Filling the plane with a sequence I am not sure if this is the right stack to ask this question, but since there is a definite fractal dimension to it, I thought I'd give it a go.

The problem I am facing is one of calculating an expensive function $f(x,y)$ in a finite 2D domain $D$. I perform the evaluation of $f$ at each point in $(x,y)$ independently from all other $(x,y)$ points.

The dumb way to parallelise the process of "scanning" $f$ over $D$ was to define a fixed rectangular grid of points covering $D$.
By creating a trivial correspondence between integers and the position in the grid, I can assign each parallel task a range of integers to work with.
I.e., each integer maps to only one point in $D$.

The solution above has a huge drawback: I cannot have a quick preview of what $f(x,y)$ looks like.
The problem is akin to the one solved by progressive image encoding: I want to be able to refine the picture of $f(x,y)$ as I add more points.

The way I am framing the problem as is: how to map integers to a series of $(x,y)$ points which never repeat while becoming denser on an average way as the integer value increases.

This made me go straight to space-filling curves or trees: I could triangulate $D$ and in each triangle iterate to the desired scale.
In this scheme, the number of points evaluated at iteration $n$ is $4\times(4-1)^{n-1}$.
For instance, at iteration 3 there are 36 new points (which is different from the points in this example), and I would easily find myself "filling in" a certain corner of the triangle before others, resulting in a rather "non-progressive" way of filling in the triangle.

So, even if this would be a way to iteratively improve the density, in each iteration the overall sampling density would not be balanced (uniform) across $D$.

Any ideas on how to make the sampling in each iteration more uniform are very welcome.

Eventually decided for a quasi-random sequence since it verifies both the integer to plane mapping and the uniform density criteria.

Would the Peano or Hilbert curve do the trick?


Difference between "intercept" and "intersect" What is the difference between intercept and intersect? Can they be used interchangeably? For example, intersecting lines and intercepting lines.

There is a temporal element to intercept which is absent from intersect: if two paths are generated respectively by two particles moving through space, the paths may intersect; however, one particle only intercepts the other if they arrive at the point of intersection simultaneously.

Technically intersection is  defined as, it is a meeting point of two variabled curve which varied  as  time ie, the  equations of curves will vary.
        But in case of intercept, one  curve  equation is  constant  and  other is  varied  as its bounded  area.


Proving the circumradius (R) for an equilateral triangle I am intrigued about how people would go about proving the formula for the circumradius in the image above. I went about it by using the cosine rule and it worked out nicely. Any other methods?

You have a right triangle with angles $30,60$ and the hypothenuse is $R$. You can get $R$ from there.

Use Pitagoras Theorem
$$R=\sqrt{\left(\frac{s}{2}\right)^2 + \left(\frac{s}{2 \sqrt{3}}\right)^2}=s\frac{\sqrt3}{3}$$


Prove that if $a\equiv b \pmod m $ , then $a \bmod m = b \bmod m$ Question

Prove that if $a\equiv b \pmod  m$ , then
$a \bmod  m = b \bmod m$

Approach
Given,

$a\equiv b \pmod m$

$\implies m\mid (a-b)$
$\implies (a-b)=m\cdot k$
$\implies a=b+m\cdot k$
Now,

$a \bmod  m$   can be written as
$\implies b+m\cdot k  \pmod m$

No idea how to move forward to get
$$b \pmod m \impliedby b +m\cdot k  \pmod m$$
Please help me out!!!
Thanks!

Hint $\, $ Show  $\ c\equiv  d\pmod{\!m}\,$  and $\,|c-d|<m\,\Rightarrow\,  c= d$

Apply that to $\,\ \underbrace{a\bmod m}_{\Large c}\!\equiv a\equiv b\equiv \underbrace{b\bmod m}_{\Large d}\,\ \pmod{\!m}$

Remark $ $ Any sequence of $m$ consecutive integers forms a complete system $S$ of representatives (normal forms) of all equivalence classes mod $m,\,$ so the above implies that congruence reduces to equality on these normal forms, i.e.  $\ a\equiv b\iff \bar a = \bar b,\,$ where $\bar n =$ unique normal-form rep in $S$ congruent to $n\,$ (which is analogous to: $ $ fractions are equivalent iff their least form rep's are equal, i.e.  least terms fractions are normal forms for equivalence of fractions).

An alternative convenient choice of normal forms for modular arithmetic arises by choosing the least magnitude reps (signed or balanced residue system), e.g. $\,\{-2,\color{#c00}{-1},0,1,2\}\pmod{\! 5}.\,$ In these signed systems the  normal form of $\,m\!-\!1\,$ is $\,-1,\,$ e.g. above $\,4\equiv\color{#c00}{-1}\pmod{\!5},\,$ simplifying common calculations such as $\,(m\!-\!1)^k\equiv (-1)^k,\,$  e.g. when casting out elevens.

I think it's obvious:

You have that $a \equiv b+mk \pmod m$.

But, $mk \equiv 0 \pmod m$ , so $a \equiv b+mk \equiv b\pmod m$.

Done.


Proof that if $A$ is a subset of $B$ and $C$ is a subset of $D$, then $A \setminus D$ is a subset of $B \setminus C$. Started my proof by assuming that $A \subset B$ and $C \subset D$. Then I proceeded by supposing that $x \in  A \setminus D$. So $x \in  A$ and $x \notin D$. Is it valid to conclude that $x \notin C$ since  $C \subset D$? I'm assuming it is always true but I'm not sure if it is a valid argument I can use for formal proofs.
Aside from that, I'm also lost as to how I should continue my proof. For now I continued with my proof mentioned above and added "since $A \subset B$, then $x \in B$. Also, $x \notin C$ so it follows that $x \in B \setminus C$. Therefore,  $A \setminus D \subset B \setminus C$. " Although, I'm not really sure if this is a valid argument. For instance, I'm having doubts with $x \in B$ and $x \notin C$ since $x$ is also not an element of $D$. Therefore, there's a possibility that $x \in B \setminus D$ (??). Any help with regards as to how I should properly and formally prove this will be a great help.
Edit: Thanks a lot to everyone who verified my approach!

You're correct stating that if $x\not\in D$, since $C\subset D$ then $x\not\in C$. If it isn't clear to you, look at the contraposition (if $x\in C$, since $C\subset D$ then $x\in D$).
In the second part, you proved that for every element $x\in A\setminus D$, you have $x\in B\setminus C$, therefore $A\setminus D\subset B\setminus C$. If $x\in B$ and $x$ is not an element of $D$ then automatically $x\in B\setminus D$.

Your argument is right. In symbols, you're essentially doing $$A\subset B\quad\text{and}\quad D^c\subset C^c\implies A\cap D^c\subset B\cap C^c$$
using the fact that $M\subset N\iff N^c\subset M^c$, which is right.


How does determinant method of solving equations work? Say you have two equations

$ax+by+c=0$

$dx+ey+f=0$

by determinant method

$\frac{x}{bf-ce}=\frac{-y}{af-cd}=\frac{1}{ae-db}$

Clearly, the solution is valid

How exactly was this derived?

Anvit, using my set up of equations, here is a way:
Multiply $ax+by=e$ by $d$ to get $adx+bdy=ed$ and multiply $cx+dy=f$ by $b$ to get $bcx+bdy=bf$. Now eliminating $bdy$ from both equations through subtraction yields $(ad-bc)x=ed-bf$ from which $x$ follows by division $ad-bc$ which is called determinant. For $y$ you can do the same thing. Can you give this a try?

EDIT: Considering $ad-bc$ and $ed-bf$ as determinants of two matrices, it ultimately gives Cramer's Rule

it is $$x=\frac{ce-bf}{bd-ae},y=\frac{cd-af}{ae-bd}$$
and $$bd-ae$$ is the determinant, see also the Cramers rule


Fubini theorem question Let $f$ and $g$ be Lebesgue measurable nonnegative functions on $\mathbb{R}$.  Let $A_y=\{x:f(x) \geq y\}$ Let $F(y)=\int_{A_y} g(x)dx$.  Prove $\int_{-\infty}^\infty f(x)g(x)dx=\int_0^\infty F(y)dy$.  I know this has to do with Fubini's theorem but I cannot prove it.

recall Fubini’s theorem and apply to couple of function $(f(x|t) ,p(t))$


For $ \frac{n(n-1)}{2}<i \leq \frac{ n(n+1)}{2} $, why is $\text{round}(\sqrt{2i})=n$ I was doing Problem 67 of Project Euler, which involves number triangle which has similar structure to Pascal's triangle (the nth row contains n numbers). While thinking of a way to represent the triangle as a single list, I realized that the number of elements in a triangle of height n was $1+2+3+...+n = n(n+1)/2$, which must mean that the last number of any row n must be the $(n(n+1)/2)^{th}$ number in the triangle. More generally, we have that for any index $i$ starting at 1, $\frac{n(n-1)}{2} < i \leq  \frac{n(n+1)}{2}$ where $n$ is the row that contains the $i^{th}$ number of the triangle. From here, I figured that taking the square root of $2i$ and rounding it would give something close to $n$, and indeed rounding the square root of $2i$ results in the pattern $1,2,2,3,3,3,4,4,4,4,5,5,5,5,5$, etc. So far, I have not been able to prove this result to myself, so any help would be appreciated, thanks!

$$\frac{(n-1)^2}2<\frac{n(n-1)}2<\frac{n^2}2<\frac{n(n+1)}2<\frac{(n+1)^2}2$$

We also know that

$$\frac{n(n-1)}2<i<\frac{n(n+1)}2$$

On finding differences, we realise that $i$ will always be closer to $\frac{n^2}2$ than $\frac{(n-1)^2}2$ or $\frac{(n+1)^2}2$

Hence $i=round(\sqrt{2i})$


Weak and pointwise convergence in a $L^2$ space Let $I$ be a measured space (typically an interval of $\Bbb R$ with the Lebesgue measure), and let $(f_n)_n$ a sequence of function of $L^2(I)$.

Assume that the sequence $(f_n)$ converge pointwise and weakly. How to prove that the pointwise limit and the weak limit are the same ?

Here's a functional analytic approach:

A weakly convergent sequence in a Hilbert space $H$ is bounded, and by the Banach-Saks theorem has a subsequence whose Cesàro averages converge strongly in $H$ to the same limit.
Almost sure convergence is preserved by taking subsequences and Cesàro averages.
So, without loss of generality you may assume that your weakly convergent sequence is actually strongly convergent.

Both strong $L^2$ convergence and almost sure convergence imply convergence locally in measure, so you only need to show that such limits are unique, which  is easy.

Assume that $f_n \rightharpoonup g$ and $f_n \rightarrow f$ a.e. Then
\begin{equation}
 | \int_{I} (f-g)| \le \int_{I} |f_n - f| + \int_{I} |f_n - g|
\end{equation}
As $f_n \rightharpoonup g$, e $1 \chi \{I\} \in L^{2}(I)$ we have
\begin{equation}
\int_{I} |f_n - g| \rightarrow 0.
\end{equation}
Now, notice that
\begin{equation}
|f_n -f| \le |f_n| + |f| < 2|f| + \varepsilon.
\end{equation}
for $n>>1$. Then by Dominated convergence Theorem
\begin{equation}
\int_{I} |f_n - f| \rightarrow 0.
\end{equation}
Hence $f=g$ a.e.


How does linear algebra over the octonions and other division algebras work? An interesting question, which has been discussed in many forms on this site, is how many results from the study of linear algebra over vector spaces carries over when we allow the scalars to form an algebraic structure more general than a field.

For example, if you stop requiring multiplicative commutativity and allow the scalars to form an arbitrary division ring, then a surprising amount of structure carries over unchanged, as discussed here: you lose the notion of the determinant and eigenvalues become more subtle, but you still have unique basis cardinality, Gaussian elimination, the Rouché-Capelli theorem, and matrix representations of arbitrarily linear maps between finite-dimensional modules. A similar story applies if you allow the scalars to form an arbitrary commutative ring. However, all hell can break loose if you drop commutativity and division: a module over an arbitrary ring does not necessary have an invariant basis number, so much of the structure of linear algebra over fields immediately falls apart.

What happens if you further generalize the scalars' division ring to be a division algebra that is not associative? How much structure do you lose if you give up associativity but preserve division?

To be concrete, consider generalized modules with the octonions as their "scalars". (A precise definition of what I mean is given in this question, together with the requirement $(rs)\cdot x = r \cdot (s \cdot x)$.)

Do these modules have a notion of invariant basis number?
If so, can arbitrary linear maps between finitely generated such generalized modules be represented by matrices? Are the matrix elements given by the usual formula $A(\hat{e}_j) = \sum_i A_{ij} \hat{e}_i$?

As Billy points out, the rule $(rs)\cdot x=r\cdot(s\cdot x)$ implies that $(rs)t$ and $r(st)$ will act identically on your module.

According to Wikipedia's multiplication table we have
$$ e_1(e_3e_5) = -e_1e_6 = e_7 \\
(e_1e_3)e_5 = -e_2e_5 = -e_7 $$
so we must have
$$ e_7\cdot x = -e_7\cdot x $$
whence (by distributivity)
$$ 2e_7\cdot x = 0 $$
and by dividing by $2e_7$,
$$ 1\cdot x = 0 $$
Since the module axioms also require $1\cdot x$ to be $x$, this means that the only module that satisfies all of your rules is $\{0\}$ -- not very exciting. Thus:

1. Yes, the basis number is invariant. Every basis contains exactly $0$ vectors.
2. Yes, if you allow $0\times 0$ matrices.

You definitely want to change your definition of what a module is, since
$(r s) * x = r * (s * x)$  for all $r, s$ in skew field $K$, $x$ in module $M$
is clearly a version of the associative law, which doesn't work for non-associative alternative algebras (like alternativ-körper / alternative skew fields like the octonions).
I'm assuming you know about the commutator
$[a, b] = a b - b a$
and the associator
$[a, b, c] = (a b) c - a (b c)$
already. Sometimes the associator is written as $\{a, b, c\}$ instead, if $[a, b, c]$ is already used for the "jacobian" $[[a, b], c] + [[b, c], a] + [[c, a], b]$. But let's not complicate things too much.
Let's say that the above mentioned left module law is part of the definition of an associative left module.
For an alternative left module we instead want (works in characteristic not 2):
$[r, s, x] + [s, r, x] = 0$,
where $[r, s, x]$ is the left "associator" $(r s) * x - r * (s * x)$, which vanishes for associative modules.
In characteristic 2 (yes there are split octonions in char 2 and other finite characteristics, not really fields, but almost, they are composition algebras) we instead have to use
$[r, r, x] = 0$
since this gives us
$0 = [r + s, r + s, x] = [r, r, x] + [r, s, x] + [s, r, x] + [s, s, x] = [r, s, x] + [s, r, x]$ as desired.
The other direction doesn't hold though. This because in characteristic 2,
a + b = 0 i.e. a = -b is the same as a = b, not forcing
a = -a => a = 0, thus the desired left alternativity
r^2 * x = r * (r * x) is not guaranteed otherwise.
Similarly for an alternative right module we use:
[x, r, s] + [x, s, r] = 0 using the right "associator" [x, r, s] = (x * r) * s - x * (r s)
or
[x, r, r] = 0.
I have know idea if the notion of an alternative "middle module":
[s, x, r] + [r, x, s] = 0 or [r, x, r] = 0
is useful in and of itself, but the middle "associator" [s, x, r] = (s * x) * r - s * (x * r) is definitely useful.
For an alternative bi-module we need more, namely the law
[x, r, s] = [r, s, x] = [s, x, r]
which is not deducible from the above.
This law together with either one of left alternativity, right alternativity or middle alternativity (plus distributivity), give us the other above laws, and a feasible notion of alternative bi-module we seek.
We could go one step further and define a central field of our alternative skew field, that our module must be a vector space over. Or, if we like, a nuclear skew field that our module must be an associative module over.
Nuclear elements n are such that for all r, s :
(n r) s = n (r s)  {left nuclear},
(r n) s = r (n s)  {middle nuclear} and
(r s) n = r (s n)  {right nuclear}.
All such elements n together form the left nucleus, middle nucleus or right nucleus of a Moufang loop or alternative algebra,
and all such elements n satisfying all these three conditions (the intersection of these nuclei) give us simply the nucleus of the Moufang loop or alternative algebra, this nucleus is always a group or associative algebra, respectively;
by using similar laws for certain nuclear scalars acting on the module, we ensure that our base skew field (if any) of certain such scalars for our alternative algebra is functioning as a base skew field for our module as a bi-vector space (left and right vector space, being an associative bi-module) over this skew field. This idea is probably pretty alien if not combined with the next one.
Central elements c are such that for all r :
c r = r c
; these form the commutative center.
The intersection of the nucleus and the commutative center form simply the center, which is always an Abelian group or commutative associative algebra, depending on context;
by using similar laws for (nuclear) central scalars acting on the module we ensure that our base field of certain such scalars for our alternative algebra is functioning as a base field for our module as a vector space over this field.
In the ordinary field case we get (for a, b in base field F; r, s in algebra K; x in module M):
(a r) * x = a (r * x) = r * (a x)
and
(a x) * r = a (x * r) = x * (a r)
for all base field scalars a, algebra scalars r, and module elements x.
Noting that a r = r a, (a r) s = a (r s), (r a) s = r (a s), (r s) a = r (s a), (a b) x = a (b x) and redundantly a x = x a.
In the skew field case we get (for a, b in base skew field F; r, s in algebra K; x in module M):
(a r) * x = a (r * x)
(r a) * x = r * (a x)
(r * x) a = r * (x a)
(a x) * r = a (x * r)
(x a) * r = x * (a r)
(x * r) a = x * (r a).
Noting that (a r) s = a (r s), (r a) s = r (a s), (r s) a = r (s a), (a b) x = a (b x), (a x) b = a (x b) = a x b and (x a) b = x (a b).
Note that the action of an associative algebra over an alternative (left, right, bi-) module may not be associative, and the action of a commutative associative algebra over an associative (left, right, bi-) module may not be commutative.
Thus e.g. the quaternions, as well as the complement of the complex numbers in the quaternions, are (non-commutative) associative bi-modules for the complex numbers.
And the octonions, as well as the complement of the quaternions in the octonions, are (non-associative) bi-modules for the quaternions.
The octonions, the octonion complement of the quaternions, and the octonion complement of the complex numbers are (non-associative) bi-modules for the complex numbers as well.
Using either the real numbers or the rational numbers (or any field in between) as the base field.
This paper is a wee little bit sloppy but still rigorous enough to give a good notion about what bimodules over octonions are, or ought to be:
https://arxiv.org/abs/2007.05375


Am I doing this counting problem correctly? $12$ people were asked which of the three television channel ch1, ch2, and ch3 they watch. How many possible results there could be? (For example, one result would be 8 people watch ch1, 10 people watch ch2, and no one watches ch3.

My answer is $12\times2^3$. Is this correct?

For each person, they could watch ch1 or not, then watch ch2 or not, and watch ch3 or not so for each person there are $2^3$ possibilities.

$12$ people are watching the television. A television can have $3$ states.

There are $3^{12} = 531441$ possibilities.

If you have zero or more non-viewers amongst these people the correct answer would be: $4^{12} = 16777216$ possibilities.

Description:

There is $c^p$ number of possibilities where $c$ equal the total number of channels and $p$ equal the total number of people watching.

Why?

no.  person 1   person 2  etc....
1       0          0
2       0          1
3       0          2
4       1          0
5       1          1
6       1          2
7       2          0
8       2          1
9       2          2
.       .          .
.       .          .

The truth table show that if there where only $2$ persons and only $3$ channels there would be $3^2 = 9$ possibilities. For larger number of people one could extend this truth table. Its increasing exponentially.

I think the answer should be :-

(1)If each person can choose only one channel :- (3)^12, as each person has three options, and each one could choose any one of the three.

(2) If a person has to choose atleast one channel :- (3C1 + 3C2 + 3C3)^12 = 7^12

(3) If a person can choose any number of channels (from 0 to 3) :- (3C0 + 3C1 + 3C2 + 3C3)^12 = 8^12


How to perform a fair coin toss experiment over phone? I was recently asked this question by my friend. Suppose the two individuals participating in a toss are not near each other, but could communicate over a telephone. How does one construct a fair coin toss experiment that is mutually agreeable to both of them? They can't agree on a function of quantities like the time or the telephone number, as these decide the winner a priori (before the experiment is conducted).

I suggested they disconnect the call and try again; whoever manages to reach the other first is the winner. But the state machine involved here is a bit complicated to get the simple (0.5,0.5) probabilities.

PS: They do not trust each other, so one of them can't toss a fair coin and convey its outcome to the other. Both of them throwing simultaneously also doesn't work, as the second person has the incentive to lie when they are communicating the results to each other.

Here's one way to do it. Let's call the two parties Alice and Bob (as is popular to do in cryptography and theoretical computer science more broadly these days).

Alice and Bob agree on a secure hash function $h$. Alice chooses a random string $r_A$ and Bob chooses a random string $r_B$. Bob tells Alice $r_B$.

Now, Alice flips a coin, call the result $x$. Alice sends $h(x,r_A,r_B)$ to Bob and asks Bob to call the toss. Let's say Bob calls $y$. Then Alice tells Bob $(x,r_A)$ and he can verify himself that $x = y$ by checking that $h(x,r_A,r_B) = h(y,r_A,r_B)$. In this way if Bob called it wrong, then Alice can prove that he was wrong.

Obviously, if Bob calls the coin flip correctly, then the two hashes match. Moreover, it's extremely hard for Alice to cheat because if Bob says "tails" for example when the coin toss was indeed "tails" but Alice wants to trick him into thinking it was "heads", she'd have to come up with a random string $r$ such that $h(H,r,r_B) = h(T,r_A,r_B)$, which is hard by the assumption that $h$ is a secure hash function and the fact that Bob chose $r_B$. Essentially, the purpose of $r_A$ and $h$ are to make Alice "commit" to her initial toss $x$. The point of $r_B$ is so that, without it, Alice might pick some $r_A$ for which she knows another string $r$ which might let her lie.

One party chooses a town in your state; the other party immediately states odd or even. Let's say the zip code ends in an even number.  So even would win and odd would lose.  This is how we have done a "coin flip" over the phone before.


We can use the Vasc's EV-method. See here the corollary 1.7 (b):

https://www.emis.de/journals/JIPAM/images/059_06_JIPAM/059_06_www.pdf

Indeed, let $a_1\leq a_2\leq...\leq a_n$, $a_1+a_2+...+a_n=const$ and $a_1^n+a_2^n+...+a_n^n=const$.

Thus, by EV $a_1a_2...a_n$ gets a minimal value in the following cases.

One of our variables is equal to $0$.

Let $a_n=0$.

In this case we need to prove that
$$(n-1)^{n-1}\sum_{k=1}^{n-1}a_k^n\geq\left(\sum_{k=1}^{n-1}a_k\right)^n$$ or
$$\frac{\sum\limits_{k=1}^{n-1}a_k^n}{n-1}\geq\left(\frac{\sum\limits_{k=1}^{n-1}a_k}{n-1}\right)^n,$$
which is Power Mean inequality;

$a_1=x$ and $a_2=...=a_n=1$, where $0\leq x\leq1$.

We need to prove that $f(x)\geq0$, where
$$f(x)=(n-1)^{n-1}(x^n+n-1)+n^nx-(x+n-1)^n.$$
But $$f'(x)=n(n-1)^{n-1}x^{n-1}+n^n-n(x+n-1)^{n-1}\geq n^n-n\cdot (1+n-1)^{n-1}=0,$$
which says that $f$ is an increasing function.

Thus, $f(x)\geq f(0)=0$ and we are done!

Maybe this can help you, it seems very similar.


What are NP-complete problems and why are they so important? I keep hearing questions about whether something is NP-complete, but they never really mention what it is. Why do people care so much about NP-complete problems?

A problem is in class NP if its solution may be verified in polynomial time, that is if the dimension of the problem is n you may be sure that for large enough n you need less than r·nk operations to verify the solution.

A problem is in class P if its solution may be found in polynomial time, instead. A problem in P is in NP by definition, but the converse may not be the case; probably the most important open question in computer science is whether classes P and NP are the same, that is P=NP.

NP-complete is a family of NP problems for which you know that if one of them had a polynomial solution then everyone of them has.

(EDITED) For the time being, only known algorithms for NP-complete problems are exponential in number of operations, so they are not practically solvable for n large.

A decision problem D is said to be NP-complete if
    1.it belongs to class NP;
    2.every problem in NP is polynomially reducable to D.

Reducable means we can convert one instance of problem A in to problem B[NP problem] then it means A is reducable to B.


How to solve: $\int \frac{1}{36x^2(1+12x)^2}\mathrm dx$ I was given the following problem: $$\displaystyle\int \frac{1}{36x^2(1+12x)^2}\mathrm dx$$ I was told to solve it using the following formula from an integral table: $$\int\frac{\mathrm du}{u^2(a+bu)^2}=-\frac1{a^2}\left[\frac{a+2bu}{u(a+bu)}+\frac{2b}a\ln\left|\frac{u}{a+bu}\right|\right]+C$$
I am not sure how to plug this in. I do see that you can pull $\frac1{36}$ out of the problem, and that we have $x^2$ and $(1+12x)^2$ in the denominator, but what do we need to set as our variables?

Let $u=6x$. Then $u^2 = 36x^2$, $2u=12x$, and $du=6 \;dx$. Your integral becomes

$$\int \frac{1}{36x^2(1+12x)^2}dx = \int \frac{1}{u^2(1+2u)^2}\cdot \frac 1 6 \; du$$

You can solve it from here using your formula.


Functional equation for distribution function I have next functional equation for some distribution:
$$\overline F_\xi^2(T) = \overline F_\xi(2T) \forall T>0$$
If suggest, that it differentiable, we can do something like this:
$$2\overline F_\xi(T) \cdot \overline F'_\xi(T) = 2\overline F'_\xi(2T)$$
$$\overline F_\xi(T) \cdot \overline F'_\xi(T) = \overline F'_\xi(2T)$$
But how to solve this differential functional equation?
where $$\overline F_\xi = 1 - F_\xi = P(\xi > T)$$
Also, I have $\mathbb E\xi=1.$

For brevity, put $f=\bar F_{\xi}$. Since $f$ is decreasing, if $f(x)=0$, a simple proof by induction shows $f(2^{-n}x)=0$ and therefore $f\le 0$ everywhere, which contradicts $\Bbb E[\xi]=1$. Hence, $f>0$ and we can define $g(t)=\ln f(t)$ so the hypothesis yields $2g(t)=g(2t)$. If we assume $f$ is analytic on $\Bbb R$, $g'(t)=g'(2t)$ and induction shows $g^{(n)}(t)=2^{n-1}g^{(n)}(2t)\Rightarrow g^{(n)}(0)=0$ for $n>1$ so the Taylor Series of $g$ reduces to $g(t)=a_0+a_1t$. Imposing the condition that $2a_0+2a_1t=2g(t)=g(2t)=a_0+2a_1t$, and equating coefficients forces $a_0=0$. The definition of $g$ shows $f(t)=e^{g(t)}=e^{at}$ so that $\xi\sim \text{Exp}(-a)$, but $\Bbb E[\xi]=-\frac 1a=1$, so $a=-1$.


Divisibility by $10^6$? Let $p_k$ be the $k^{th}$ prime number.

Find the least $n$ for which $(p_1^2+1)(p_2^2+1) \cdots (p_n^2+1)$ is divisible by $10^6$.

I have no idea where to start on this problem. Any help would be appreciated.

For each prime above $2$, $p^2+1$ is even. For $2^2+1=5$ so we only need to look for the first $5$ odd primes so that $5\mid p^2+1$.

$p^2\equiv-1\pmod{5}\iff p\in\{2,3\}\pmod{5}$. Scanning the primes: $3,7,13,17,23$ are the first $5$ so that $p\in\{2,3\}\pmod{5}$. Therefore, the product
$$
(2^2+1)(3^2+1)(5^2+1)\dots(23^2+1)
$$
will be divisible by $10^6$. If some $p^2+1$ is divisible by $25$, you may need fewer terms. Indeed, $7^2+1=50$, so we only need
$$
(2^2+1)(3^2+1)(5^2+1)\dots(17^2+1)
$$

if you want to solve $p_k^2+1\pmod 5=0$

$p_k^2+1-5\pmod 5=0$

$p_k^2-4\pmod 5=0$

$p_k^2-4=5q$

find the prime such that

$(p_k-2)(p_k+2)=5q$

so $p_k=5k+2$

or $p_k=5k-2$


Prove using Rolle's theorem that $x^2$ has $1$ root I'm having a hard time using Rolle's theorem in this proof and my professor said that it will likely be an exam question similar to the one above.

Let $f(x) = x^2$. Now $f(0) = 0$, and $f$ is strictly increasing on $(0, +\infty)$.
Suppose there is another $a > 0$ such that $f(0) = f(a) = 0$. Then by Rolle's theorem there exists a point $c \in (0, a)$ such that
$$f'(c) = 0,$$
but this is clearly impossible, because as we said earlier $f$ is strictly increasing and therefore $f'(x) > 0$ on $(0, +\infty)$. The same argument can be applied to the interval $(-\infty, 0)$ and we conclude that $0$ is the only root.

Hint: Consider the function $x^2$ over the interval $[-a,a]$

  Rolle's theorem: If a real-valued function $f$ is continuous on a closed interval $[a, b]$, differentiable on the open interval $(a, b)$, and $f(a) = f(b)$, then there exists at least one $c$ in the open interval $(a, b)$ such that
  $$f'(c) = 0.$$


learning/teaching approach to rigorous math with the goal of improving I will state this now: yes, this is a subjective question. But I feel the answers people give may benefit students.

I want to get better at doing non trivial proofs. Real analysis is standard material some students have to learn. I have, broadly speaking, two choices for a textbook: 1) something like "baby Rudin", where I may struggle for long periods of time on problems, often having little direction and no progress, or 2) something easier, where smaller theorems are developed in the book and where there are less sophisticated jumps of reasoning (demonstrated, required) in the proofs.

What are the preferences of educators/students for using one over the other? Do you have reason to believe one is more successful at actually helping students improve? How do you determine success?

My beliefs:
I tend to think the easier book may help me develop my own techniques (I could be wrong). I see more examples and I get more confidence in doing basic things. But more advanced material will probably be less accessible, so I know that at some point I will have to just deal with the sort of terseness that occurs in Rudin. Anyway, I can keep going back and forth about this.

If you haven't had much experience with writing proofs, I would suggest starting with something like Spivak's Calculus or, in fact, some abstract algebra, which most students find more accessible than analysis.

"Mathematics is no more computation than typing is literature." - John Allen Paulos

There are two things that alienate people from math: 1. mindless computation. 2. obscure proofs. While both are important, the emphasis must be on discovery, exploration, and understanding. Oliver Heavenside once said, "I don't reject my food because I don't understand digestion," when defending his use of certain mathematical ideas he didn't fully understand. Exploration is a prerequisite to proofs. To extend the analogy; How can we determine what is happening in digestion if we are unaware the stomach exists? Few things help people understand better than Geometry, Symmetry, and Patterns. Rigorous math requires motivation which for me occurs when the context and importance are apparent. I agree with your assessment that the "easier" book is better since it will likely provide more opportunity to contextualize and thereby motivate you in some direction that may be far removed from the proofs you would learn in the "more advanced" book.


Unique Properties of Matrix I am hoping you guys can point me in the right direction for what I should go read more on.

Goal: I have a sample data set of 3D matrices.  I want to determine if a new data point (3D matrix) is within the bounds of the original data set.

Is there some sort of unique property or some way to characterize these matrices?

I thought maybe I could approximate a characterization with the first 4 moments of the distribution, but I am wondering if there is something smarter and more exact out there that I just dont know about.

I am not a mathematician but have studied similar sounding things in grad school economics and stats classes.

It is not clear to me what a data point in terms of a 3D matrix is.  Could you clarify or given an example?


If $\lim a_n/n=a>0$ then $\lim a_n=\infty$ Prove that:
  if
  $$lim_{n\to\infty} \ \frac{a_n}{n} = a>0 \ \ \ \ \ \ \ (1)$$
  then
  $$lim_{n\to\infty} \ a_n = \infty \ \ \ \ \ \ \ (2)$$

My proof:
Using a definition of limit, (1) yields:
$$\forall\epsilon>0 \ \ \exists N \ \ \forall n>N \ \ \left|\frac{a_n}{n} - a\right|<\epsilon$$
which is equivalent to:
$$\forall\epsilon>0 \ \ \exists N \ \ \forall n>N \ \ n(a-\epsilon)<a_n<n(a+\epsilon) \ \ \ \ \ \ \ (3)$$
To prove (2) wee need to show that
$$\forall M \ \ \exists N \ \ \forall n>N \ \ a_n>M \ \ \ \ \ \ \ (4)$$
So, let $M$ be any positive real number. Then let's take $N=\frac{M}{a-\epsilon}$. From (3) we know that $a_n>n(a-\epsilon)>N(a-\epsilon)=M.$ So indeed (4) holds.

Is this proof correct?

Edit:

Ok, I rethought this once again and I think proof above is indeed wrong. So let's take a second try:

Using a definition of limit, (1) yields:
$$\forall\epsilon>0 \ \ \exists N \ \ \forall n>N \ \ \left|\frac{a_n}{n} - a\right|<\epsilon$$
which is equivalent to:
$$\forall\epsilon>0 \ \ \exists N \ \ \forall n>N \ \ n(a-\epsilon)<a_n<n(a+\epsilon) \ \ \ \ \ \ \ (3)$$
To prove (2) we need to show that
$$\forall M \ \ \exists \tilde{N} \ \ \forall n>\tilde{N} \ \ a_n>M \ \ \ \ \ \ \ (4)$$
So, let $M$ and $\epsilon$ be any positive real numbers.

From (3) we know that it exists $N$ s.t. $\forall n>N \ \ a_n>n(a-\epsilon)$.

Now let's take $\tilde{N}=max\left(N,\lceil\frac{M}{a-\epsilon}\rceil\right)$. We will show that $\forall n>\tilde{N} \ \ a_n>M$.

We have now two cases: $N\geq\lceil\frac{M}{a-\epsilon}\rceil$ or $N<\lceil\frac{M}{a-\epsilon}\rceil$.

Let's consider the first case. We have from (3)

$$\forall n>\tilde{N}=N \ \ a_n>n(a-\epsilon)>N(a-\epsilon)\geq \lceil\frac{M}{a-\epsilon}\rceil(a-\epsilon)\geq M$$

Now let's consider the second case. Because $\lceil\frac{M}{a-\epsilon}\rceil>N$, it holds from (3) that

$$\forall n>\lceil\frac{M}{a-\epsilon}\rceil \ \ a_n>n(a-\epsilon)>\lceil\frac{M}{a-\epsilon}\rceil(a-\epsilon)=M$$

No it is not correct. You did not define $ε$ in your proof. If you fix that you will also realize another error in your later reasoning.

Yes it looks more or less fine to me.

More briefly you could say something along these lines: definitely $a_n>\frac{an}{2}$ and since $b_n=n$ grows to infinity, so does $\frac{an}{2}$ and so does $a_n$.


Use Ito's formula to determine the stochastic differential equation satisfied by $V_t$ A stochastic process $V_t$ is defined by
$$V_t  =\sqrt{t(t+W_t^2)}$$
$W_t$ is the Wiener process and $t$ denotes the time ($t > 0$).
Use Ito's formula to determine the stochastic differential equation satisfied by $V_t$.

I have taken the known process to be just $W$. Then $a(x,t) = 0$ and $b(x,t)=1$.
The process $V$ is:
$$V_t = h(W_t,t)\ \text{where}\ h(x,t) = \sqrt{t(t+x^2)}$$

I then calculated the partial derivatives of $h(x,t)$ and substituted all my values into Ito’s formula.

However, no matter how many times I try, I don’t seem to get the correct answer and I am unsure as to why.

Set $$f(t,x) := \sqrt{t \cdot (t+x^2)}. \DeclareMathOperator \sgn{sgn}$$

By the chain rule,

$$\begin{align*} \frac{\partial}{\partial x} f(t,x) &= \frac{xt}{f(t,x)} \tag{1} \\
\frac{\partial^2}{\partial x^2} f(t,x) &= \frac{t}{f(t,x)} - \frac{x^2 t^2}{f(t,x)^3}\tag{2} \\
\frac{\partial}{\partial t} f(t,x) &= \frac{2t+x^2}{2f(t,x)} \tag{3}\end{align*}$$

In order to find the corresponding SDE, we have to express the expressions on the right-hand side in terms of $f(t,x)$ and $t$. From the definition of $f$, it follows that

$$x^2 = \frac{f(t,x)^2}{t}-t \tag{4}$$

If we plug $(4)$ into $(2)$ and $(3)$, we get

$$\begin{align*}
\frac{\partial^2}{\partial x^2} f(t,x) &= \frac{t}{f(t,x)} - \left( \frac{f(t,x)^2}{t}-t \right) \frac{t^2}{f(t,x)^3} = \frac{t^3}{f(t,x)^3} \\
\frac{\partial}{\partial t} f(t,x) &= \frac{t}{f(t,x)} + \left( \frac{f(t,x)^2}{t}-t \right) \frac{1}{2 f(t,x)} = \frac{1}{2} \frac{t}{f(t,x)} + \frac{f(t,x)}{t}. \end{align*}$$

Similarly,

$$\begin{align*} \frac{\partial}{\partial x} f(t,x) &= \frac{t}{f(t,x)} \sgn(x) \sqrt{\frac{f(t,x)^2}{t}-t} = \sgn(x) \sqrt{t- \frac{t^3}{f(t,x)^2}}. \end{align*}$$

Now if we apply Itô's formula, i.e.

$$V_t = f(t,W_t) = \int_0^t \frac{\partial}{\partial x} f(s,W_s) \, dW_s + \int_0^t \left( \frac{1}{2} \frac{\partial^2}{\partial x^2} f(s,W_s) + \frac{\partial}{\partial t} f(s,W_s) \right) \, ds,$$

we find that

$$V_t = \int_0^t \sgn(W_s) \sqrt{s} \sqrt{1- \frac{s^2}{V(s)^2}} \, dW_s + \int_0^t \left( \frac{V(s)}{s} + \frac{1}{2} \frac{s}{V(s)}+\frac{1}{2} \frac{s^3}{V(s)^3} \right) \, ds.$$

Note that the diffusion coefficient depends on the Brownian motion $(W_t)_{t \geq 0}$. Since $V_s$ does not contain any information whether $W_s$ is positive or negative, we cannot expect to express $\sgn(W_s)$ in terms of $V_s$. This means that we cannot write $(V_t)_{t \geq 0}$ as the solution of a stochastic differential equation.

I took a slightly different route

$$S_t = V_t^2 = t(t+x^2)$$
$$x^2 = \frac{V_t^2}{t} - t$$

$$dS_t = 2 V_t dV_t=\frac{\partial S_t}{\partial t} dt + \frac{\partial S_t}{\partial x} dW_t + \frac{1}{2} \frac{\partial^2 S_t}{\partial x^2} dt$$

$$\frac{\partial S_t}{\partial t} = 2t + x^2$$
$$\frac{\partial S_t}{\partial x} = 2tx$$
$$\frac{\partial^2 S_t}{\partial x^2} = 2t$$

and so

$$\begin{align*}
2 V_t dV_t &= \left(t + \frac{V_t^2}{t}\right)dt + 2t\left(\frac{V_t^2}{t} - t\right)^\frac{1}{2} dW_t + tdt\\
&= \left(2t + \frac{V_t^2}{t}\right)dt + 2\left(tV_t^2 - t^3\right)^\frac{1}{2} dW_t
\end{align*}$$

$$dV_t = \left(\frac{t}{V_t} + \frac{V_t}{2t}\right)dt + \left(t-\frac{t^3}{V_t^2}\right)^\frac{1}{2}dW_t$$

I'm slightly disconcerted that my $dt$ coefficient looks so different to saz's. I'll need to do some rearranging to check whether or not they actually agree...

Edit: As pointed out in the comments I missed the implicit absolute value in the square root and left out the second order term in $dS_t$. Blast.


Is $Cr^a\hat r$ always a conservative vector field? Is the vector field $\vec r(r, \theta, \phi) = Cr^a\hat r$, where $C, a\in \Bbb R$ are constants and $r \in \Bbb R^+ \cup \{0\}$ is the radial component, always a conservative vector field?  I really just want to know if I have a force field of this type can I always find a potential function?

Yes. The curl in spherical polars involves $\theta$ and $\phi$ derivatives of the radial component, but not $r$ derivatives.

Hence $\nabla \times \vec r = 0$ and therefore the vector field is conservative.

Added: As the grad of a scalar field in polar coordinates is given by

...it is easy to integrate your vector field to find a corresponding potential scalar field.


How can I calculate $\lim\limits_{x\rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e}{x}$? How can I calculate this limit?
  $$\lim_{x\rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e}{x}$$

I thought about L'Hospital because case of $\frac{0}{0}$, but I don't know how to contiune from this point..

I just would like to add one idea to what's already been said. When using L'Hospital, never be mechanic. And I don't mean just to test that you have the right kind of indetermination before each time you apply the rule (checking hypotheses is a must before using any theorem/property). But it is also useful to watch in detail the function you are taking limits to and try to figure out where the difficulty is, so you can work apart with that specific expression. Taking derivatives of everything is not always of help; and it's better if you take the control of all those formulas instead the other way around.

For instance, the first time you use L'Hospital here you get the limit
$$\lim_{x\to 0} \quad (1+x)^{\tfrac1x}\cdot \frac{x-(1+x)\log(1+x)}{x^2(1+x)}.$$
I put the first factor separate on purpose, because that one tends to $e$, while in the fraction both numerator and denominator go to $0$. Since the first factor bring no trouble, most of the times the better thing will be to leave it there and work separately with the quotient. Even more, the $(1+x)$ in the denominator is not what makes it $0$, so we could actually move it away too (and maybe cancel it with another $(1+x)$ from $(1+x)^{\tfrac1x}$).

So it's better to work with
$$\lim_{x\to 0} \frac{x-(1+x)\log(1+x)}{x^2},$$
where we can use L'Hospital again: once you know where this tends to, you can put it together with the rest.

If you go on, you'll have again a similar situation, and so it will be important to think how you write things and what part you leave aside, too.

Of course, never forget to collect all the partial results at the end in order to give the solution of the original problem. (Yes, it's very easy to make such mistakes when you're nervous or short of time. =S)


Determining whether a polynomial is monotonic on a given interval $I$ Consider a regression/data fitting problem with polynomials.
We have an extra constraint that the solution must be monotonic. So I have the following questions:

  For a given polynomial, how to efficiently determine if it is monotonic (at least in a given interval)?
  Furthermore, how to impose the monotonic constraint (again at least in a given interval)?

Thanks a lot in advance!

A polynomial $p(x)$ is monotonic on an interval $I$ iff its derivative $p'(x)$ is everywhere nonnegative or everywhere nonpositive on $I$, or equivalently if all of the roots of $p'(x)$ in the interior of the interval have even order.

As an manually workable example, consider a generic cubic polynomial $$p(x) := Ax^3 + Bx^2 + Cx + D.$$ Differentiating gives $$p'(x) = 3A x^2 + 2B x + C,$$ and (henceforth provided that $A \neq 0$) this has no real roots if $$\Delta := (2B)^2 - 4(3A)(C) = 4(B^2 - 3 A C)$$ is negative and a real double root if $\Delta = 0$. If $\Delta > 0$ then $p'$ has two single roots.

So, $p$ is monotonic on $\mathbb{R}$ iff $\Delta \leq 0$. On an interval $I$, $p$ is monotonic iff either (1) $\Delta \leq 0$ or (2) $\Delta > 0$ and neither of the roots of $\Delta$ (which can be computed with the quadratic formula) are contained in the interior of $I$.

Travis is right, hobs is mistaken I'm afraid. When the roots of the derivative have even multiplicity but are real, there exist values of which the derivative is strictly negative and values of which it is strictly positive. So your polynomial is not monotonic. I think hobs thinks about the matrix of which the derivative is the characteristic equation. Then the roots are the eigenvalues, and the matrix is definite.


A question regarding Frobenius method in ODE Suppose $b(x),c(x)$ are real functions analytic at $0$. Let $b(x)=\sum_{i=0}^\infty b_ix^i, c(x)=\sum_{i=0}^\infty c_ix^i$ on $(-R,R)$. Suppose $r$ is a double root of  $r(r-1)+b_0r+c_0=0$. It is well known that the differential equation $$x^2y''+xb(x)y'+c(x)y=0$$ has a solution of the form $$y_1=x^r(1+\sum_{i=1}^\infty a_ix^i),$$ where the series $\sum_{i=1}^\infty a_ix^i$ has radius of convergence $\ge R$ (e.g., see Tyn Myint-U, Ordinary Differential Equations). Another solution is of the form $$y_1\ln x + x^r(\sum_{i=1}^\infty A_ix^i).$$ Most books (including Tyn's) mention without proof that $\sum_{i=1}^\infty A_ix^i$ also has radius of convergence at least $R$.

Let $I(s)=s(s-1)+b_0s+c_0$. Then $A_i, i\ge 1$ satisfy the following recursive relation
$$I(r+i)A_i=-\sum_{k=0}^{i-1}A_k[(r+k)b_{i-k}+c_{i-k}]-2ia_i-\sum_{k=1}^{i}b_ka_{i-k}$$
where $A_0=0$. (Note that $2r=1-b_0$).

Proving that the power series $\sum_{k=1}A_kx^k$ has radius of convergence at least $R$
directly seems to be hopeless. I tried comparison test as in Coddington's book Introduction
to Ordinary Differential Equations, but with no success.

Does anyone have a proof, or a reference where a proof is given?

TCL, you are absolutely correct and the proof of Theorem 4.5 in my book is not complete. That $h_2$ has the desired radius of convergence can be shown similarly as for $h_1$; the only trick is to change the recurrence for $j\le m$ to avoid the division by $0$ at $j=m$. The details can be found in the proof of Theorem 4.13 which covers the present situation by the discussion at the end of this section on p139.

I have added this information to the errata which can be found on my webpage.

For record, I will give a direct proof here. First note that a power series $\sum_i d_ix^i$ has radius of convergence at least $R$ if and only if for any fixed positive number $t$ less than $R$, the sequence $|d_i|t^i, i\ge 0$ is bounded. (To prove, show that $\limsup |d_i|^{1/i}\le 1/t.$)

Let $s$ be such a number less than $R$. Let $B=\sum_{i=0}^\infty |b_i|s^i, C=\sum_{i=0}^\infty |c_i|s^i$. Since $\sum_{i=0}^\infty ia_ix^i$ has the same radius of convergence as $\sum_{i=0}^\infty a_ix^i$, the sequence $2ia_is^i,i=0,1,2\cdots$ is bounded, say by $D$. $\sum_{k=0}^i b_ka_{i-k}s^i$ is the $i$-th term of the product series of $\sum_{i=0}^\infty b_is^i$ and $\sum_{i=0}^\infty a_is^i$, so the sequence is bounded by a number say $E$. Choose $N$ such that $$\frac{(|r|+i)B+C}{|I(r+i)|}\le \frac{1}{2},$$ $$\frac{D}{|I(r+i)|}\le \frac{1}{4},$$ $$\frac{E}{|I(r+i)|}\le \frac{1}{4},$$ for all $i\ge N$. Choose $F\ge 1$ such that $|A_i|s^i\le F$ for $i\le N$.

Suppose $|A_j|s^j\le F$ for $j<i$. Then by the recursive relation, \begin{eqnarray*}|A_i|s^i &\le& F\frac{(|r|+i)B+C}{|I(r+i)|}+\frac{D}{|I(r+i)|}+\frac{E}{|I(r+i)|}\\
&\le& \frac{F}{2}+\frac{1}{2}\\
&\le& F
\end{eqnarray*}
This proves by induction that $|A_i|s^i, i=0,1,2\cdots$ is bounded by $F$.


About the use of Set algebra I'm stocked with the following exercise:  Show that $A\cap(B\setminus C) =(A\cap B)\setminus (A\cap C) $.

Proof (Attempt): $A\cap (B\setminus C)= A\cap(B\cap C^c) = (A\cap B)\cap C^c $ since intersection is associative. But then... Iv'e got no idea on a following step, or even if this is the right way to get the proof. So, any help would be grateful.

$A\cap (B\setminus C)= A\cap(B\cap C^c) = (A\cap B)\cap C^c=(A\cap B)\setminus(C)=(A\cap B)\setminus(C\cap A)$, since you're removing some elements from $A\cap B$ (the last step), so we take care of elements of $C$ in $A$.

If $x \in A \cap (B -C)$, then $x$ is in $A$ and $B$, but not in $C$. Then certainly $x$ is not in $A \cap C$, so $x \in A \cap B-(A \cap C)$. Conversely, if $x \in A \cap B-(A \cap C)$, then $x$ is in both $A$ and $B$, but not in $C$. Hence $x$ is in $A$ and $x$ is in $B-C$.


What difference does "for every $\epsilon$" vs. "for any $\epsilon$" make for the definition of convergence? The definition in my advanced calculus textbook of convergence for sequences is:

A sequence $\{a_n\}$ is said to converge to the number $a$ provided that for every positive number $\epsilon$ there is an index $N$ such that
$$|a_n - a| < \epsilon$$ for all indices $n \geq N$.

Say we replace "for every positive number $\epsilon$" with "for any positive number $\epsilon$". I'm wondering: what difference would that make on the definition of convergence?

I believe your question to be a language one, hence there is no difference in the formulation.

Well, .....
If for every positive $\epsilon$ this is true, then for any positive $\epsilon$ we look at will be true because this is true for every possible $\epsilon$
And if for any positive $\epsilon$ this is true, then the only way that can be true is if it true for every positive $\epsilon$; if there were any positive $\epsilon$ that it wasn't true for, than that $\epsilon$ would be an exception to it being true for any positive $\epsilon$.
......
So even though in english "any one you pick" and "every one you pick" (might) have conceptual contextual differences in a literal linguistic sense, in the world of mathematics (and even the world of practical english applications) they have no practical difference.
SO, they are the same.
....
I pity any non-native english speaker who tries to comprehend the above.


Meaning of Math Symbols What is the best way to learn the meaning of the math symbols beyond the alphabet or the cardinal numbers?

I understand some

ϵ>0 means epsilon is greater than zero

∃δ>0 means there exists delta greater than zero

How should I say these?

∀x,y∈R, δf∘g=min{δ,δ1}, and f∘g

Symbols carry meaning contextually. Sure, they have a literal "English" counterpart, but the sentence "plums deify" is pretty much nonsense, unless you're readying a fictional work in which plums are intelligent and have organized some kind of religion. In order to separate these kinds of nonsense sentences from the ones that really do make sense, you have to learn the context.

So to that end, I think your best bet is to figure out where those symbols you are looking at are coming from, and study some of that material. If I had to take a guess as to what you would be looking at when you encountered something like that, I'd venture to say it's some kind of elementary analysis text. So maybe try out a book like Ross's Elementary Analysis. It is very beginner friendly, and will teach you all the necessary notation used in the foundations of analysis.

Juast read and learn math and you'll see that there is nothing to learn ;)


Use Blumenthal's 0-1 law to solve the following From Durett: $B(t)$ represents a Brownian motion,

"$(i)$ Suppose $f(t) > 0$ for all $t>0$. Use Blumenthal's $0-1$ law to conclude that $\lim \sup_{t \rightarrow 0^+} \frac{B(t)}{f(t)} = c$, $P_0$ a.s., where c is a constant in $[0,\infty]$.

$(ii)$ Show that if $f(t) = \sqrt{t}$ then $c=\infty$, so with probability one Brownian Paths are not Holder continuous of order $1/2$ at $0$"

I understand that the random variable $\lim \sup_{t \rightarrow 0^+} \frac{B(t)}{f(t)}$ is in the germ sigma field thus we can apply Blumenthal's $0-1$ law to all events involving it.

$(ii)$ We need to show that $P_0(\lim \sup_{t \rightarrow 0^+} \frac{B(t)}{\sqrt{t}} \leq s) = 0$, $\forall s \in \mathbb{R}$. By Blumenthal's $0-1$ law this is equivalent to $P_0(\lim \sup_{t \rightarrow 0^+} \frac{B(t)}{\sqrt{t}} \leq s) < 1$, $\forall s \in \mathbb{R}$

$P_0(\lim \sup_{t \rightarrow 0^+} \frac{B(t)}{\sqrt{t}} \leq s) = \lim_{n \rightarrow \infty} P_0(\sup_{t \leq 1/n} \frac{B(t)}{\sqrt{t}} \leq s) \leq P_0(\frac{B(1/n)}{\sqrt{1/n}} \leq s) = \Phi(s) < 1 $

Where the last equality comes from the fact that $\frac{B(t)}{\sqrt{t}} \sim N(0,1)$

(i) I can't find a similar way to deal with this one, any idea ?

If every set in a sigma field has probability 0 or 1 then any function measurable with respect to it is almost surely constant. To prove this just note that the distribution function $F$ of this random variable has the property that $F(x)$ is 0 or 1 for every $x$. It is fairly easy to show that if $a=\inf \{x:F(x)=1\}$ then $F(x)=0$ for $x<a$ and 1 for $x \geq a$. Hence the function is almost surely equal to $a$.

Here is an easy way of seeing that $c=+\infty$ almost surely, in the case that $f(t) = \sqrt t$.

Indeed, note that $B_t/\sqrt{t}$ has a $N(0,1)$ distribution for all $t$, hence $P(B_t/\sqrt t>M)=c>0$ for all $M>0$ and $t>0$.

By "reverse Fatou's lemma", it holds that $$P\bigg( \limsup_{t\to 0} \frac{B_t}{\sqrt{t}} >M\bigg) = E \big[ \limsup_{t \to 0} 1_{\{t^{-1/2}B_t >M\}}\big] \geq \limsup_{t \to 0} P\big( B_t/\sqrt{t}>M)=c>0. $$Now apply Blumenthal to conclude that the probability must be $1$.


Write 1000 using three 13s The brother of a classmate, who is in elementary school, got the following homework, but he can't figure it out. So his brother shared it to see if someone in his class could help out with it:
"Write the number 1000 with 3 numbers 13". Only the standard arithmetic operations of addition, subtraction, multiplication, and division are allowed.
How does one solve this problem?

If you can break up the $13$'s:
$(13-1\times 3)^{1\times3} = 1000$.


Two expressions with similar arrangement of coefficients, does it imply that the variables are same? If $a|x|^2 + b|x| + c = a|y|^2 + b|y| +c$, then is it always $|x|=|y|$ ?

Or similarly,

if  $a|x|^n + b|x|^{n-1} + c|x|^{n-2} = a|y|^n + b|y|^{n-1} + c|y|^{n-2}$, then is it $|x| = |y|$ ?

More generally, does it happen that,

when,
$a|x| + b|y| + c|z|  =  a|m| + a|n| + c|p|$

[where, a,b,c are constants. Others are variables]

then,  $|x|=|m|$ , $|y| = |n|$ and $|z| = |p|$ ?

In other words, you're asking whether the functions
$$ x \mapsto ax^2+bx+c \\
(x,y,z) \mapsto ax+by+cz $$
are injective. No, they are not, in general.

$ax^2+bx+c$ can be injective, namely if (but only if) $a=0$ and $b\ne 0$.

$ax+by+cz$ is never injective no matter what $a$, $b$, and $c$ are.

In the first case, the answer is clear: yes. But not only. You have in general two solutions. Let be:
$$s=x-y$$
$$t=x+y$$
$$a(x^2-y^2)+b(x-y)=0$$
$$a(x-y)(x+y)+b(x-y)=0$$

Then $ast+bs=0$, $s(at+b)=0$ That bring us two solutions: $s=0$, $x=y$ and $t=\frac{-b}{a}$, $y=\frac{-b}{a}-x$


Is the surface of a donut distinguishable from pac-man's world? Pac-man's world is topologically like the surface a donut. Pac-man's world is also locally flat. For example, the interior angles of a small triangle will always add up to 180 degrees.

Conversely, when I picture a donut in my mind, it's not very flat. Dissimilar to pac-man's world. Yet I'm having trouble coming up with a way to distinguish between the donut's surface and pac-man's world from the inside of the space.

(When I say 'donut' I mean a literal donut embedded in 3d space, with a hole through the middle and all that, but from the perspective of ants walking on the surface trying to figure out if they're pac-man or not.)

(Maybe the fact that the inside diameter is smaller than the outside diameter?)

Your picture of the surface of a donut is correct: it is curved.

Not all metrics on a topological surface will have the same local properties.

For example, although a donut surface does indeed have flat metrics, it also has nonflat metrics, and the donut surface is one of them. 2d entities living on the surface of a donut could indeed distinguish their donut from pac-man's world, by making little measurements of length and area. For instance, they could measure the radius and area of a small circle on their world, and for almost any circle they chose, they would discover that the formula $A=\pi r^2$ fails, so their world is not flat.

More specifically, if they take a small circle centered at a point on the inside of the torus, $A > \pi r^2$ (the surface is negatively curved in the inside region). On the other hand, if they take a small circle centered at a point on the outside of the torus, $A < \pi r^2$ (the surface is positively curved in the outside region).

The portal thing where Pac-man goes to the mid-left side of the screen and appears on the mid-right side of the screen would also be explained by a donut world.


Can I calculate the product over the primes efficiently and with high precision? I want to calculate the following product : $$\prod_{p\ prime\ ,\ p\le 10^{10}} 1-\frac{1}{p}$$

I know the approximation formula $$\frac{e^{-\gamma}}{\ln(10^{10})}$$ where $\gamma$ is the Euler-Mascheroni-constant. The result should be good to $12$ decimal digits.

  Is there an efficient way (not brute force by determining all primes, which takes long with PARI/GP) to calculate this product with high precision ?

Call the product $P$, and take logs to get $\log(P)=\sum_{i=1}^n\log(1-1/p_i)=-\sum_{i=1}^n\frac{1}{p_i}-\frac{1}{2}\sum_{i=1}^n\frac{1}{p^2_i}+...\approx-\sum_{i=1}^n\frac{1}{p_i}-\gamma$

So to get a good approximation, use ever-better approximations of $\sum_{i=1}^n\frac{1}{p_i}$. As an example, the Meissel Merten constant is defined as $M:=\lim_{n\rightarrow\infty}\sum_{i=1}^n1/p_i-\ln\ln(n)\approx 0.26149$.

If you're still getting error, you can look into the error for finite sum approximations of $\gamma=\sum_{m=2}^\infty (-1)^m\frac{\zeta(m)}{m}$.


Finding the volume bounded by a cylinder and a plane I have been given the following equations:

$$x^2 + z^2 = 9$$
$$ x = 0 $$
$$ y = 0 $$
$$ z = 0 $$
$$ x + 2y = 2 $$

and have been asked to find the volume of the bounded region. I understand the principle of the integration and how to apply it in this scenario, but I tend to have trouble determining the bounds of integration for a three dimensional region such as the one described above. I have tried sketching the cylinder and the plane, but to no avail; I'm consistently unable to determine the bounds. In general, how would one go about determining the integral bounds for a region such as the above?

Sketching is indeed a good idea when you deal with non-trivial boundaries. In this case, the main boundaries are the coordinate planes $x=0$, $y=0$, and $z=0$. Then, by inspection, we notice that the given plane $x = 2- 2y$ makes the bound region (in $x$-$y$ plane) in the first octant, which means that our boundaries become

$x\in[0, 2-2y]$, $y\in [0, 1]$ and $z \in [0, \sqrt{9-x^2}]$.

The latter boundary appears due to the fact that the cylinder is oriented along the $z$-axis.

I guess practice of working with this type of problems bring essential experience and then each new problem will look easier and easier.

Hope this helps.

Apologies in advance, for bad English. You need to start off, by integrating left, integrating right and combining. For example, z component vector, you can integrate first in this case (because it is involved in the least equations). Then, you may go about, triple integrate the other ones (for example, x component first then y component). You will easily be able to determine the integral bounds after this, as you've integrated all three components.


A car goes from point A to point B exactly in three hours with $x(t)={59\over 3}t^2-t^3$ .Find the average velocity A car goes from point A to point B exactly in three hours.We assume that the car is moving in a straight line and that the distance of $x(t)$ from A at time $t$ ( where time is measured in hours and the distance in kilometers ) is given by the function $x(t)={59\over 3}t^2-t^3$.

$i)$what is the average velocity on this route?

$ii)$what is the instantaneous velocity of the car at the time of reaching to point B ($t=3$)

Could anyone give me any ideas ir hints for these two questions?

i) Average velocity is simply $$\frac{\Delta x}{\Delta t}=\frac{x(3)-x(0)}{3-0}$$

ii) The instantaneous velocity $v(t)$ is the derivative of $x(t)$ w.r.t. to $t$. Then the instantaneous velocity at point B (where $t=3$) is just $v(3)$.

First of
$$v(t)=\frac{dx(t)}{dt}$$
Which means in your case:
$$v(t)=\frac{118}{3}t -3t^2$$

I am assuming x(t) is given in units of $[x]=m$ and that means $v(t)$ will be in $[v]=\frac{m}{sec}$. If you want to find $v(3_{[hours]})$ you have to translate $t$ to units of $[t]=sec$ and then assert $t=\text{Translated Value}=t'$.

For the Average speed:
$$v_{Average}=\frac{1}{t_1-t_0}\int_{t_0}^{t_1} v(t)dt$$

In this case:
$$v_{Average}=\frac{1}{t'-0}\int_{0}^{t'} v(t)dt$$
$$\int v(t)dt=x(t)+c \implies v_{Average}=\frac{1}{t'-0}\cdot \left[x(t')-x(0)\right]$$


A polynomial $P(x)$ of degree $5$ with lead coefficient one,increases in the $(-\infty,1)$ and $(3,\infty)$ and decreases in the interval $(1,3)$ A polynomial function $P(x)$ of degree $5$ with leading coefficient one,increases in the interval $(-\infty,1)$ and $(3,\infty)$ and decreases in the interval $(1,3)$. Given that $P(0)=4$ and $P'(2)=0$, find the value of $P'(6)$.

I noticed that $1,2,3$ are the roots of the polynomial $P'(x)$.$P'(x)$ is a fourth degree polynomial. So

Let $$P'(x)=(x-1)(x-2)(x-3)(ax+b)$$
Now I expanded $P'(x)$ and integrated it to get $P(x)$ and use the given condition $P(0)=4$, but I am not able to calculate $a,b$.

As $P(x)=x^5+\ldots$, we have $P'(x)=5x^4+\ldots$. This gives us $a=5$.
Moreover, the facts that $P$ is decreasing on $(1,3)$ and $P'(2)=0$ imply that $x=2$ must in fact be a multiple root of $P'$. We conclude that $2a+b=0$, so $b=-10$.

The polynomial of degree 5, P(x) has a leading coefficient 1, has roots of multiplicity 2 at x=5 and x=0, and a root of multiplicity 1 at x=-2 find a possible formula for P(x).


What is the probability that Alex applies for his first sick leave on the fifth day? Alex will work on a consulting project for SALT Solutions for $5$ days.

During these $5$ days, the probability that Alex applies for sick leave on a particular day is the same, which is less than $0.5$.

The probability that Alex applies for his first sick leave on the second day is $0.21$.

The event that Alex applies for sick leave on a particular day is independent of the event that Alex applies for sick leave on other days.

What is the probability that Alex applies for his first sick leave on the fifth day?

What in the world is this question even saying? What does it mean that the probability that Alex applies for sick leave on a particular day is the same, which is less than $.5$? Is this saying that the probability for applying for each sick day is $.21$ independent of every other day since he does that on the second day? If so, is this just a geometric random variable with $p = .21$ so the answer is $$.21(1-.21)^4?$$

You are correct in saying that it is a geometric random variable. The first part where it says that the probability that he applies for sick leave on any of his days is the same simply satisfies the requirement for geometric distributions that the probability of the outcome occurring on each trial is the same (you can ignore the part that says that it is less than 0.5 since the question later gives you the exact probability of success which is 0.21). Since the probability of him applying for his first sick leave on any day is the same it means that the probability of him applying for sick leave on day "n" is the same as him applying for sick leave on day 2 which equals 0.21. It all really boils down to this being a tricky or poorly written question.

Day 2 means that the person didn't get sick on day one but got sick on day = .21 = p(1-p) => Quadratic Equation solve for p. p=.3 q=.7
That is why it states that p is less than .5

Solution

(.3).70^4 = .07203


Complex power series : Radius of Convergence Could anyone please suggest me how to deal with these questions (Complex Variable) :
Note that all problems are in $\mathbb{C}$.

3.1 Determine the radius of convergence $\rho$ of each of the following series :
$$\sum_{n=0}^\infty \frac{z^n}{n!}, \sum_{n=0}^\infty \frac{z^n}{n^2}, \sum_{n=0}^\infty \frac{z^n}{n}, \sum_{n=0}^\infty n!z^n $$
I use the ratio test and obtain the result : $\infty, 1, 1, 0$. , respectively.

The problem is : 3.2 For those with $\rho = \infty$, can you conclude anything about "convergence at infinity" ? I do not know what is "convergence at infinity". I search both in the text books and the internet already, but it seems no clues.

3.3 Find the radius of convergence of the power series :
$$\sum a_nz^n,$$
where $a_0 = 0, a_1 = 1, a_n= a_{n-1} + a_{n-2}$ for all $n > 1$.

I do not know how to handle this series since $a_n$ is defined recursively, so most of method for determining radius of convergence cannot apply.

I think the solution is provided by Pp.. in his comments but if you didn't get it then first follow this link  from Wikipedia which is about this series and its generating function. The second hint as provided by Pp.. is that you need to find the singularities of the generating function of this power series which is $f(z) = \frac{1}{1-z-z^2}.$ As is obvious the singularities are $\frac{\sqrt{5} \pm 1}{2}$. The minimum of which is $\frac{\sqrt{5} - 1}{2}$ and that is the radius of convergence.

PS: The other way to find the radius of convergence is through continued fraction you can go for that as well but for that you need to know what continued fraction are.

EDIT: Please look into this link.
Hope it helps.


Where does the notation $S^n$ for the $n$-dimensional sphere originate? Just as in the title. The notation is somewhat surprising in its use of a superscript rather than a subscript, but I gather there is a history of using $M^n$ to refer to an $n$-dimensional manifold. Still, it is perhaps a bit awkward in leading us to say things like "$S^1 \times S^1$ is not isomorphic to $S^2$". The notation could perhaps be understood in the context of a smash product, but this is likely anachronistic.

Anyway, enough speculation. The question is simple: where does the use of $S^n$ to denote the $n$-dimensional sphere originate?

We know that on one hand mathematical notation was developing evolutionarily, sometimes even ad hoc. On the other hand, mathematics is huge, requires a lot of notation, so much notation that it is hard to keep it short. Thus, at least in practice, consistency is often not attained across different times and domains.


Ring Theoretical Method of Solving a Math Olympiad Problem These paragraphs are from Steve Olson's book Count Down: Six Kids Vie for Glory at the World's Toughest Math Competition.

On page 170, the author said:

  The sixth and last problem on the Forty-second Olympiad ---
  by tradition the hardest of all --- looked deceptively straightforward to the competitors.

  Let $a > b > c> d$ be positive integers and suppose that $ac +bd =(b + d + a -c)(b + d -a + c)$. Prove that $ab + cd$ is not
  prime.

On page 174:

  Gabriel's answer to problem six demonstrated his power as a
  mathematician. "Gabe's solution was overkill," says Stankova,
  "but he solved the problem the way a mathematician would solve it." In his solution he used a mathematical idea called a ring --- a set of mathematical objects, any two of which can be added or multiplied to yield another member of the set.

Edit
On page 210:

  Gabriel's use of imaginary numbers in problem six was directly linked to the famous equation $e^{\pi i} = -1$. The number omega ($\omega$)
  is defined as $\omega = e^{2\pi i/3}$. So $\omega^2 = e^{2\pi i/3}\times e^{2\pi i/3}=e^{2\pi i/3+2\pi i/3}=e^{4\pi i/3}$ (because the exponents of $e$ can be added together when the two numbers are multiplied). By the same token, $\omega^3 =e^{6\pi i/3}=e^{2\pi i}=e^{\pi i}\times e^{\pi i}= -1 \times -1 = 1$.
  Thus the set of numbers $1, -1, \omega, -\omega, \omega^2$, and $-w^2$ are related in a particular way. If you multiply any two of them together, you get another member of the set. Gabriel used the powerful properties of this group to crack problem six.

The original problem: https://www.imo-official.org/problems.aspx (2001)

My Question: How did he solve the problem in a ring theoretical method?

Gabriel's original solution.
Posted with the permission of the author.


Solution for PDE $f f_x = a f_t + b$ Does anyone know the solution to this PDE, with $f = f(x,t)$

$$f f_x = a f_t + b$$

the boundary conditions are: $f(0,t) = 0$, $f(L,t) = \text{const}_1$, $f(x,0) = \text{const}_2$

Let us solve $af_t - ff_x = -b$ over $0\leq x \leq L$, $t>0$ with the prescribed initial and boundary conditions.
In some cases, the solution can be deduced from the method of characteristics.
Here, there are three families of characteristic curves. We have the curves
\begin{aligned}
&x = \frac{b}{2a^2} (t - t_0)^2 &&\text{where}\quad f = -\frac{b}{a}(t-t_0), && t_0>0\\
&x = \frac{b}{2a^2} t^2 - \frac{c_2}{a}t + x_0 &&\text{where}\quad f = -\frac{b}{a}t + c_2, && 0\leq x_0\leq L\\
&x = \frac{b}{2a^2} (t - t_0)^2 - \frac{{c}_1}{a}(t - t_0) + L &&\text{where}\quad f = -\frac{b}{a}(t-t_0) + c_1, && t_0>0
\end{aligned}
under the requirements $a\neq 0$, $b>0$. The solutions deduced from these relations are valid provided that they are unique (i.e. characteristic curves carrying different informations do not intersect). In the present configuration, characteristics with different information almost always intersect somewhere. Therefore, the nature and expression of the solution (shock, rarefaction, etc.) will be highly dependent on the values of the parameters.

$ff_x=af_t+b$

$af_t-ff_x=-b$

Hint:

Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:

$\dfrac{dt}{ds}=a$ , letting $t(0)=0$ , we have $t=as$

$\dfrac{df}{ds}=-b$ , letting $f(0)=f_0$ , we have $f=f_0-bs=f_0-\dfrac{bt}{a}$

$\dfrac{dx}{ds}=-f=bs-f_0$ , letting $x(0)=g(f_0)$ , we have $x=\dfrac{bs^2}{2}-f_0s+g(f_0)=g\left(f+\dfrac{bt}{a}\right)-\dfrac{ft}{a}-\dfrac{bt^2}{2a^2}$


Series convergence ques I was just solving another question regarding this topic and just wanna check if I am doing it right.

So the question is whether the given series converges or not, and justify.

$$\sum_{n = 1}^{\infty} \frac{(\ln{n})^2}{ n^2}$$

I have tried using ratio and comparison test, but that leads me nowhere.
For comparison test I compared the term with $\frac{1}{n^2}$. Am I doing that wrong?

Thank you so much :)

Recall that $f(s) = \sum_{n=1}^{\infty} n^{-s}$ converges when $s>1$.

Then we observe that $f''(s)= \sum_{n=1}^{\infty} \frac{(\ln(n))^{2}}{n^s}$, with the same region of convergence.

Hence $f''(2) = \sum_{n=1}^{\infty} \frac{(\ln(n))^{2}}{n^2}$.

$\sum_{n=1}^{\infty}\frac{(ln(n))^2}{n^2}< \sum_{n=1}^{\infty}\frac{ln(n) \sqrt{n}}{n^2}<\sum_{n=1}^{\infty}\frac{ln(n)}{n^{\frac{3}{2}}}<\sum_{n=1}^{\infty}\frac{1}{n^{\frac{3}{2}}}$

This is a p-series with p>1 so it converges. Thus $\sum_{n=1}^{\infty}\frac{(ln(n))^2}{n^2}$ converges as well.


is a^(m)b^(n) | m >= 99 and n>=999 a regular language? I've been stuck on this problem for a while. Say we have the following language?

a^(m)b^(n) | m >= 99 and n>=999

I'm trying to use the pumping lemma to prove that it isn't regular because intuitively to me it seems like it isn't. Can anyone please give me a concrete answer?

The language is regular here is a regular expression that fits for the language:

$$a^*(\underbrace{aa\dots}_{\text{99 times}})b^*(\underbrace{bbb\dots}_{\text{999 times}})$$

The language is the language accepted by the regular expression /aa...aa*bb...bb*/ where the number of a and b hiding in the dots is adjusted to needs.


How many substrings of length m can be formed from a string of length n? Simple counting problem but can't wrap my head around it

ABCD -- Substrings of length 2 : AB, BC, CD. Similarly ABCDE and length 3 -- ABC, BCD, CDE

I see m choose n won't work -- we'll get all subsequences too

Consider how many of the characters in the string are valid "starting characters" for the substrings you desire. For instance, in "ABCD," only "A", "B", and "C" are valid starting characters for substrings of length 2.

In general, in a string of length $n$, if you want a substring of length $m$, you can only choose the first $n - m + 1$ characters as starting characters.

It's the permutation topic.

1) length of main string : n

2) Select "k" number of charactered in ordered so,

Number of k-substring =

( (n!)/(n-k)! ) = (n-k+1) number of k-substrings in given string of length n.


number of infinite sets with different cardinalities I would like to know whether the number of infinity sets of different cardinalities is uncountable or countable. Is the proof simple?
Thanks.

Technically, neither one is true: the class of cardinalities is too large to form a set.  (However, having to choose between "countable" and "uncountable", certainly "uncountable" seems better: there is at least an uncountable set of cardinalities of sets.)

I think this is equivalent to a previous question on this site, but anyway: please see Fact 20 of

http://math.uga.edu/~pete/settheorypart1.pdf

and let me know if you have any further questions.

For any set $X,$ the set $P(X)$ of all subsets of $X$ has a bigger cardinality than $X$ itself (for $X$ is finite this is easy, for $X$ infinite you need a clever argument from Cantor, obtaineble in any set theory text). You can do this again, obtaining:

card(X) < card(P(X)) < card(P(P(X))) < ...


how to prove $\lim_{x\to \infty}\frac{n}{n^3+1}=0$ using definition of limits? $$\lim_{n\to \infty}\frac{n}{n^3+1}=0$$

I know that to prove this i have to find a $N \in \Bbb N$

$\forall   \epsilon>0$ s.t

if $n>N $ then $|\frac{n}{n^3+1}-0|< \epsilon$ i tried it some wayes but i couldnt find an answer

Using that $n\leq n^2$ and that $0\leq 1$ we get

$$\frac{n}{n^3+1}\leq \frac{n^2}{n^3+1}\leq \frac{n^2}{n^3}=\frac{1}{n}$$

I see it this way. You can divide both numerator and denominator by n to get

lim(n-->inf) 1/(n^2+1/n)

1/n --> 0 in the denominator, so the trend would be 1/n^2, which as n approaches infinity, the limit approaches 0.


If I can prove $Y$ from "$X$ is true" and from "$X$ is false", can I prove $Y$ without using $X$ at all? Suppose I have a statement $X$, for which I do not know whether it is true or false. And suppose further that I want to prove a statement $Y$:

I first assume that $X$ is true, and I construct an involved argument that shows that $Y$ follows.
I then assume that $X$ is false, and another involved argument shows that $Y$ still follows.

I have then proven $Y$ (assuming excluded middle). Now, is there any kind of meta-theorem in mathematics that tells me that there should be a proof of $Y$ that is not mentioning $X$ at all (whatever that means formally)?
You can think about this as follows:

Was it necessary to have formulated $X$ in order to give a proof for $Y$, or could we have proven $Y$ right away without the "halt in between"?

For example, the Riemann hypothesis seems to be such an important "halt in the middle" for some statements.
I know, this is a rather informal question.
I do not want to formulate it in terms of mathematical logic and proof theory, simply beacuse I do not know enough about these subjects.
Maybe the question has a trivial counterexample if we go formal enough.
But then again, maybe someone with enough knowledge still sees an interesting variant of the question.
I am also happy to receive quite technical answers.

It is hard to say about what does a proof "contains" in here, indeed if $T∪\{a\}⊢x$ and $T∪\{¬a\}⊢x$ then $T⊢x$ because in any model of $T$ either $a$ or $¬a$ are true, hence $T⊨x⇒T⊢x$, but how do you define "not mentioning"?
One interpretation you can think of is if $a$ is not in the language of $T$, that is:
Given a theory $T$ in the language $\cal{L}$, and given a language $\cal{L}'\supset\cal{L}$, is it possible to have a formula $a$ in the language of $\cal{L}'$ such that $T∪\{a\}⊢_{\cal{L}'}x$ and $T∪\{¬a\}⊢_{\cal{L}'}x$ and $T⊢_{\cal{L}}x$ is false?
***In this interpretation the answer is false:
Let $\cal{L}$ be empty and $\{f,g\}=\cal{L}'$ be 1-ary functional symbols, and let $a$ be the following theory:
$$(f\text{ is injective}∧f\text{ is not surjective})∧(g\text{ is injective}∧g\text{ is surjective})$$
And let $x$ be:
$$∃z,y\ (z≠y)$$
Notice that $a$ does prove $x$, as well as $¬a$ proves $x$, but $x$ is not tautology.***
***As Noah said, this example is wrong, $¬a$ does not prove $x$.
Although the above answer is wrong, I leave this post up and not deleting it because it is yet another possible interpretation of the question.

Proposition: The set $c\mathbb R = \{cx : x\in\mathbb R\}$ is closed.
You can prove this if you assume $c=0,$ and also if you assume $c\ne0.$ But I will be surprised if you can do it without considering those two cases.


Find out the general solution of $\tan x + \cot 2x = 2$. I want to solve for $x$ in

$$\tan x + \cot 2x = 2.$$

I tried to write in terms of $\sin x $ and   $\cos x$ but couldn't get the answer. Any help is appreciated.

$$\tan x+\cot2x=\dfrac{\sin x}{\cos x}+\dfrac{\cos2x}{2\sin x\cos x}=\dfrac{2\sin^2x+\cos2x}{2\sin x\cos x}=\csc2x$$  as $\cos2x=1-2\sin^2x$

$$\implies\sin2x=\dfrac12$$

Can you take it from here?

$\tan x+\cot 2x=2\implies(\sin x/\cos x)+(\cos2x/\sin2x)=0$.
Taking LCM,
$(\sin2x\sin x+\cos2x\cos x)/\sin2x\cos x=2$
{Using identity: $\cos(a-b)=\cos a\cos b+\sin a\sin b$}
$(\cos x/\sin 2x\cos x)=2\implies \sin2x=1/2=\sin(\pi/6) $
$2x=n\pi+(-1)^n (π/6)\implies x=nπ/2 +(-1)^n (π/12)$


Why do we use base $e$ in population growth questions? I know that we need base e to differentiate but I don't see what makes this formula work.

$$ P = P_0 e^{rt} $$

where the 0 refers to initial population, $r$ the rate, and $t$ the time.

Changing the base changes the curve, so why does base e work? I mean $r$ and $t$ are pretty straightforward numbers so there's no fancy constants (other than $e$). Why is it not base $2$ or something else?

Two issues: (1) Why is base $e$ used?; and (2) "Changing the base changes the curve." is wrong if you do things right.

Suppose the population doubles every $30$ years.  Then what is the population after $180$ years?

Notice that $180/30=6$, i.e. $6$ is the number of $30$-year periods and thus the number of doublings.  So the population will be $P_0\cdot2^6 = P_0\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2 = 64 P_0$.

What is the population after $t$ years?  The number of $30$-year periods is $t/30$, so that's the number of doublings, and it is $P_0 \cdot 2^{t/30}$.

Notice that base $e$ is nowhere used above.

But what is the instantaneous rate of change at time $t=0$?  It is
$$
\left.\frac{dP}{dt}\right|_{t=0} = P_0 \frac d {dt} 2^{t/30} = P_0 \cdot 2^{t/30} (\log_e 2)\cdot 1 {30}.
$$
If you're not dealing with instantaneous rates of change, then you don't need $e$.

Suppose now we're told the initial population is $P_0$ and it's growing at $P_0/40$ per year.  How long will it take to double?  It follows that
$$
P= P_0 e^{t/40}. \tag 1
$$
If $t$ is the doubling time, then $e^{t/40}=2$, so $\dfrac t {40} =\log_e 2$ and the doubling time is $t=40\log_e 2 = 27.725887\ldots\text{ years}$.  Again, we need $e$ only because instantaneous rates of change are involved.

Notice that $\dfrac d {dt} 8^t = (8^t\cdot\text{constant})$.  If the base had been $6$ rather than $8$, the constant would be different.  Only when the base is $e$ is the constant $1$.  That is what is "natural" about $e$.

Above we found that the doubling time is $27.725887\ldots$ years, and the first argument above shows that
$$
P = P_0 \cdot 2^{t/27.725887\ldots}. \tag 2
$$

Is $(1)$ different from $(2)$?  No.  They're the same.  Changing the base does not change the curve if $t$ is multiplied by the constant appropriate to the base in each case.

I've seen students write things like $P = P_0 e^{-(\ln 8) t}$ and then fail to do the simplification that says this is $P = P_0\left(\frac 1 8 \right)^t$.  If $t=2$, then you have $P=P_0\cdot\frac 1 {64}$, but sometimes students say $\ln 8 = 2.079$ and $e^{-(2.079)\cdot 2} = 0.015638804272$, and fail to notice that that is close to $1/64$, and think that by adding lots of digits they're making it very accurate.  Those later digits are garbage. Notice that $1/64 = 0.015625$ and compare that with that previous number.

$\displaystyle P = P_0 e^{-(\ln 8) t}$ is the same as $\displaystyle P = P_0\left(\frac 1 8 \right)^t$, so changing the base doesn't change the curve if things are done correctly.

Note that $e = \displaystyle \lim_{n\to \infty} \left(1+\dfrac{1}{n}\right)^n$, and this is the continuous growth or decay problem, then you take the limit as $n$ to infinity and get $e$.


Name for a function that can handle arbitrary number of arguments I am writing a proof, which uses functions that can handle an arbitrary number of parameters like summing, or averaging. In other words, I'm referring to functions like:

$f(x_1, ... x_n) = \sum_i x_i$, or

$f(x_1, ... x_n) = 1/n \sum_i x_i$, or even

$f(x_1, ... x_n) = \sum_i i \cdot x_i$

Clearly, $n$ can be arbitrary and these definitions would still make sense. I'd like to call these functions a name. Is there an existing name for such functions?

As an example, I'd like to be able to say: "Let $f$ be a ______ function that acts on some (or all) of the elements of $S$" (So here $f$ would be any function that accepts any number of elements from $S$ as arguments". What's the right way to word this?

As hinted by Fabio Somenzi and JonathanZ, in the context of computer science you can call these kind of functions variadic or aggregating.

By the way, you can easily define this kind of function in a mathematically rigorous way. The intuition is that a function $f$ would have an arbitrary arity, i.e. $f$ would accept any number of elements from a set $S$ as arguments. From a rigorous prospective, you can say the domain of $f$ is the set $S^*$ (denoted also by $S^{<\omega}$) of finite sequences of elements of $S$, i.e.

\begin{align}
S^* = \bigcup_{n \in \mathbb{N}} \overbrace{S \times \dots \times S}^{n \text{ times } S}.
\end{align}

So, for all $n \in \mathbb{N}$, $s_1, \dots, s_n \in S$ iff $(s_1, \dots, s_n) \in S^*$: in other words, $S^*$ contains all and only the tuples of elements of $S$, for every (finite) length.

Strictly speaking, the function $f$ defined in this way is unary (with respect to the domain $S^*$), but it perfectly translates the idea that $f$ has an arbitrary arity (with respect to $S$). How powerful the language of mathematics is!

The term for the number of arguments a function takes is arity and for arbitrary (but finite) arguments is the somewhat unimaginative n-ary.


Is an irreducible, asymmetric adjacency matrix diagonalizable? I am wondering whether the following statement is true or not. My computer has so far not been able to provide me with any counterexamples, nor have I been able to find much in the way of arguments to support it. I might just be missing something obvious, but I'm hoping that someone is able to help me. The statement is:

Let $D$ be a simple, directed graph with adjacency matrix $A\in\{0,1\}^{n \times n}$. If $D$ is strongly connected, then $A$ is diagonalizable.

Kind regards,
Pepijn

Edit: somehow managed to forget the essential requirement in the body of the question.

Take $$ A =
\begin{pmatrix}
0 & 1 & 0 \\
1 & 1 & 1 \\
1 & 0 & 1
\end{pmatrix} %A={{0,1,0},{1,1,1},{1,0,1}}$$
The directed graph of this matrix is strongly connected,which implies $A$ is irreducible.
But this matrix is not diagonalizable, since $0$ is an eigenvalue with multiplicity $2$, and $\operatorname{rank} A=2$.


Pre-calculus algebra logarithm question I don't understand how to solve this equation. Been struggling with it and don't know how to start:
$$\log_2x=8+9\log_x2$$
Can someone please help me out?

$\log_2x-8-9\log_x2=0$

Using $\log_x2=\frac{1}{\log_2 x}$ and setting $u=\log_2 x$ we get:

$u-8-\frac{9}{u} \Longleftrightarrow u^2-8u-9=0 \Longleftrightarrow (u-9)(u+1)=0$

This gives $\log_2x=-1 \Longrightarrow x=\frac{1}{2}$ or $\log_2x=9 \Longrightarrow x=512$.

Thanks guys for the hints, however I found another way to solve it and thought I would share it here:
Let $z=log_2x$ and $y=log_x2$
thus we have $x=2^{z}$ and $x=2^{y}$ therefore $2^z=2^{y} , z=y$

plugging that into our equation we get:

$z=8+9y , z=8+9z, z=-1, y=-1$ thus $ x=\frac{1}{2}$


Comparing genus of domain and image for maps between Riemann surfaces. I've been asked to show that if $R$ and $S$ are compact, connected Riemann surfaces, and $f: R \to S$ is holomorphic then $g(R) \ge g(S)$ (g is the genus).

Now surely this fact follows from the fact that $f$ is an open map hence $f(R)$ is clopen hence $f(R) = S$ and so the surfaces are homeomorphic hence they have the same genus? But this shows in fact $g(R) = g(S)$ which is stronger? Have I missed something obvious here?

Thanks

Well, I reiterate use the Riemann-Hurwitz relation:

$$\chi(R)=n\chi(S)-B$$ where $B$ is the sum of all the ramifications.
Where $\chi(S)=2-2g(S)$ etc is the Euler characteristic.
Now if $g(S)=0$ then the result is clear. So assume that $g(S)\geq 1$ and thus
$\chi(S)\leq 0$ this means that

$$\chi(R)=n\chi(S)-B \leq \chi(S)$$

And $\chi(R) \leq \chi(S)$ gives $g(S)\leq g(R)$.

pull the holomorphic forms from $S$ back to $R$ and show they are independent.

Note that the genus is equal to the number of independent holomorphic 1-forms. That the pull-back of independent one forms is independent is obvious. So $g(S)\leq g(R)$.


Localizing a ring of invariants Let $A$ be a domain and $G$ a finite group of automorphisms of $A$. I define $$A^G=\{a\in A\mid\sigma(a)=a ,\forall\sigma\in G\}.$$ Furthermore let $S\subset A$ be multiplicatively closed such that $\sigma(S)\subset S$ for all $\sigma\in G$ and we write $S^G=S\cap A^G$. I want to show that $$(S^G)^{-1}A^G\cong(S^{-1}A)^G.$$ (Atiyah and Macdonald, Introduction to Commutative Algebra, Exercise 12, Chapter 5.)

First off, I'm not really sure what is meant by $(S^{-1}A)^G$ here, so it would be nice if somebody could try to guess what would be the most natural interpretation of this.

Regarding the general procedure: I only know that $A$ is integral over $A^G$, which would give me $(S^G)^{-1}A$ being integral over $(S^G)^{-1}A^G$, though I'm not sure if this would be helpful even if I knew what $(S^{-1}A)^G$ was. So depending on this, maybe a little hint would also be nice.

Thanks in advance.

The following works without demanding that $A$ is a domain. There is a trick using the finiteness of $G$ that reminds one of the norm in Galois theory: Take $\frac{a}{s} \in (S^{-1}A)^{G}$, i.e., $$\frac{a}{s} = \sigma\left(\frac{a}{s}\right) = \frac{\sigma(a)}{\sigma(s)}$$ for all $\sigma \in G$. We can write $$\frac{a}{s} = \frac{a \prod_{\sigma \in G \setminus \{\rm id\}} \sigma(s)}{\prod_{\sigma \in G} \sigma(s)},$$ having the advantage of the denominator $s'$ being $G$-invariant. Let $a'$ be the numerator. Since now $$\frac{a'}{s'} = \frac{\sigma(a')}{s'}$$ for all $\sigma \in G$, for each $\sigma$ there is some $u_{\sigma} \in S$ such that $$u_{\sigma} s' (a' - \sigma(a')) = 0,$$ and multiplying all those $u_{\sigma}$ gives a $u \in S$ working for all $\sigma \in G$ simultaneously, $$u s' (a'-\sigma(a')) = 0.$$ Finally, using the same trick again, let $u' := \prod_{\sigma \in G} \sigma(u)$. Then $u' \in S^{G}$ and $$u's'(a'-\sigma(a')) = 0$$ for all $\sigma \in G$. Setting $\overline{a} := u's'a'$, for all $\sigma$ we have $$\sigma(\overline{a}) = \sigma(u')\sigma(s')\sigma(a') = u's'\sigma(a') = \overline{a},$$ so $\overline{a} \in A^G$ and hence $$\frac{a}{s} = \frac{a'}{s'} = \frac{u's'a'}{u's'^2} = \frac{\overline{a}}{u's'^2} \in (S^G)^{-1}A^G,$$ as desired.

The injectivity of the canonical map $(S^G)^{-1}A^G \rightarrow (S^{-1}A)^G$ can also be dealt with using this "norm trick".

There is a natural injection $(S^G)^{-1} A^G \hookrightarrow (S^{-1}A)^G$. To show that this is an isomorphism, you must show that every invariant fraction is a fraction of invariants.

Here are some hints.

So assume $\sigma \cdot \frac{a}{s}= \frac{a}{s}$. This is equivalent to $s \sigma (a) = a \sigma (s)$.

Since $G$ is a finite group, there is some $n$ such that $\sigma^n=id$. Then note that

$$
\frac{a}{s} = \frac{\sigma^{n-1}(s) \sigma^{n-2}(s)\ldots \sigma(s)a }{\sigma^{n-1}(s) \sigma^{n-2}(s)\ldots \sigma(s)s} = \frac{\sigma^{n-1}(s) \sigma^{n-2}(s)\ldots \sigma(a)s }{\sigma^{n-1}(s) \sigma^{n-2}(s)\ldots \sigma(s)s}
$$

By induction one can show that the numerator is $\sigma^{n-1}(a)\sigma^{n-2}(s) \ldots \sigma(s) s$.

Now its not difficult to see that both the numerator and denominator are invariant.


Kernels and Cokernels of differentials on a bounded long exact sequence. Given a bounded long exact sequence on a smooth projective variety where objects are direct sums of $\mathcal{O}(n)$ for different integer $n$'s i.e. $\bigoplus_i\mathcal{O}(n_i)$. Is there anything we can tell about the kernels and cokernels of the differentials? Like are they necessarily generated by extensions of line bundles $\mathcal{O}(n)$ or they can be very complicated vector bundles?

Question: "Like are they necessarily generated by extensions of line bundles O(n) or they can be very complicated vector bundles?"
Answer: Hartshorne Corr. II.5.18 says: If $X \subseteq \mathbb{P}^n_A$ is projective with $A$ noetherian, it follows any coherent sheaf $F$ is a quotient
$$ f_1:\oplus_{i(1)=1}^{N(1)} \mathcal{O}(n(1)_{i(1)}) \rightarrow F \rightarrow 0.$$
If $A$ is noetherian it follows $X:=Proj(A[x_1,..,x_n]/I)$ with $B:=A[x_i]/I$ noetherian. Hence if $F$ is a coherent module it follow any submodule $E \subseteq F$ is coherent. It follows $\oplus \mathcal{O}(n(1)_i)$ is coherent hence $ker(f_1)$ is a coherent module. Hence there is a surjection
$$ f_2: \oplus_{i(2)}^{N(2)} \mathcal{O}(n(2)_{i(2)} \rightarrow ker(f_1)$$
and you get a long exact sequence of $\mathcal{O}_X$-modules
$$S1.\text{   }\cdots \rightarrow  \oplus_{i(2)}^{N(2)} \mathcal{O}(n(2)_{i(2)} \rightarrow  \oplus_{i(1)=1}^{N(1)} \mathcal{O}(n(1)_{i(1)}) \rightarrow F \rightarrow 0.$$
If $E$ is a finite rank locally free sheaf, it follows $E^*$ is coherent, hence there is a surjection
$$f: \oplus_{i=1}^m \mathcal{O}(n_i) \rightarrow E^* \rightarrow 0$$
Dualize to get
$$0 \rightarrow E \cong E^{**} \rightarrow \oplus_i \mathcal{O}(-n_i) \rightarrow \cdots .$$
Hence any finite rank locally free module can be realized as a submodule of a direct sum $\oplus_i \mathcal{O}(-n_i)$.
The "syzygy theorem" says that such a sequence is finite when $X$ has an open cover $U_i:=Spec(A_i)$ where $A_i$ is a polynomial ring over $k$.
In such cases we get an exact sequence  with a finite number of terms
$$0 \rightarrow E_{k+1} \rightarrow \oplus_j \mathcal{O}(n(k)_j) \rightarrow \cdots \rightarrow E_1 \rightarrow 0.$$
Example: Projective space is such a variety: $\mathbb{P}_k^N$ has an open $D(x_i) \cong \mathbb{A}^N_k$. Hence on projective space it follows any coherent sheaf has a finite resolution of direct sums of invertible sheaves. A similar result holds for the grassmannian variety.
What is true is that by HH.II.5.18 for any coherent sheaf $F$ you always get a long exact sequence where the terms in the sequence are finite direct sums of  linebundles. Note that for the grothendieck group $K_0(X)$ if there is a long exact sequence
$$ 0 \rightarrow F_k \rightarrow \cdots \rightarrow F_2 \rightarrow F_1 \rightarrow 0$$ you get a relation in $K_0(X)$:
$$[F_1]=[F_2]-[F_3]+ \cdots +(-1)^k[F_k].$$
We get
$$[\mathcal{O}(n)]=[\mathcal{O}(1)^{\otimes n}]=[\mathcal{O}(1)]^{ n}$$
hence if the $F_2,F_3...$ are direct sums of linebundles it follows
$$[F_1]\in  \{ \sum_i [\mathcal{O}(n_i)]:\text{$n_i\in \mathbb{Z}$} \} \subseteq K_0(X).$$
If $F$ has a finite resolution of invertible sheaves for any finite rank vector bundle $F$ on $X$ it follows $K_0(X)$ is generated by the invertible sheaves $\mathcal{O}(n)$ for $n \in \mathbb{Z}$. Hence there is a surjection
$$ \rho: \mathbb{Z}[t, t^{-}] \rightarrow K_0(X)$$
defined by $\rho(f(t^i)):=f([\mathcal{O}(i)]), i=1,-1$, and an isomorphism
$$I1.\text{  }K_0(X) \cong \mathbb{Z}[t, t^{-1}]/(ker(\rho)).$$
There is an isomorphism $K_0(X)_{\mathbb{Q}}\cong CH^*(X)_{\mathbb{Q}}$
where $CH^*(X)$ is the Chow group of $X$ and $CH^*(X)$ is "infinite dimensional". But this does not immediately contradict the isomorphism I1.
Question: "Is there anything we can tell about the kernels and cokernels of the differentials? Like are they necessarily generated by extensions of line bundles O(n) or they can be very complicated vector bundles?"
Answer: As shown above: Any coherent module is the cokernel of such a map. The kernel of such a map is always coherent. If $E$ is finite rank and locally free, it follows $E$ can be realized as a subsheaf of such a finite direct sum. In the case of projective space it follows any coherent sheaf $E$ has a finite resolution by finite direct sums of invertible sheaves and you get an equality
$$ K_0(\mathbb{P}^N_k)\cong \mathbb{Z}\{[\mathcal{O}(n)]:n \in \mathbb{Z} \}.$$
The projective bundle formula confirms this calculation: It gives
$$K_0(\mathbb{P}^N_k)\cong \mathbb{Z}[t]/(t^{N+1})$$
where $t:=1-[\mathcal{O}(-1)]$.
https://en.wikipedia.org/wiki/Hilbert%27s_syzygy_theorem
Comment: "This doesn't answer the question as it is not known whether this resolution will end or not. If you re-read the question it is about a bounded long exact sequence on such objects."
Answer: It is well known the grothendieck group $K_0(X)$ is not generated by invertible sheaves in general, hence the answer to your question is: Not in general. For projective space it is true as shown above.


Difference between pure and mixed strategy In game theory, what is the difference between pure and mixed strategy?
I arrived to that mixed strategy related with probability of the other agent do the action.
I'm not sure that the result is correct.
and even this result is correct, that is not enough.

The set of strategies available to a player is a full list of the choices he can make.

There are two major conventions to interpret this definition. The first one, following natural language, is that there is a set of choices that the player can make. These "native" options are called pure strategies. F.i., in a Prisoners' dilemma, you can stay mum (cooperate) or incriminate your partner (defect).

A more extensive interpretation is that a player can inject randomness in his choice, by selecting a randomising device to pick one of his pure strategies. F.i., he may decide to toss a coin and then cooperate/defect based on getting heads/tails. The options to randomise over pure strategies are called mixed strategies. The set of pure strategies can be associated with the subset of mixed strategies that pick a specific pure strategy with probability one.

The mathematical motivation for mixed strategies is that they are necessary to prove existence of a value in zero-sum two-player finite games or existence of equilibrium in finite games. An intuitive motivation is that sometimes your best option is making yourself unpredictable: f.i., think of playing hide-and-seek --- if you choose your hiding spot randomly, you give the seeker no clue where to look for.

A player is considered to have a number of options and that the game will be played multiple times.  A pure strategy says the player will always select the same option.  A mixed strategy will choose different options different times, usually based on a random selection with some probabilities.  For example, in rock-paper-scissors, a pure strategy would be to always play rock.  A mixed strategy would be to play rock $1/3$ of the time, paper $1/2$ of the time, and scissors $1/6$ of the time based on rolling a die.  Some games make a pure strategy better than any mixed strategy, some make a mixed strategy optimal.  For rock-paper-scissors you should play all options equally often.  I don't understand anything after your question mark.


Pointwise and uniform convergence of $\sum\limits_{n=1}^{+\infty}(-1)^n\frac{x}{x+e^{-nx}}$ $\quad$ $x\in R$ How can I prove pointwise and uniform convergence of $\sum\limits_{n=1}^{+\infty}(-1)^n\frac{x}{x+e^{-nx}}$ $\quad$ for $x\in \mathbb{R}$?

uniform convergence: the pointwise convergence is on $E=(-\infty,0]$.

$f_n(x)={\frac{x}{x+e^{-nx}}}$

sup$_{E}|{\frac{x}{x+e^{-nx}}}|=f_n(\frac{-1}{n})={\frac{1}{1-ne}}$ general term of a divergent series.So there is not uniform convergent in E.
But if I consider intervals as $(-\infty,a],a<0$ there is uniform convergence?

You can only try to use Leibnitz criterion if $x<0$ because otherwise $|a_n|$ is not decreasing and convergent to zero. In any case, for $x >0$, the series is divergent as the general term does not go to zero.

Series is not even point wise convergent . Look at $x=1$.


Real $2\times 2$ matrix $X$ such that $X^2 + 2X= -5I$ Find a real $2\times 2$ matrix $X = \left(\begin{matrix} a& b\\ c & d\end{matrix}\right)$ such that $X^2 + 2X = -5I.$

With this question I'm kinda lost with the $2X$ part but a full explanation or a little bit of a lead would be greatly appreciated.

For a $2\times 2$ matrix $X$ its characteristic polynomial has the form
$$\chi_X(\lambda)=\lambda^2-\operatorname{tr}(X)\lambda+\det X$$
and by the Cayely-Hamilton theorem we have so
$$X^2-\operatorname{tr}(X)X+(\det X)I=0$$
so to answer the question it suffices to construct $X$ such that $\operatorname{tr}(X)=a+d=-2$ and $\det X=ad-bc=5$. Take for example
$$X=\begin{pmatrix}-1&4\\-1&-1\end{pmatrix}$$

Expanded you get that

\begin{matrix}
        5 + a (2 + a) + b c & b (2 + a + d) \\
        c (2 + a + d) & 5 + b c + d (2 + d)  \\
        \end{matrix}

equals 0.

Solving using Mathematica 9

X = {{a, b}, {c, d}};
Id = IdentityMatrix[2];
Solve[ X.X + 2 X == -5 Id, {a, b, c, d}]

Atleast these are possible answers:

{a -> -1 - 2i, b -> 0, c -> 0, d -> -1 - 2i},
{a -> -1 - 2i, b -> 0, c -> 0, d -> -1 + 2i},
{a -> -1 + 2i, b -> 0, c -> 0, d -> -1 - 2i},
{a -> -1 + 2i, b -> 0, c -> 0, d -> -1 + 2i}


Find the number of series Find the number of series $(a_1,..., a_{2n})$ that have terms from ${\{0,...9\}}$ so that:

$$
11|\sum_{i=1}^{n}a_i-\sum_{i=n+1}^{2n}a_i
$$

(this is not a homework)

There is a similar problem (Problem 6) in the IMO 1995.

For a fixed $n$, let $ A_r $ be the number of sequences of $n$ terms, whose sum is $r \pmod{11}$. Then, we want the value of $ \sum A_r ^2 $.

As pointed out by ABC, the generating function $f(x) = \left ( 1 + x + x^2 + \ldots + x^9 \right) ^n $ gives us the number of sequences of $n$ terms, with a total sum of $r$. We find the sum of those coefficients with the power mod 11, we use the typical counting trick. Let $\omega$ be a primitive 11th root of unity. We know that $ 1 + \omega + \omega^2 + \ldots + \omega^{10} = 0 $. For $ i \not \equiv 0 \pmod{11}$, we have $ f( \omega^i) = ( -\omega^{10i})^n$

We have $$ 11 A_r = \sum_{i=0}^{10} \omega^{-ri} f( \omega^i) = 10^n + \sum_{i=1}^{10} \omega^{-ri} (- \omega^{10i})^n = 10^n + (-1)^n \sum_{i=1}^{10} \omega^{(10n-r)i}. $$

In particular, this shows that for $r \neq 10n \pmod{11}$, we have $A_r = \frac{10^n-(-1)^{n}}{11}$. Otherwise, for $R \equiv 10n \equiv -n \pmod{11}$, we have $A_R = \frac{10^n+10 \times (-1)^n}{11}$.

Hence, we can now find the sum as

$$ \sum_{r=0}^{10} A_r^2 = 10 \left (\frac{10^n-(-1)^n}{11} \right)^2 +  \left( \frac{10^n+10\times (-1)^n}{11} \right)^2$$

Note: This solution is nice 9 because 2 less than 11, which makes $f(\omega^i)$ easy to evaluate. For a much more general case, the equations would not be pretty.

Since the second term is larger than the first term

-Sum = (2n * (2n+1))/2 - 2*(n*(n+1))/2
= n*(2n+1) -n*(n+1)
=n*n

(Sorry not used to site so plain english).

So 11 divides the series iff 11 divides n.


Find the radius of convergence for the Taylor series of $f(z)$ at $z=0$. Let , $$f(z)=\frac{z}{\sin (z^2+z)}.$$
$(1)$ Define $f(0)$ so that $f$ becomes analytic  at $z=0$.
$(2)$ Find the radius of convergence for the Taylor series of $f(z)$ at $z=0$.

First part is done. If we define $f(0)=1$ then $f$ becomes analytic.
Now Taylor series expansion is $f(z)=\sum_{n=0}^{\infty}a_nz^n$ , where , $a_n=\frac{f^{(n)}(a)}{n!}$. From here how we find $a_n$ ?

$$\lim_{x\to 0}\frac{x}{\sin(x+x^2)}=\lim_{x\to 0}\frac{x}{x+x^2}\cdot\frac{x+x^2}{\sin(x+x^2)}=1$$
and the radius of convergence of the Taylor series in $x=0$ is given by the distance from the origin of the closest singularity. $\sin(z)$ vanishes only for $z\in \pi\mathbb{Z}$, hence:

$$ \rho = \frac{-1+\sqrt{1+4\pi}}{2}=1.34162771851\ldots$$

Hint: Let us firstly compute $f'(0).$ We have $f'(0)=\lim_{z\to 0} \frac{f(z)-f(0)}{z}=\lim_{z \to 0} \frac{z-\sin(z^2+z)}{z \sin(z^2+z)}.$ Since $\sin(z^2+z)=z^2+z+o(z)$ then $a_1=f'(0)=1.$
We have $$f'(z)=\frac{\sin(z^2+z)-z(2z+1)\cos(z^2+z)}{\sin^2(z^2+z)}.$$ Try to find recurrence formula to get the expression of $f^{(n)}(z)$ for each $z$ different from $0$ and take the limit after  that of $z$ to $0.$