tightening" around a specified point.

It's tempting to call this "point" a number, but that's unacceptably vague for a mathematician. What is this "point", for one thing? Certainly not a rational number (the only kind of number we understand, so far). One approach is to say that the only reason we believe this "point" exists is because we have a sort of oracle that tells us, given any rational, whether it's "too big" (it's square is bigger than $2$) or "too small". It's by the use of this oracle that we can find better and better approximate square roots of $2$: make a guess, then make it a little bigger or a little smaller depending on if its square is smaller or bigger than $2$. Thus we might say that any time we have such an "oracle", we can claim to have found one of these mysterious idealized "numbers", which can be approached but never precisely given a value. This is essentially a Dedekind cut.

Hopefully I've convinced you of two things:

There is no obvious reason (other than constant mindless drilling in mandatory education with symbols like "$\sqrt\cdot$" and "$\dots$") to suspect anything resembling the real numbers exists or is worth discussing.
Nevertheless, with some reflection such a reason can be found (otherwise the real numbers would never have been developed, of course!), and it leads very naturally to the various constructions of the reals.

Now, if you accept (1), then the answer to your question is simple. If it's not obvious that the reals exist, then postulating their existence is absurd. Even if you're okay with the idea of just writing down and investigating some random axioms, there would be no reason at all to suspect these axioms describe the real world, or anything interesting at all, in any useful way. That last point is crucial. The constructions of the real numbers are the only reason to think that real numbers bear any relation to reality at all. Proving that a complete ordered field exists in ZFC, as pointed out in the other answers, is neat, but it's not really the most important reason to construct the real numbers, since merely existing doesn't imply that a structure is interesting or is a valid model for real-world quantity.

And consider (2). As we saw above (in a very summarized way), by the time you've given the question enough thought to convince yourself that there's any such thing as a real number, you're half way to rigorously constructing the real numbers anyway, so you may as well finish the job.

There is a calculus textbook, "Mathematical Analysis I" by V. Zorich, that starts with $\mathbb R$ and its properties taken for granted. It then derives $\mathbb N$ and $\mathbb Q$ from $\mathbb R$, and proceeds to actual calculus.

Usual construction of $\mathbb R$ from $\mathbb Q$ is given as a supplemental material, if I recall correctly. So, in a sense, such a construction is not actually crucially important.


matlab problem - removing frequencies after FFT, signal processing I want to stress that this is not a coding problem, my problem is that i don't fully understand the mathematics surrounding the subject and that's why I believe I have a problem.

I was given an assignment to choose a song (I chose the song Stricken by disturbed), perform fast fourier transform on the song, this will shift us from time domain to frequency domain.

Now I was asked to remove the first $10000$ frequencies. if $f(t)$ describes the signal of the song, and $\hat f(u)$ describes the fourier transform of the signal, then define $\hat g(u)$ such that $g(u)=0$ for all $u \in [0,10000]$ and $\hat g(u)=\hat f(u)$ otherwise.

After I removed the first $10000$ frequencies, I was asked to do something else that the teacher kept hidden from us and wanted us to figure out ourselves what we need to do, then perform an inverse fourier transform on $\hat g(u)$ to get $g(t)$. If the hidden action was performed correctly, than $g(t)$ should be strictly real.

So the goal of this exercise is to find out what the hidden action is.

Here is what I think:

The problem with removing the first $10000$ frequencies is that the magnitude is no longer even.

We know for sure, that if $f(t)$ is strictly real, then $|\hat f(u)|$ is even.
For example, this is a graph of the magnitude of the fourier transform as a function of frequency, of the song i sampled

Now if I change the first $10000$ values of the fourier transform, the magnitude of transform as function of frequency looks like

You can clearly see that now the magnitude is not an even function. So there is no way that the inverse transform will be a real function.

However, if we now also remove the last $10000$ frequencies, the magnitude should be an even function.

And the graph shows that it is. However, when I calculate the inverse transform, it is not a real function. So something is wrong or missing.

Here is the code that I wrote:

As you can see, after invoking the inverse fourier transform, i get a function that is not real. Why? What am I missing?

You are actually zeroing out the last $10001$ samples of the FT: length(490000:500000) = 100001. However, that's not the only problem. Remember that the first sample in the FFT is DC and that there is only one value for it. Thus, you want to zero out the last $9999$ samples of the FFT:

N = 500e3;
Nz = 10e3;
x = rand( N, 1 );
X = fft( x );
Xz = X;
Xz(1:Nz) = 0;
Xz(end-Nz+2:end) = 0;
xz = ifft( Xz );
whos xz
  Name           Size              Bytes  Class     Attributes

  xz        500000x1             4000000  double

sum( imag( xz(:) ) )

ans =

     0

The trick is only a complex conjugate spectrum will give back real values after doing inverse fft.

You need to arrange your frequency using fftshift in matlab, then you can see the maximum peak will be starting frequency. try to make zero on either side. it will give real values after ifft.

Note: make sure your signal length is a $2^n$, if not padded it zeros to make the length $2^n$.

Ex: if signal length is $1000$ make it $1024$ ($2^{10}$) by padding $24$ zeros at end.


Can we define a joint probability distribution over different sample spaces / probability spaces? I am trying to understand the concept of a joint probability distribution. According to the definition of a (say) bivariate probability distribution, the two random variables $X$ and $Y$ should be defined on the same probability space.
http://en.wikipedia.org/wiki/Joint_probability_distribution

Are there such things as joint probability distributions over different sample spaces or different probability space?

Of course you can. The joint probability has Cartesian product of each own's sample space as joint sample space and product $\sigma-$algebra as joint events space. The joint probability, obviously, is the product of probabilities of $X$ and $Y$.


What useful properties does usual summation have, but alternatives do not? (Cesaro, etc) Before I really ask my question, I want to give my train of reasoning. Suppose we have some method of summation (as I understand, assigning a number to a series) that satisfies some or all of regularity, linearity, or stability, as defined in Hagen von Eitzen's answer to this question.

Suppose this method also assigns $1-1+1-1+\dots=1/2$---say it's Cesaro summation, or whatever. Evidently this method cannot admit rebracketing, because $(1-1)+(1-1)+\dots=0$ while $1+(-1+1)+(-1+1)\dots = 1$.

However, we can rebracket series that converge in the usual sense, even though they do not give the same results after rearrangement. Right? I think the limits of the partial sums $S_N = \sum^N_{k=1} a_k$ and $S_M = \sum_{k=1}^M(a_{2k-1}+a_{2k})$ must be the same---suppose that $S_N$ converges, and then evaluating for finite values of $N$ and $M$ shows that $S_M$ is a subsequence of $S_N$, so converges to the same limit. But $S_M$ is a rebracketing of $S_N$.

Therefore if a series converges, we enjoy the power to rebracket, which we do not have with Cesaro summation. However, in both cases, the axioms show we can still sum term by term, etc, so Cesaro summation keeps some nice things. What things doesn't it keep? Rebracketing appears to be an example.

So: What does a convergent series grant us that a series that is simply Cesaro summable (or otherwise) does not?

===============

I had some additional questions, which I found answers to here: Can we show that $1+2+3+\dotsb=-\frac{1}{12}$ using only stability or linearity, not both, and without regularizing or specifying a summation method?, but I'm leaving this paragraph anyway because comments refer to it.

As I was reading this link: http://www.nottingham.ac.uk/~ppzap4/response.html, where a physicist explains why some steps in a simple derivation he made were behind-the-scenes justified, one thing he says is that his manipulations do not contradict the three axioms, given in Hardy's Divergent Series. They seem to encapsulate linearity and stability, but not regularity.

if $\sum a_n = A$ then $\sum ka_n = kA$
if $\sum a_n = A$ and $\sum b_n = B$ then $\sum (a_n + b_n) = A+B$
$\sum\limits_{n=0}a_n = A$ if and only if $\sum\limits_{n=1} a_n = A - a_0$

I'm sure that from axiom 3 (stability) we can obtain that if $\sum a_n = A$, then $\sum c_n = A$, where $\{c_n\}$ is just $\{a_n\}$, but with maybe infinitely many $0$'s thrown between the $a$'s. For instance $c_1 = a_1$, $c_2 = a_2$, but $c_i=0$ for $3 \leq i \leq 7$, and then $c_8 = a_3$, and so on.

Assume $S_1 = 1-2+3-4+\dots = 1/4$ in whatever sense we consider.
Let $Z = 1+2+3+4+\dots$. It is not summable by any linear/stable method. But he writes

$$S_1 + 4Z = (1+0) + (-2 + 4\cdot1) + (3+0) + (-4 + 4\cdot2)=1+2+3+4+\dots=Z$$
$$-3Z = S \Rightarrow Z = -1/12$$

which requires stability. I suppose obtaining a "correct" value was a coincidence, then. He claims in the link they do not contradict the axioms, but the axioms don't hold in the first place.

The answer to your question has to do with topology. The notion of limit coming from topology as applied to the standard topology of the real line coincides with the more elementary $\epsilon$-$\delta$ definition. Hence all of the theory applicable to the notion of limit of a sequence in a topological space is applicable to the familiar notion of limit defined by $\epsilon$-$\delta$ definition of limit.

You may find learning about topology and limits in topological spaces interesting, and even more so, combining that with knowledge from order theory about orderings of a countably infinite set.

Edit:
To expand upon this, a topology for a set $X$ is a collection of subsets of that set, $\tau$, which are defined to be open such that:

1) $\emptyset,X\in\tau$, or in other words, the empty set and the entire set are in the topology.

2) Given any collection of open sets, $U_i\in\tau,i\in I$, the union $\bigcup_{i\in I}U_i\in\tau$. (Closed under arbitrary unions)

3) Given any finite collection of open sets, $U_n\in\tau$, $n\in\mathbb{N}_N$, the intersection, $\bigcap_{n\in \mathbb{N}_N}U_n\in\tau$. (Closed under finite intersections)

The pair $(X,\tau)$ is called a topological space, $\tau$ is the topology of the space, and $X$ the point set.

Because most topologies or interest are infinite and usually uncountably infinite, it is important to have ways of describing a topology in simpler terms. This leads one to consider a basis for the topology, as well as the related idea of a subbasis.

A basis for a topology, $\tau$ is a subset of the topology, $\mathcal{B}\subset\tau$, such that:

$\forall V\in\tau,\exists U_i\in\mathcal{B}$ such that $\bigcup_{i\in I}U_i=V$.

The point is: every open set in the topology $\tau$ can be represented as a union of open sets from the basis $\mathcal{B}$. The notion of subbasis takes this idea further:

A subbasis for a topology $\tau$ is a subset of $\mathcal{S}\subset\tau$ such that:

$\forall V\in\tau,\exists U_{(i,n_i)}\in\mathcal{S}$ such that $\bigcup_{i\in I}\bigcap_{n_i\in\mathbb{N}_{N_i}}U_{(i,n_i)}=V$

The point being that every open set of in the topology $\tau$ can be represented as a union of finite intersections of subbasis elements. The collection created from a subbasis, $\mathcal{S}$ by taking intersections only is the corresponding basis for that topology $\mathcal{B}$ (note: more than one basis may exist for a given topology, and likewise, more than one subbasis may exist for a given basis).

To give a familiar concrete example of a topology via a basis. We consider the point set $X=\mathbb{R}$, the real numbers. The standard topology for the real numbers has basis:

$\mathcal{B}=\{(a,b)|a<b;a,b\in\mathbb{R}\}\cup\{(-\infty,b)|b\in\mathbb{R}\}\cup\{(a,\infty)|a\in\mathbb{R}\}$

Aka, the set of all open intervals is the basis for the standard topology of the reals.

The familiar definition of limit of a sequence says that a sequence $(a_n)_{n\in\mathbb{N}}$ converges to a point $x$ if and only if $\forall\epsilon>0$, $\exists N\in\mathbb{N}$ such that $\forall n>N$, $|x-a_n|<\epsilon$.

Rephrasing the above in terms of open intervals containing $x$, aka open neighborhoods of $x$. The above definition says that the sequence $(a_n)_{n\in\mathbb{N}}$ converges to the point $x$ if and only if the sequence is eventually entirely within that neighborhood. Explicitly, $(a_n)_{n\in\mathbb{N}}$ converges to the point $x$ if and only if

$\forall(a,b)\in\mathcal{B}$ such that $x\in(a,b)$, $\exists N\in\mathbb{N}$, such that, $\forall n>N$ $a_n\in(a,b)$.

The topological definition of limit of a sequence generalizes this to contexts where there is no analog of intervals, and in a way that is independent of the basis:

In a topological space, $(X,\tau)$ a sequence of points, $(x_n)_{n\in\mathbb{N}}$, $x_n\in X$, is said to converge to a point $x\in X$ if:

For every open set $U$ containing $x$, there exists $N\in\mathbb{N}$ such that for every $n>N$, $x_n\in U$.

This is actually equivalent to the above description involving intervals in the reals, but is applicable in any topological space. Hence, with this definition, the notion of limit of a sequence is topological. Or in other words, depends solely on the topology of the space in question.

What does all of this have to do with the convergence or divergence of an infinite series (and its value)? The answer is in the definition of infinite series:

The infinite series, $\Sigma_{i=1}^{\infty}a_i$, is said to converge to the value $S$ of the limit of the sequence of partial sums, $\lim_{n\rightarrow\infty}S_n=\lim_{n\rightarrow\infty}\Sigma_{i=1}^na_i$, provided this limit exists. Hence the converge of an infinite series depends on the existence of the limit of the sequence of partial sums, which in turn depends on the topology of the space in question ($\mathbb{R}$ in the present case).


Odds of being correct X times in a row Is there a simple way to know what the chances are of being correct for a given number of opportunities?

To keep this simple: I am either right or wrong with a 50/50 chance. What are the odds that I'll be correct 7 times in a row or 20 or simply X times?

... and can the answer be put in simple terms ;)

Assuming the outcome of "answering" any question is independent of the outcome of answering other questions:

Yes, you'll be right $n$ times in a row with probability: $\left(\dfrac{1}{2}\right)^n$

If $X$ and $Y$ are independent events, then the probability that both $X$ and $Y$ occur is the product of the probabilities of $X$ and $Y$. For each question, you have a $50/50$ chance of answering correctly, which translates to a probability of $\frac 12$: for "randomly guessing" one question, you'll have a probability of $\frac 12$ that you'll be correct.

So answering correctly two questions in a row has probability: $$\frac 12 \times \frac 12 = \left(\frac 12\right)^2 =\frac 14$$

The probability of answering correctly three questions in a row: $$\frac 12 \times \frac 12 \times \frac 12= \left(\frac 12\right)^3= \frac 18$$

$$\vdots$$

The probability of answering correctly $n$ questions in a row: $$\underbrace{\frac 12 \times \frac 12 \times \cdots \frac 12}_{\large n \;\text{factors}}= \left(\frac 12\right)^n$$

You guys are making it too complicated. It goes like this:

double your last...

1st time: 1 in 2   -  (win a single coin toss)
2nd time: 1 in 4   -  (win 2 consecutive coin tosses) . . .etc.
3rd time: 1 in 8
4th time: 1 in 16
5th time: 1 in 32
6th time: 1 in 64 . . . etc . . . 1 in 2 to the nth. If you wanna get complicated with it then if 2 events have the same odds, there's no such thing as "independence" it's all in your head. Just add number of attempts together (nth)<--(this is an exponent). If the odds aren't 50/50 or if there are multiple events multiply the odds together. 1 in 4 should be represented as the fraction 1/4, 2 in 17 as 2/17 and so on... An example would be: "What would the odds of winning both a 1 in 3 bet and a 1 in 6 bet?" 1/3 X 1/6 = 1/18 or just multiply the 3 and 6 to make it easy. If you wanna know the odds of winning one and losing another, you don't do anything different, just flip the losses upside-down (inverse fractions) Make sense? "What are the odds of winning a 1 in 4 bet and a 1 in 6 bet, then losing a 1 in 4 bet?" 1/4 X 1/6 X 4/1(or just 4)= 1/6. I'm not going to bother with probability vs odds syntax since both boil down to the same thing. Just make sure that a preposition is never the word that you end your sentence in.


Proof by Strong Induction: $n = 2^a b,\, b\,$ odd, every natural a product of an odd and a power of 2 Can someone guide me in the right direction on this question?

Prove that every $n$ in $\mathbb{N}$ can be written as a product of an odd integer and a non-negative integer power of $2$.

For instance: $36 = 2^2(9)$ , $80 = 2^4(5)$ , $17 = 2^0(17)$ , $64 = 2^6(1)$ , etc...

Any hints in the right direction is appreciated (please explain for a beginner).
Thanks.

To prove something by strong induction, you have to prove that

  If all natural numbers strictly less than $N$ have the property, then $N$ has the property.

is true for all $N$.

So: our induction hypothesis is going to be:

  Every natural number $k$ that is strictly less than $n$ can be written as a product of a power of $2$ and an odd number.

And we want to prove that from this hypothesis, we can conclude that $n$ itself can be written as the product of a power of $2$ and an odd number.

Well, we have two cases: either $n$ is odd, or $n$ is even. If we can prove the result holds in both cases, we'll be done.

Case 1: $n$ is odd. Then we can write $n=2^0\times n$, and we are done. So in Case 1, the result holds for $n$.

Case 2: If $n$ is even, then we can write $n=2k$ for some natural number $k$. But then $k\lt n$, so we can apply the induction hypothesis to $k$. We conclude that ...and you should finish this part...

So we conclude that the result holds for all natural numbers by strong induction.

Proof:

By the fundamental theorem of algebra, every integer $N$ can be uniquely factored as $\prod^{n}_{i=1}p_{i}^{a_{i}}$. Now, mark $2=p_{1}$, note $a_{i}$ can take value of $0$. You got the theorem.

For the "inductive" proof, suppose for $n<k$ this is true. For $n+1$ its factors must be in previous $n$ numbers. Hence $n+1=\prod n_{i}$. Decompose $n_{i}$ by induction hypothesis you get the statement.


Show that the following set of six matrices is a group Show that the following set of six matrices is a group
  $$\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix},\begin{bmatrix}0 & 1\\-1 & -1\end{bmatrix},\begin{bmatrix}-1 & -1\\1 & 0\end{bmatrix},\begin{bmatrix}1 & 0\\-1 & -1\end{bmatrix},\begin{bmatrix}-1 & -1\\0 & 1\end{bmatrix},\begin{bmatrix}0 & 1\\1 & 0\end{bmatrix}.$$

The only property that is not immediately apparent is closure. The question before this one involved the symmetric group $S_3$ and based off another post, I believe this to be necessary for this question. However, I am not really sure how to make the connection. Any help would be appreciated.

Perhaps the least tedious way is to show we have some identification (I am speaking of an isomorphism, here) with some known group. As the comments above suggest, the most likely candidate (based on your matrices' orders) is $S_3$, or (equivalently), the symmetry group $D_3$ of the equilateral triangle.

The latter group is most succinctly summed up as:

$D_3 = \langle a,b| a^3 = b^2 = e; ba = a^2b\rangle$.

With sufficient patience, one can show this determines a group of $6$ elements:

$D_3 = \{e,a,a^2,b,ab,a^2b\}$ and the rules $a^3 = b^2 = e$ and $ba = a^2b$ are sufficient to determine all $36$ entries in a multiplication table; for example:

$(ab)(a^2b) = a(ba)ab = a(a^2b)ab = (a^3)(bab) = e(bab) = (ba)b = (a^2b)b = a^2(b^2) = a^2e = a^2.$

Then it is a matter of finding some matrix $A$ of your $6$ we can map to $a$, and another $B$ that maps to $b$, and showing that the matrix multiplication of these two matrices satisfies our $3$ rules:

$A^3 = I\\B^2 = I\\BA = A^2B.$

If that is so, then the isomorphism we would get is:

$A^kB^l \mapsto a^kb^l$, and we would get a "parallel" multiplication table for your $6$ matrices corresponding to the multiplication table for $D_3$ (thus showing closure).

As pointed out above, we have for $A = \begin{bmatrix}0&1\\-1&-1\end{bmatrix}$:

$A^2 = \begin{bmatrix}0&1\\-1&-1\end{bmatrix}\begin{bmatrix}0&1\\-1&-1\end{bmatrix} = \begin{bmatrix}-1&-1\\1&0\end{bmatrix}$, so that:

$A^3 = A^2A = \begin{bmatrix}-1&-1\\1&0\end{bmatrix}\begin{bmatrix}0&1\\-1&-1\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix} = I.$

If we take $B = \begin{bmatrix}1&0\\-1&-1\end{bmatrix}$, we have:

$B^2 = \begin{bmatrix}1&0\\-1&-1\end{bmatrix}\begin{bmatrix}1&0\\-1&-1\end{bmatrix} = \begin{bmatrix}1&0\\0&1\end{bmatrix} = I$, so there's $2$ out of $3$.

The third rule is tedious to verify, but here goes:

$BA = \begin{bmatrix}1&0\\-1&-1\end{bmatrix}\begin{bmatrix}0&1\\-1&-1\end{bmatrix} = \begin{bmatrix}0&1\\1&0\end{bmatrix}$, while:

$A^2B = \begin{bmatrix}-1&-1\\1&0\end{bmatrix}\begin{bmatrix}1&0\\-1&-1\end{bmatrix} = \begin{bmatrix}0&1\\1&0\end{bmatrix}$,

so, indeed $BA = A^2B$.

Multiply every matrix with every other matrix (that are 36 multiplications) and check if the resulting matrices are in your group.
That should proof closure.


Why does Rudin say "the rational number system is inadequate as a field"? In the INTRODUCTION of chapter 1 of Baby Rudin, he says

  The rational number system is inadequate for many purposes, both as a field and as an ordered set.

Addition and multiplication of rational numbers are commutative and associative, and multiplication is distributive over addition. Both 'zero' and 'one' exist.

Plus, as I recall, rational numbers are the smallest subfield of $\mathbb{C}$. So exactly what does Rudin mean by “inadequate as a field”?

Rudin isn't questioning $\mathbb{Q}$'s status as a field — he's questioning it's suitability for doing algebra and analysis.

For the purposes of algebra, $\mathbb{Q}$ is not a very nice field to work with: for example, many (most?) polynomials don't even have a root, let alone factor completely, and dealing properly with this deficiency is the subject of the frontiers of mathematical research!

(for comparison, while not every polynomial over $\mathbb{R}$ has a root, it has a very simple relationship with the complex numbers $\mathbb{C}$, a field over which every nonconstant polynomial has a root)

Similarly, for the purposes of analysis, $\mathbb{Q}$ is not a very nice ordered set; for example, they form a totally disconnected space, making them nearly useless for capturing even basic familiar geometric notions such as continuity.

The real numbers are a subfield of the complex numbers. Not every sequence of rational numbers converges to a rational number: rational numbers aren't a complete set.


How to reduce 0-1 knapsack to knapsack-like problem with overflow? Consider a knapsack-like problem where there is a set of items, and each item has a cost $c_i$ and value $v_i$. The goal is to find a subset $S$ that minimizes $\sum_{i\in S}c_{i}$ with the constraint that $\sum_{i\in S}v_{i}\geq b$) where $b$ is some arbitrary value.

This is obviously very similar to the knapsack problem, but I haven't been able figure out a way to reduce the knapsack problem to it using a polynomial time transformation on the inputs due to the inequality constraint. Maybe I'm just confused on how reductions are done.

The 0-1 knapsack problem is defined as follows:
Maximize $\sum_{i=1}^nv_ix_i$ subject to $\sum_{i=1}^nw_ix_i \leq W$, $x_i\in\{0,1\}$ where $n$ is the number of items, $\mathbf{v}$ are item values, $\mathbf{x}$ indicates whether an item is put into the knapsack, $\mathbf{w}$ are the weights of items, and $W$ is the maximum weight the knapsack can hold.

Since both are optimization problems, you are talking about the NP-hard knapsack problem, not the NP-complete one, because only decision problems (or decision version of optimization problems) can be NP-complete. See the wiki page for Knapsack problem for definitions.

The standard reduction when working with NP-hard problems is the Turing reduction, not the Many-one reduction which is used inside NPC. The key difference are:

whether you are allowed to call the oracle which solves the knapsack-like problem more than once (Turing reduction: yes, Many-one reduction: no)
when you are allowed to call the oracle (Turing reduction: not limited, Many-one reduction: only at the end of computation)
whether you can modify the results returned by the oracle before returning (Turing reduction: yes, Many-one reduction: no).

As @dtldarek said, it doesn't seem easy to find a many-one reduction between the two problems, where only transformation on the inputs are allowed, but it's possible to solve either one with multiple calls to an oracle solving the other (thus a Turing reduction).

Solving Knapsack-like using Knapsack: Note that for increasing total cost C, the maximal total value is non-decreasing. This property allows binary search:

Binary search for the total cost C (in range $[0,\sum c_i]$), and check (using the oracle to Knapsack) whether it's possible to achieve a total value $\ge b$. Choose either half according to the result. The smallest C can be found with small modification to binary search.

Solving Knapsack using Knapsack-like: Similar property holds: for increasing total value, the minimal cost (weight) is non-decreasing. So a similar algorithm is easy to find.

Okay, let

$$m_i = \frac{1}{c_i}$$

$$q_i =\frac{1}{v_i}$$

$$x_i = 0, 1$$

Thus our problem now has become:

Maximize $mx$ such that $qx \le 1/b$

If $1/b$ is our bag size. $M$ is the cost. $Q$ is the size of the items this is equivalent to knapsack.

If you are allowing negative values please let me know. It will require a change to this.


Undergrad looking for study material/advice for applied mathematics. I am an undergraduate math student (junior) who is looking to get a masters degree in Applied Math. I like pure math, but I want to use my education to get a great-paying job. Here are a few questions that I have been dying to ask someone with experience:

From my understanding Applied Mathematics is highly dependent on computer programming and statistics, is that right? If so, I have zero background in programming. What should I do to kick start my programming abilities? Books, programs, hobbies, etc.
Is the job market good for applied mathematicians? Also, are the salaries good?
Do applied mathematicians work crazy hours? I've heard they do.
What potential careers do applied mathematicians have? And what is the most common? Are there any cool one's you've heard of?
What general advice do you have for and undergraduate mathematician seeking a degree and a career in applied mathematics?

Are you really just asking for commonplace/cliché?

It is not true.
There are plenty of researches and rankings on this argument. But, are you really studying maths for salary or future job?
Does not make any sense
See 2.
Look at the arguments teached in your uni and see if you like it!

Remember, carrers are not based on what you have studied!

Many HRs does not even know what you have studied.
At junior level they are just looking for higly motivated, smart, fast learning (and son on) people.

However basic programming knowledge will help you.


Applications of Differential Calculus (KUMON Level O) I don't understand what I need to do to solve this problem:

  Show that equation $$\frac{x^2 - 3}{x-1}=0$$ has $2$ real solutions in $[-2,2]$.

I've found the derivative which is:

$$f'(x) = \frac{x^2 - 2x + 3}{(x-1)^2} > 0$$

The function is discontinuous at $x=1$.

I don't know what to do next.

Just solve
$$\frac{x^2-3}{x-1}=0,$$
there is no need for calculus.

$Y’$ is alway greater than $0$ in the domain thus the function is always increasing in the domain at approaching $1$ from the negative side it is equal to infinity for the positive side it is equal to negative infinity and $f(-2)$ is smaller then $0$ so that is one real solution because the function is increasing  so it goes from some thing negative to infinity it will pass by $0$ and from negative infinity to posit if something it will go by $0$ too. thus there is two real solutions


About finding the inverse of a matrix I am solving a linear algebra problem and this matrix came up from a system of linear equations.

$A = \begin{pmatrix}
1 & 2 & \cdots & n \\
1 & 2^2 & \cdots & n^2 \\
\vdots &  & &\vdots &\\
1 & 2^n & \cdots & n^n
\end{pmatrix}
$

I do not know how to check if my system has an unique solution or not (or the matrix is invertible) in this case.

The way to go is to run, say, LU-factorisation using one of broadly available codes allowing evaluation of the matrix condition ratio. Run the program for n=2,3,4... to see the pattern. You will probably see that the condition ratio is growing with each next n (in numerical sense, the matrix looks "bad" to me in general). As soon as the condition ratio reaches the value of about 10**16, the matrix (and all the others for n larger than the found one) is singular in numerical sense, because the computer operates with double precision variables, ie. it stores 16 significant digits only.


Find the sum of all the number formed by 2,4,6, and 8 without repetition.Number may be of any digit like 2, 24, 684, 4862. Find the sum of all the number formed by 2,4,6, and 8 without repetition.Number may be of any digit like 2, 24, 684, 4862.

My Approach:

single digit no formed = 2,4,6,8

sum= 2+4+6+8= 20

two digit= 24,26,28,42,46,48,62,64,68,82,84,86

sum= 660

three digit no=246+264+426+462+624+642=2664

268+286+628+682+826+862=3552

248+284+428+482+824+842=3108

468+486+648+684+846+864=3996

sum of all 3 digit nos =13310

Similarly for all 4 digit numbers.

There are:

four $1$-digit numbers, whose average is $5$: sum = $20$
twelve $2$-digit numbers, whose average is $55$: sum = $660$
twenty-four $3$-digit numbers, whose average is $555$: sum = $13320$
twenty-four $4$-digit numbers, whose average is $5555$: sum = $133320$

To see why these averages are correct, you can pair off each number uniquely with its 'complement' obtained by exchanging $2$ with $8$ and $4$ with $6$. The average of each of these pairs is $5\dots5$.

Consider an example by taking digits as $3,4,6,8$.
total combinations possible$=4\times 3\times 2=24$ numbers possible
no of digits$=4$
hence $\dfrac{24}{4}$ each digit comes $6$ times in ones, tens and hundred's place
at ones place=$6\times (3+4+6+8)=126\equiv6$.
at ten's place$=6\times (3+4+6+8)+12(\text{carry})=138\equiv 8$.
At hundred's place$=6\times (3+4+6+8)=139\equiv 9$
so, $3$ will come at thousands place and $1$ at ten thousands place,
Hence, the number is $139864$.


Prove the laplace transform of $\sinh(at)$? My problem today is as above. Here is what I have done:

Use integral definition of laplace transform to get

$$\int_0^\infty \sinh(at)\exp(-st)dt$$

$$= \lim_{b \to \infty}\int_0^b \sinh(at)\exp(-st)dt$$

By the definition $\sinh(at) = \frac{1}{2}(\exp(at)-\exp(-at))$

We can write

$$= \lim_{b \to \infty}\frac{1}{2}\int_0^b (\exp(at)-\exp(-at))\exp(-st)dt$$

$$= \lim_{b \to \infty}\frac{1}{2}\int_0^b \exp((a-s)t) - \exp((-a-s)t)dt$$

$$\lim_{b \to \infty}\frac{1}{2}\left[\left(\frac{1}{a-s}\exp((a-s)b) - \frac{1}{-a-s}\exp((-a-s)b)\right) - \left(\frac{1}{a-s} - \frac{1}{-a-s}\right)\right]$$

But taking the limit as $b \to \infty$ yields infinity terms. What have I done wrong here?

Hint.  If $s>|a|$ then $\exp((a-s)b)$ tends to $0$, not $\infty$, if $b\to\infty$.

I'm not quite sure if I did it right but I think I did. Just correct me if I'm wrong.


Are there dictionaries in math? Consider the following dictionary in the programming language Python:

D = {'A': 1, 'B': 2, 'C': 3}

It is saying that the value of A is 1, the value of B is 2, and the value of C is 3. It also has the property that D['A'] = 1 etc.

How would I write such thing in math? I was thinking about

$$D = \{A = 1, B = 2, C = 3\}.$$

However, I am not sure if this is the right or best way to do such thing.

I would like to use the structure for taking sums: e.g. 'AAAA' is interpreted as $1+1+1+1$ etc. What kind of notation should I use?

A dictionary is just a function $\mathrm{Dict}\colon \mathrm{Keys} \rightarrow \mathrm{Values}\cup\{\epsilon\}$ where $\epsilon$ is a "null character" with the understanding that $\epsilon\notin\mathrm{Values}$.

For example, let $\mathrm{Keys}=\{A,B,C,...,Z\}$, and $\mathrm{Values}=\mathbb{Z}$. Then, in your case,

$$
\mathrm{Dict}(x)=\begin{cases}
1 & \text{if }x=A\\
2 & \text{if }x=B\\
3 & \text{if }x=C\\
\epsilon & \text{otherwise}
\end{cases}
$$

A dictionary is a partial function from the key space $K$ to the value space $V$, where $K$ is the set of all hashable objects, and $V$ is the set of all Python objects.

A partial function from $K$ to $V$ is a subset of $K\times V$ with the condition that for each $k\in K$ it contains at most one pair that has $k$ as first element. The difference to a function is that a function needs to have exactly one pair with $k$ as first element.

$K\times V$ is the set of pairs $(k,v)$ with $k\in K$ and $v\in V$.


Couldn't we have defined the material conditional differently? I've been mulling this over lately, and I can't seem to understand why exactly the material conditional wasn't defined in a completely different way.

Obviously, the motivation behind our current definition of the material conditional is, amongst others, so that we don't end up with a definition that makes most restricted quantifiers impossible to work with. For example, take two triangles. A triangle is called equilateral if the lengths of its three sides are the same, whereas a triangle is defined to be isosceles if two of its three sides have the same length. We know then, by definition, that any equilateral triangle is also an isosceles triangle.

With this in mind, we can set up the following implication where $P(T)$ and $Q(T)$ are open sentences over the domain S of all triangles:
$$P(T): T \; \textit{is equilateral.} \; \text{and} \; Q(T): T \; \textit{is isosceles.}$$

$$P(T) \implies Q(T)$$

This implication should then be required to hold true for all triangles $T$ that are equilateral, so we could write:

$$\forall \, T, \, P(T) \implies Q(T)$$

but we run into the problem of "what if T isn't equilateral?" If we choose to define the truth values of the material conditional any other way than we currently have, say by making the implication false whenever $P(T)$ is false, we'd end up with a false implication. But surely, our implication (which we know to be true by definition) can't be proven wrong by looking at triangles that the implication doesn't even deal with in the first place. Therefore, we must require this implication to be vacuously true.

However, couldn't all of this have been avoided by simply stating:

$$\forall \, T: \textit{T is equilateral}, \, P(T) \implies Q(T)$$

and assigning a false truth value to the material conditional whenever its antecedent is false? That way, we'd avoid situations like:

$$\forall x \in \mathbb{R}, \; x>2 \implies x^2>4$$
$$x=-1$$
$$\therefore (-1)^2>4$$

...situations which we know to be untrue.

I'm probably not really getting the big picture here, but it seems to me that a simple specification of the properties of a quantified statement fix the issue of quantifier incompatibility. Could somebody enlighten me?

Cheers.

Your inference is wrong.

If we start with the reasonable law :

  $∀x \in \mathbb R,x > 2 \to x^2 > 4$

and we instantiate it with $x=-1$, we get :

  $-1 > 2 \to (-1)^2 > 4$.

But we cannot infer the false : $(-1)^2 > 4$ because, in order to do it, we have to use modus ponens, and to do this we have to assert the false : $-1 > 2$, and we have no reason to assert it.

Consider the statement: If it is raining, then it is cloudy.

  Symbolically: $Raining\implies Cloudy$

What does this mean? First, what it does not mean: It does not mean that rain somehow causes cloudiness, or that cloudiness somehow causes rain. Neither is the case. It means only that it is not both raining and not cloudy.

  Symbolically: $\neg[Raining \land \neg Cloudy]$

There is no suggestion of causality -- of one thing causing another -- or of time-based data about a historical relationships between raining and cloudiness.

Everything about material implication -- the truth table for $\implies$, and the detachment and contra-positives rules -- all make sense when you understand this. Simply stated, the rule is:

  $A\implies B\equiv \neg[A\land \neg B]$

EDIT:

The Truth Table

$\begin{array}{c|c|c|c}
A&B&A\implies B\\
\hline
T&T&T\\
T&F&F\\
F&T&T\\
F&F&T
\end{array}$

The Intuition

In the first row, we have $A\implies B$ and $A$ being true. Of course, $B$ must also then be true, as indicated.

In the second row, we $A$ being true and $B$ being false. Then $A\implies B$ must be false, as indicated.

Note that when $A\implies B$ is true and $A$ is false, as you would expect, $B$ is indeterminate, i.e. $B$ could be either true or false, as indicated in the the last two rows.

We also need $A\implies B$ to be true when $A$ is false to prove vacuous truth, an essential method of proof in mathematics.


Probability that a sequence repeats itself Given an infinite sequence $a_n$ of uniformly random integers $0$ to $9$, what is the probability there exist an integer $m$ such that the sequence $a_1$ to $a_m$ is equal to that from $a_{m+1}$ to $a_{2m}$?

What if we restrict to two symbols, or $k$ symbols?

I can't give an exact answer, but an interesting connection and some numerical results.

Some dabbling with short examples shows that the events for different integers $m$ are nearly but not exactly independent. For instance, some sequences that don't repeat at $m=3$ repeat at $m=2$ but not at $m=1$, whereas for sequences that do repeat at $m=3$ repetition at $m=2$ implies repetition at $m=1$. However, the effect from the dependence appears quite small, so

$$
\prod_{m=1}^\infty(1-k^{-m})
$$

should be a good approximation to the probability that there is no repetition. Incidentally, if $k$ is a prime power, this is the probability that a large random square matrix over $\mathbb F_k$ is invertible; see Probability that a random binary matrix is invertible. In that answer, I computed the product for $k=2$ to be approximately $0.288788$, which would make the probability for a binary sequence to exhibit repetition approximately $0.711212$.

Here are numerical results for $k=2$ for repetitions up to $m=17$. Each row contains the number of strings of length of $2m$ that repeat, the number of strings that don't repeat, the proportion $p_m$ of strings that repeat, and a value $\alpha_m=-\log_2(1-(1-p_m)/(1-p_{m-1}))$ that would be $m$ if repetition at $m$ were independent of previous repetitions (here's the code):

$$
\begin{array}{rrrlr}
m&\text{repeating}&\text{non-repeating}&\text{proportion}&\alpha_m\\\hline
1 & 2 & 2 & 0.5 & 1.00000\\
2 & 10 & 6 & 0.625 & 2.00000\\
3 & 44 & 20 & 0.6875 & 2.58496\\
4 & 182 & 74 & 0.7109375 & 3.73697\\
5 & 738 & 286 & 0.720703125 & 4.88753\\
6 & 2972 & 1124 & 0.7255859375 & 5.83794\\
7 & 11924 & 4460 & 0.7277832031 & 6.96450\\
8 & 47768 & 17768 & 0.7288818359 & 7.95290\\
9 & 191214 & 70930 & 0.7294235229 & 8.96725\\
10 & 765136 & 283440 & 0.7296905518 & 9.98483\\
11 & 3061104 & 1133200 & 0.7298240662 & 10.98340\\
12 & 12245530 & 4531686 & 0.7298904657 & 11.99044\\
13 & 48984342 & 18124522 & 0.7299235761 & 12.99397\\
14 & 195941804 & 72493652 & 0.7299401015 & 13.99640\\
15 & 783776080 & 289965744 & 0.7299483567 & 14.99761\\
16 & 3135122038 & 1159845258 & 0.7299524820 & 15.99838\\
17 & 12540523572 & 4639345612 & 0.7299545438 & 16.99901\\
\end{array}
$$

The agreement with $0.711212$ is OK but not marvelous. As suggested by the $m=3$ example, the dependence slightly increases the probability of repetition because a repetition implies correlations between the conditions of repetition at lower values of $m$; but this effect is strongest at $m=3$ and becomes negligible at higher values of $m$, where $\alpha_m\approx m$ shows that almost exactly the expected proportion of sequences repeat for the first time. Thus we can get an accurate estimate of the limit probability from $p\approx p_{17}+(p_{17}-p_{16})\approx0.7299566056$, which should be accurate to eight or nine digits.

I checked OEIS for the sequences of both the repeating and the non-repeating counts; no hit.

P.S.: Note that as t.b. pointed out in a comment under the answer linked to above, by the pentagonal number theorem the above product is given by

$$
\begin{align}
\prod_{m=1}^\infty(1-k^{-m})&=\sum_{j=-\infty}^\infty(-1)^jk^{-j(3j-1)/2)}\\&=1-k^{-1}-k^{-2}+k^{-5}+k^{-7}-k^{-12}-k^{-15}+k^{-22}+k^{-26}-\dotso\;,
\end{align}
$$

which leads to an interesting pattern of digits in its representation in base $k$:

$$
\begin{align}
\prod_{m=1}^\infty(1-2^{-m})&=0.01001001111011100000010000111111110111110000000000100\ldots_2\;,
\\
\prod_{m=1}^\infty(1-10^{-m})&=0.89001009999899900000010000999999998999990000000000100\ldots_{10}\;.
\end{align}
$$

(I'm not an expert on this (I'm a newbie), but I suddenly hit something while reading the question, it may be wrong)

Assuming $_m$ is a constant:

It will depend on how big $_m$ is. If the $_m = 1$, then the probability would be $\frac1{10}$. If $_m = 2$, then $\frac1{100}$. (ie if you had $19$ as the first two, there are $100$ possibilties of the second two, so the probability of $19$ coming up is $\frac1{100}$.

Therefore, the probablity is $$\frac1{10^m}$$

If $_m$ can be any number:

Then..... I don't know how to work that out. (Unless I suddenly have a brainwave...)

(This is my first answer, I like mathjax.)


Is it possible to simplify $\psi^{(2)}(\frac18)$ or $\psi^{(2)}(\frac pq)$? Is it possible to simplify $\psi^{(2)}(\frac18)$, where $\psi$ denotes the polygamma function?
  Or more generalized, $\psi^{(2)}(\frac pq)$ and $\psi^{(2n)}(\frac pq)$?

Background
Noticing there is a formula for $\psi(\frac pq)$, where $p,q\in\mathbb{N}$ and the following formulae:
$\psi^{(2)}(\frac12)=-14\zeta(3)$, $\psi^{(2)}(\frac13)=-26\zeta (3)-\frac{4\pi^3}{3\sqrt{3}}$, $\psi^{(2)}(\frac14)=-2\pi^3-56\zeta(3)$ and $\psi^{(2)}(\frac16)=-182\zeta (3)-4\sqrt{3}\pi^3$, it is natural to ask if there are general formulae.
Attempt
$$\psi^{(2)}\left(\frac pq\right)=\int_0^1\frac{\ln^2 t}{t-1}t^{p/q-1}dt\\=q^3\int_0^1\frac{\ln^2 u}{u^q-1}u^{p-1}du$$
I tried to use contour integration to solve this integral but failed.

This is an answer showing that why $\psi^{(2n)}\left(\frac pq\right)$ has a closed-form when $q=2,3,4,6$ and conjecturing it does not have a general closed-form when $p=5$ or $p\ge7$. (Assuming $L(s,\chi)$ is not a closed-form)
1.$L(2n+1,\chi)$ has a closed-form when $\chi(-1)=-1$
(Where $L$ denotes the $L$-series)
First, denote $f(x,\chi)$ the analytic continuation of $\displaystyle\sum_{n=0}^\infty \chi(n)x^n$. In this question, I showed that $f(x)=-\chi(-1)f\left(\frac1x\right)$.
Noticing $$\int_0^1 x^{n-1}(-\ln x)^{s-1}\frac{dx}{\Gamma(s)}=n^{-s},$$
one can prove that it is valid to change the position of $\sum$ and $\int$:
$$L(2s+1,\chi)=\int_0^1\sum_{n=0}^\infty\chi(n)x^{n-1}(-\ln x)^{2s}\frac{dx}{\Gamma(s)}\\
=\frac{1}{2\Gamma(s)}\int_0^\infty\frac{f(x)}{x}(-\ln x)^{2s}dx$$
Since $f(x)$ can be expressed in form $\frac{P(x)}{Q(x)}$, where $P$ and $Q$ are polynomial funtions, the integral has a closed-form according to Residue Theorem.
2.$L(n,\chi)$ has a closed-form when $\chi$ is a principal character
If $n$ is a prime, it is trivial.
If $n=p_1^{\alpha_1}\cdots p_{\tau}^{\alpha_\tau}$, it is always possible to add and subtract some terms to make the series $\zeta(n)$. Furthermore,  $L(n,\chi)=\frac pq\zeta(n)$
3.It is possible to deduce $\psi^{(2n)}\left(\frac pq\right)$ from the values of $L$-series when $q=2,3,4,6$.
For simpleness, I will discuss $\zeta\left(n,\frac pq\right)$ instead of $\psi$ here.
$2^nL(n,\chi_{2,1})=\zeta(n,\frac12)$,
$3^nL(n,\chi_{3,1})=\zeta(n,\frac13)+\zeta(n,\frac23)$, $3^nL(n,\chi_{3,2})=\zeta(n,\frac13)-\zeta(n,\frac23)$
$4^nL(n,\chi_{4,1})=\zeta(n,\frac14)+\zeta(n,\frac34)$, $4^nL(n,\chi_{4,2})=\zeta(n,\frac14)-\zeta(n,\frac34)$
$6^nL(n,\chi_{6,1})=\zeta(n,\frac16)+\zeta(n,\frac56)$, $6^nL(n,\chi_{6,2})=\zeta(n,\frac16)-\zeta(n,\frac56)$
Solving this simultaneous equation gives the values of $\zeta\left(n,\frac pq\right)$. (The values of $L$-series in LHS is known when $n$ is odd)
4.Conjecturing it does not have a general closed-form when $p=5$ or $p\ge7$
This post shows that $\varphi(n)>2$ when $n>6$. It is also obvious that $\varphi(5)>2$.
There is $\frac12\varphi(k)+1$ values known in $L(2s+1,\chi)$, and we have $\varphi(k)$ variables. Since $\varphi(k)>2$ for $k=5$ and $k>6$, $\varphi(k)>\frac12\varphi(k)+1$. It is not possible to deduce $\zeta(2s+1,\frac pq)$ via $L$-series.

I find this formula with Maple,consider the sum and we obtain an expression of Psi(1,1/8).


Hitting time and its distribution ÷I'm reading an italian book about casual process (Probabilità e modelli aleatori of Enzo Orsingher). At pag 105 there's the probability of the stopping time $T_\beta$.
$$P\{T_\beta \leq t\}=P\{\max_{0\leq s\leq t} B(s) \geq \beta\}=$$

$$=2P\{B(t) \geq \beta\} = \sqrt{\frac{2}{\pi}} \int_{\frac{\beta}{\sqrt{t}}}^{\infty} \exp\{-\frac{w^2}{2}\}dw$$
Because the book used the last result :
$$P\{T_\beta \leq t\}=P\{\max_{0\leq s\leq t} B(s) \geq \beta\}= 2P\{B(t) \geq \beta\}$$

Now I don't undersatnd why $T_\beta$ has the inverse gaussian density
$$f_{\beta}(t)= \frac{|\beta|}{\sqrt{2\pi t^3}}\exp\left\{-\frac{\beta^2}{2t}\right\}$$
for $t >0$.
Where did it find?

$P(B(t) > \beta) = P(B(t) > \beta | T_\beta > t ) \cdot P(T_\beta > t )
+ P(B(t) > \beta | T_\beta \leq t ) \cdot P(T_\beta \leq t )  \\
= 0 + \frac{1}{2} P(T_\beta \leq t ) \\
\Longrightarrow P(T_\beta \leq t ) = 2  P(B(t) > \beta) $.

$ P(B(t) > \beta | T_\beta \leq t ) = 1/2$ because the Brownian motion is symmetric conditioned on the hitting point (independent increase).

So, $P(T_\beta \leq t ) = 2  P(B(t) > \beta) = 2 - 2 P(B(t) \leq \beta) \\
= 2 - 2 P( \frac{ B(t)}{ \sqrt{t} } \leq \frac{ \beta }{ \sqrt{t} } )
= 2 - 2 \Phi (\frac{ \beta }{ \sqrt{t} } ) $

where $\Phi$ is the CDF of standard Gaussian distribution.
So $f_{T_\beta} (t) = \frac{d}{d t}( 2 - 2 \Phi (\frac{ \beta }{ \sqrt{t} } ) ) $.


what is the nth-derivative in 0 of $\frac{e^x}{1-x}$ Using $$f^n(0)=n! .a_n$$
I have$$\frac{e^x}{1-x}=\sum{\left(\frac{x^n}{n!}\right)}\sum{x^n}=\sum_{n \geq 0}\left({\sum_{k=0}^{n}{\frac{n!}{k!n!}}}\right)$$
How to get the combination?

The General Leibniz rule tells us that for two smooth functions $f,g$, the $n$th derivative of $h(x) = f(x)g(x)$ is

$$h^{(n)}(x) = \sum_{k=0}^n {n \choose k} f^{(n-k)}(x)g^{(k)}(x)$$

The $n$ derivative of $e^x$ at $x=0$ is $1$, and the $n$th derivative of $\frac{1}{1-x}$ at $x=0$ is $n!$ by considering the power series. Hence, the $n$th derivative of $h(x) = \frac{e^x}{1-x}$ at $x=0$ is

$$h^{(n)}(x) = \sum_{k=0}^n {n \choose k} (n-k)! = \sum_{k=0}^n \frac{n!}{k!} = n! \sum_{k=0}^n \frac{1}{k!}$$

Let $$y=\frac{e^x}{1-x},$$ or

$$y(1-x)=e^x.$$

By differentiation,

$$y'(1-x)-y=e^x,\\
y''(1-x)-2y'=e^x,\\
y'''(1-x)-3y''=e^x,\\
y''''(1-x)-4y'''=e^x,\\
\cdots$$

and obviously

$$y^{(n)}(1-x)-ny^{(n-1)}=e^x.$$

Evaluating at $0$, we get the recurrence

$$y_n=ny_{n-1}+1$$ with $y_0=1$.

To solve it, let $z_n:=\dfrac{y_n}{n!}$. (By the way, the $z_n$ are precisely the Taylor coefficients.) We have

$$n!z_n=n(n-1)!z_{n-1}+1$$ or

$$z_n=z_{n-1}+\frac1{n!}.$$

Hence,

$$\frac{y_n}{n!}=z_n=\sum_{k=0}^n\frac1{k!}.$$


Cartesian product of sets of sets If I have two sets A, B, which are sets of sets. Is there any abbreviated notation to write $$ \{ C \times D | C \in A \wedge D \in B\} $$ ?

You can use $S = \{C \times D\;|\;C \in A \land D \in B\}$ and use $S$ or define $S[A:B]$ to mean that.


Double Partial Derivatives of sum of variances of dependent random variables I have the following function
$$f(α)=Var[αX+(1−α)Y]=Var(αX)+Var[(1−α)Y]+2α(1−α)Cov(X,Y)$$

Partial derivative of this function w.r.t α leads us to following result
$$f′(α)=2αVar(X)−2(1−α)Var(Y)+2Cov(X,Y)(1−2α)=0$$

$$ α=\frac{Var(Y)−Cov(X,Y)} {Var(X)+Var(Y)−2Cov(X,Y} $$

This α is the minimum value of the function but in order to prove that it is minimum value, one has to prove that double partial derivative of the function  f(α) is greater than Zero  i.e.  f′′(α) > 0 .
But upon solving for f′′(α) I get

$$f′′(α)=Var(X)+Var(Y)-2Cov(X,Y)$$

How can conclude that this f′′(α) > 0 . Please help on this.

$$
0\le\def\Var{\operatorname{Var}}\Var(X-Y)=\Var(X)+\Var(Y)-2\operatorname{Cov(X,Y)}\;.
$$

Indeed this is the most you can show; you cannot show $f''(\alpha)\gt0$, since for $X=Y$ $f$ is constant.

As the variances is always a non-negative and Var(X-Y) is always greater than zero and our equaion leads to the proof that we wanted i.e.
$$ f′′(α) > 0$$
So it has been proved that α denotes the minimum value of the function.


how can we generate random numbers using skew normal distribution in multivariate case how can we generate random numbers using skew normal distribution in multivariate case

If you are interested in theory, as @Nate Eldredge suggested, follow the url.

As written there also, you could use R in practice to generate random numbers using a skew normal distribution.

Load the library sn:

library(sn)

Or if not installed, first install it via install.packages('sn').

Then you can generate any random number with given parameters with the rsn function.

rsn(n=100, location=1.256269, scale=1.605681, shape=5)

Will generate 100 random numbers from the distribution with given parameters. If you would like to plot the histogram of your generated values, use higher sample size (e.g. 1.000 or 10.000), like:

hist(rsn(n=10000, location=1.256269, scale=1.605681, shape=5))

Edit: this answers a different question.

Supposing that the mean is $\mu$ and the variance is $\Sigma$, calculate the Cholesky decomposition $\Sigma =LL^*$, generate a vector $v$ of $\operatorname{rank} \Sigma$ independent Gaussians, and output $Lv + \mu$.


Can we *ever* use certain log/exp identities in the complex case? This article on Wikipedia points out that certain identities for the log and exponential functions which are familiar from the real case require care when used in the complex case. Failures in the following identities are pointed out

$$\log{z^w} \equiv w \log {z}\\(zw)^{\omega}\equiv z^{\omega}w^{\omega}\\e^{zw} \equiv (e^z)^w$$

The article alludes to the fact that the identities do not always work, even if we consider them as an assertion about equalities of sets (and consider the functions as multivalued).

Now, how can we work with these identities without leading ourselves into error? Is the answer simply never to employ them in the presence of a complex base or exponent? Occasionally I see uses of these identities; here is one example:

Here is a scanned page from Schaum's outline on complex variables which shows how to evaluate $\int_{0}^{\infty}\frac{x^{p-1}}{1+x}dx$ with a keyhole contour (for $0<p<1$). As usual, on the return path, we have to parameterize the path as $z = e^{2\pi i}x$, $x\in \mathbb{R}$, rather than simply $z=x$. Now when the author exponentiates this term, he uses $(x e^{2\pi i})^{p-1}=x^{p-1}e^{2\pi i(p-1)}$, employing the second identity above. What justifies this, given that the identity does not hold in general?

By definition $z^w = e^{w \log(z)}$ (for whichever branch of $\log(z)$ you are using), so
$$ \log(z^w) = w \log(z) + 2 \pi i n$$
for some integer $n$.   If, for example, you are using the branch of $\log$ that has imaginary part in $(\alpha, \alpha + 2 \pi]$, you choose $n$ so that the imaginary part is in that interval.

Similarly, $(zw)^\omega = e^{\omega \log(zw)}$, and $\log(zw) = \log(z) + \log(w) + 2 \pi i n$ for suitable integer $n$, so
$$(zw)^\omega = e^{\omega(\log(z) + \log(w) + 2 \pi i n)} = e^{2 \pi i n \omega} z^\omega w^\omega $$

And
$$(e^z)^w  = e^{w \log(e^z)} = e^{w (z + 2 \pi i n)} = e^{2 \pi i n w} e^{wz}$$

EDIT: In Schaum's example, it is misleading to write $z = e^{2 \pi i} x$ (which would be the same as $x$), it's really $e^{(2 \pi - \epsilon) i} x$ where $\epsilon > 0$ is arbitrarily small.
Then $\log(z) = \log(x) + (2 \pi - \epsilon) i$ (for the branch of log with imaginary part in $[0, 2 \pi)$), and $$z^{p-1} = e^{(p-1)\log(z)} = e^{(p-1)\log(x) + (p-1)(2\pi - \epsilon) i} = x^{p-1} e^{(p-1)(2\pi - \epsilon) i}$$

Can we ever use certain log/exp identities in the complex case?

Two answers:

Can we ever use them?  Yes.
Can we blindly use them?  No.

For example: That computation from Schaum's can be done along those lines.  By applying these identities carefully.  Not by applying them blindly.


Laplace Transform of tsin(at) using only the definition Hello I' am stuck on how to get the final result of the laplace transform of $f(t)=tsin(at)$using (a is a constant) only the definition of $$\int_0^{\infty}f(t)e^{-st}dt$$,

I know $sin(at)= {1 \over 2i} (e^{iat} - e^{-iat})$

using that I get

$F(s)$=${1 \over 2i}$$[\int_0^{\infty}te^{iat-st}dt-\int_0^{\infty}te^{iat-st}dt]$

using integration by parts for each part I get

Part I

$u_1=t$

$du_1=dt$

$v_1$$=$$e^{iat-st}\over(ia-s)$

$dv_1$$=$$e^{iat-st}$

$te^{iat-st}\over(ia-s)$-$\int$$(e^{iat-st})\over (ja-s)$dt

Part 2

$u_2=t$

$du_2=dt$

$v_2$$=$$e^{-iat-st}\over(-ia-s)$

$dv_2$$=$$e^{-iat-st}$

$te^{-iat-st}\over(-ia-s)$-$\int$$(e^{-iat-st})\over (-ja-s)$dt

It is here where I am stuck at in getting the final result.

You have answered it correctly, all you have to do is just pluging in the limits of integral. The first integral is
\begin{align}
\left.\frac{te^{-(s-ia)t}}{ia-s}\right|_0^\infty-\frac{1}{ia-s}\int_0^\infty e^{-(s-ia)t}\,dt&=0-\left.\frac{e^{-(s-ia)t}}{(ia-s)^2}\right|_0^\infty\\
&=\frac{1}{(ia-s)^2}
\end{align}
And the second integral is
\begin{align}
-\left.\frac{te^{-(s+ia)t}}{ia+s}\right|_0^\infty+\frac{1}{ia+s}\int_0^\infty e^{-(s+ia)t}\,dt&=0-\left.\frac{e^{-(s+ia)t}}{(ia+s)^2}\right|_0^\infty\\
&=\frac{1}{(ia+s)^2}
\end{align}
Then the answer is
\begin{align}
F(s)=\frac{1}{2i}\left(\frac{1}{(ia-s)^2}-\frac{1}{(ia+s)^2}\right)=\frac{2as}{\left(a^2+s^2\right)^2}
\end{align}

its easier if you try doing it by laplace transform of derivatives method


Algorithms for "pleasing" drawings of planar graphs, possibly on sphere What algorithms exist to draw planar graphs without edge crossings in a way that they are easy to interpret by humans?
There are multiple algorithms that can handle any planar graph, such as Schnyder's algorithm or Chrobak-Payne.  These typically produce a result that is not very useful.
Here's an example.
The graph corresponding to this polyhedron,

looks like this with Schnyder's algorithm:

It looks like this with whatever Mathematica has built in:

What seems to work better in practice is using the Tutte embedding.  For this graph it gives:

I chose this graph specifically because it demonstrates in what way the Tutte layout is often not very good: the points in the middle are squeezed together.
Another problem is that it only works for 3-vertex-connected graphs. It may be possible to work around this sometimes, e.g. the Maple help says that Maple will try to construct a similar but 3-connected graph to use for computing the layout.  It does not detail the procedure though.
Note that this graph clearly has a nice drawing, e.g.

One could do even better by manual rearrangement.  (This drawing is obtained by manual adjustment of a layout from Mathematica's GraphData database.)
Questions:
My goal is to plot planar graphs in a "visually pleasing" way, i.e. in a way that humans can interpret them easily.

What algorithms are there that are useful for this?

The Tutte layout often works well. How can one make it useful for non-3-connected graphs?  (References would be nice.)

Are there algorithms that lay out the graph in 3D, on the surface of a sphere?  I am hoping to solve the problem of Tutte squeezing the middle points together.

Are there any algorithms that use curved edges to make the visualization clearer? The problem with straight line drawings is that edges sometimes make a very small angle. It may be that there is no way around this, but to use curved edges.

Notes:
I am aware that a simple force-directed layout algorithm often produces good results in 3D. E.g. for the above polyhedral graph we get:

For others, it doesn't work well at all.  Here's the edge list of a difficult instance:
{{1, 2}, {1, 6}, {1, 7}, {2, 3}, {2, 6}, {2, 7}, {2, 8}, {3, 4},
 {3, 7}, {3, 8}, {3, 9}, {4, 5}, {4, 8}, {4, 9}, {4, 10}, {5, 9},
 {5, 10}, {6, 7}, {7, 8}, {8, 9}, {9, 10}, {3, 5}, {4, 2}, {6, 8}}

Here's another difficult one:

(Yes, it is planar, though this particular drawing is not.)

I previously implemented a messy but practically useful method: start with any planar drawing, then use some randomized algorithm that shuffles the coordinates around without allowing the edges to intersect. You can see the results here.  This method is clearly not very general, and it has no proper theoretical foundation. I am looking for something better than such heuristic approaches.

Update:
It seems that the problems of the Tutte embedding, i.e. that the ratio of the shortest and longest edge lengths is often very large, can be at least partially remedied by using weighting when computing barycenters.

As a comparison, this is what the unweighted version gives for the same graph:

(Update: The weighted Tutte layout I mention above is now implemented in my IGraph/M package for Mathematica. You can try the interactive example from above in the documentation notebook.)

I don't have enough reputation to comment so I'm putting this as an answer.
Any planar graph can be made 3-connected in linear time by adding edges.
So we can make the input 3-connected and remove extra edges later.
If I recall correctly, Tutte's method works by fixing the outer face and essentially simulating "springs" (ie pretending edges are springs, then outputting a "stable" system of springs).
Maybe trying different cycles for the outer face would be beneficial.
As a side note, there are methods to draw planar graphs within grid points taking up small "area".
There are also methods that let you specify the clockwise order of incident edges to a vertex.


Proving limit of quadratic using formal definition without explicitly given f(x) How would you prove the following statement?:
Assume that $\lim_{x\to1}{f(x)=6}$.
Prove using the epsilon-delta definition that $\lim_{x\to1}[5f(x)^2+ 3f(x)] = 198$.

I set up the following:
We have $\forall \epsilon_1$ in the reals, $\exists \delta_1$ such that $0<|x-1|<_1 ⇒ |f(x)-6|<\epsilon_1$
We want a $\delta$ such that $|5f(x)^2 + 3f(x) - 198|<\epsilon$
This becomes $|5f(x)+33| |f(x)−6| <\epsilon$, then, we have $|f(x)−6| <\frac{\epsilon}{|5f(x)+33|}$
Now, I am not sure how to continue and how to implement the $\delta = \operatorname{min}\{..., ...\} $thing

HINT:
If $|f(x)-6|<\varepsilon_1$ then $|f(x)|\le |f(x)-6|+6\le 6+\varepsilon_1$.
And, in particular, if you assume that $\varepsilon_1<1$, then $$|5f(x)+33|\le 5|f(x)|+33\le 5(6+\varepsilon_1)+33 < 68.$$

Note that\begin{align}5f^2(x)+3f(x)-198&=5f^2(x)-180+3f(x)-18\\&=5\bigl(f^2(x)-36\bigr)+3\bigl(f(x)-6\bigr)\\&=5\bigl(f(x)+6\bigr)\bigl(f(x)-6\bigr)+3\bigl(f(x)-6\bigr)\\&=\bigl(5f(x)+33\bigr)\bigl(f(x)-6\bigr).\end{align}So, take $\delta_0>0$ such that$$|x-1|<\delta_0\implies\bigl|f(x)-6\bigr|<1.$$Then, if $|x-1|<\delta_0$, you have $5<f(x)<7$. But\begin{align}5<f(x)<7&\iff25<5f(x)<35\\&\iff58<5f(x)+33<68\\&\implies\bigl|5f(x)+33\bigr|<68.\end{align}Now, given $\varepsilon>0$, take $\delta_1>0$ such that$$|x-1|<\delta_0\implies\bigl|f(x)-6\bigr|<\frac\varepsilon{68}$$and then$$|x-1|<\min\{\delta_0,\delta_1\}\implies\bigl|5f^2(x)+3f(x)-108\bigr|<\varepsilon.$$


Calculate: $\int 3^{3^{3^{x}}}3^{3^{x}}3^{x} dx$ How to calculate following integration?

$$\int \left(3^{3^{3^{x}}}3^{3^{x}}3^{x}\right)\ dx$$

Substituting $u=3^x, du=\ln 3\cdot 3^x dx$, we get

$$\frac{1}{\ln 3}\int 3^{3^u}3^u du$$

If you keep doing this, can you see how to finish?

in the given problem,
all u need to do is
Substitute $t=3^{3^x}$
then you will get ur answer


Probability / Permutation of large number of events. This seems like a very silly and simple question, but I'm here to learn. So here it is:

How can I compute the probability of getting heads x times when tossing a coin a large number of times?  For example, what is the probability of getting heads 181 times when tossing a coin 300 times?

I understand how to do this using permutations, factorials, and traditional (elementary) probability methods. My problem is that my calculator won't go that high. I'm unable to use (300!)/[(181!)(119!)] because I get an overflow error.

The question I'm solving doesn't specifically ask for the probability, but I'd like to learn. Thanks!

In your case we have $\binom{300}{181}\cdot2^{-181}\cdot2^{-(300-181)} = \binom{300}{181}\cdot2^{-300}$ as the correct formula, where we need to compute $\binom{300}{181} = \frac{300!}{181!\cdot119!}$.

How do you compute such large binomials? You can use the following formula:

$$\binom{n}{k} = \prod_{i=1}^k \frac{n+1-i}{i}$$

Since $n$ and $i$ are on the same order all terms stay manageable. To keep the product managable as a whole you can repeatedly halve the partial product until it gets below 1, keeping count of how many halvings you did, making sure never to do more than $300$, as well as however many are left to do at the end to finish the multiplication by $2^{-300}$.

All in all it would be a couple hundred steps, cumbersome to do by hand, but trivial for a computer.

This is just a summary of comments, below, so that the OP can vote on it.

If you're going to use a computer, as orlp suggests, the approach is trivial:

In Mathematica:

Binomial[300, 181] (1/2)^300

$\frac{11795883462622100885471935681885637931873470140170002831274647962254384085185135026
   25}{1591434356511317254897223194069826688321459682551512695809484726058110390440106801
   7057792} \approx 0.0000741211$

This approach can be used on any computer through WolframAlpha, and will yield as many digits as you could ever need.  Even the "hard" problem ("the probability of hitting exactly 234658 heads out of 1259923") posed by orlp is given to arbitrary number of digits quickly without the need to write iterative code or confronting other problems...

And since you also want to know the probability of getting at least $180$, well again this is trivial:

Sum[Binomial[300, i] (1/2)^300, {i, 180, 300}]

$\frac{40372780344505313991018191943947190320930407268761546691659955107726023855608420480
   429}{127314748520905380391777855525586135065716774604121015664758778084648831235208544
   136462336} \approx 0.00031711$,

which can be accessed on WolframAlpha instantly.

(I don't know how one would ever answer this ancillary question using a hand calculator and the methods suggested by orlp, but am eager to hear suggestions.)


Show that there exists $i$ and $j$ with $1 \leq i<j \leq 51$ satisfying $a_i$ divides $a_j$. Let $a_1<a_2<...<a_{51}$ be the given distinct natural numbers such that $1 \leq a_i \leq 100$  for all $i=1,2,....,51$. Then show that there exists $i$ and $j$ with $1 \leq i<j \leq 51$ satisfying $a_i$ divides $a_j$.

There are $25$ primes less than $100$. So we must take the set $\{a_1,...,a_{51}\}$ such that the set does not contain all $25$ primes. Because if the set contains all $25$ primes then any number other than these prime would divisible by one of the prime. But from here I can not proceed further. Please help me to solve this.

We can divide the numbers $1$ to $100$ into $50$ different sets. First, put each odd number into a different set. Then, for some set with an odd number $n$, also place all numbers of the form $2^k n$ in the set, where $2^k n$ is between $1$ and $100$. Note that for any pair of numbers in the same set, one divides the other (as one of them will be $2^a$ times the other for some $a$).

Now by the pigeonhole principle, some set must contain $2$ numbers from $a_i$ since there are 50 sets. But then one of these divides the other, finishing the problem.

Thinking in terms of prime numbers is not the way to attack this question.  Instead:

Hint: Partition $\{1,2,\dots,100\}$ into sets of the form $\{m,2m,4m,8m,\dots\}$ where $m$ is odd.  There are 50 odd numbers between 1 and 100 inclusive, so ...


Is there a difference between $T:\mathbb{R}^n \rightarrow \mathbb{R}^k$ and $T:F^n \rightarrow F^k$ in Linear Algebra? Can someone please tell me what is the difference between $T:F^n \rightarrow F^k$ and $T:\mathbb{R}^n \rightarrow \mathbb{R}^k$, I know that are both made to define a Linear Transformation but I never understood what is the $F$ stands for.

Thank you in advance.

$\mathbb{R}$ refers to real numbers, but $F$ is a general field.

Therefore, $T:\mathbb{R}^n\rightarrow \mathbb{R}^k$ is a special case of $T:F^n\rightarrow F^k$, where $T$ transforms vectors from the vector space $\mathbb{R}^n$ to the vectors from the vector space $\mathbb{R}^k$.

In general, any set of object that have the following properties form a field:

0- Operations ($+$) (addition) and ($*$) (multiplication) are defined on the objects.

For $x,y,z\in F$, the following properties should hold:

1- Closure under addition and multiplication: $x+y, x*y\in F$

2- Associativity of addition and multiplication: $x+(y+z)=(x+y)+z$ and $x*(y*z)=(x*y)*z$

3- Commutativity of addition and multiplication: $x+y=y+x$ and $x*y=y*x$

4- Existence of additive and multiplicative identity elements: The should exist $0,1\in F$ such that $0+x=x$ and $1*x=x$

5- Existence of additive inverses and multiplicative inverses

6- Distributivity of multiplication over addition

Read more about fields here.

I guess $F$ is an arbitrary field. The real numbers $F=\mathbb{R}$ is just one example of a field.

Other fields you know, are $F=\mathbb{Q}$ and $F=\mathbb{C}$. But there are also other fields, for example finite fields (fields with only finitely many "scalars" in them).