Chinese Remainder Theorem RSA I want to solve the following modular quadratic equation:

$x^2 \equiv 188 \pmod {437}$ using the fact that $437$ can be factorized by the primes as: $19⋅23$.

So far I have done:

$$x^2 \equiv 188 \pmod{19} \Rightarrow x^2 \equiv 17 \pmod {19}\\
x^2 \equiv 188 \pmod {23} \Rightarrow x^2 \equiv 4 \pmod {23}$$

$$x^2 \equiv 7 \pmod{ 19} \Rightarrow x \equiv \pm 6 \pmod {19}\\
x^2 \equiv 4 \pmod {23} \Rightarrow x \equiv \pm 2 \pmod {23}$$

And the last step. But I do not understand how to combine:

$$x \equiv  6 \pmod {19} \text{ and } x \equiv  2 \pmod {23} \Rightarrow x \equiv \text{"insert-her"} \pmod {437}\\
x \equiv -6 \pmod {19} \text{ and } x \equiv  2 \pmod {23} \Rightarrow x \equiv \text{"insert-her"} \pmod{ 437}\\
x \equiv  6 \pmod {19} \text{ and } x \equiv -2 \pmod {23} \Rightarrow x \equiv \text{"insert-her"} \pmod {437}\\
x \equiv -6 \pmod {19} \text{ and } x \equiv -2 \pmod {23} \Rightarrow x \equiv \text{"insert-her"} \pmod {437}$$

Please help me fill the "insert-her".

In general, we have distinct primes $p,q$; we have integers $a \in [0,p)$ and $b \in [0,q)$; and we want $x$ such that $x \equiv a$ mod $p$ and $x \equiv b$ mod $q$.

To solve this equation, the first step is to calculate $r = p^{-1}$ mod $q$. Assuming we have done that, the rest is easy: just set $x = (b-a)rp + a$. Then $x \equiv a$ mod $p$, obviously; and $rp \equiv 1$ mod $q$, so $x \equiv (b-a)rp + a \equiv (b-a)+a \equiv b$ mod $q$.

In practice, you will usually want $x$ to satisfy $0 \le x < pq$; this can be done by reduction mod $pq$.

So the only thing left is how to calculate $r = p^{-1}$ mod $q$. This is a well-known application of the Euclidean algorithm, which you can read about in this Wikipedia article.

In your example, we have $p=19, q=23$; and applying the Euclidean algorithm will give you $r=19^{-1}$ mod $23 = 17$. So, for instance, if $x \equiv 6$ mod $19$ and $x \equiv 2$ mod $23$, we have $x = (2-6)\cdot 17 \cdot 19 + 6 = -1286 \equiv 25$ mod $437$.

You can find the solution using this calculator just fill in $$ x=6 \mod 19$$ and $$
x=2 \mod 23
$$
Your solution is $25$ and indeed $188 \equiv 25^2 \pmod {437} $


Maximal distance between points on a line Two points A and B are on different sides of a line.  Find a point Y on the line such that the absolute value of the difference from Y to A and Y to B is maximal.

My thoughts are as follows.  Let's say that point Y is on a line but is not collinear with A and B.  Then I could draw a triangle by joining the 3 points.  For this triangle, segment AB is less than the sum of AY and BY.  Then AB is greater than the absolute value of (YA-YB).  How can I prove that this difference is maximal?

Hint: Find the function that yields the difference in the distances from $A$ to $Y$ and $B$ to $Y$ using the distance formula. This function will be of one variable since we know $A$, $B$, and the equation for the line. Then find the minimum or maximum of this function since we're looking for the greatest absolute value.

(0,0) because that would make both lines the maximum size they could be


Is $\rho$ equivalence relation and, if not, specify all the ways to expand it to the equivalence relation. On the set $\Bbb Z \times\Bbb Z$, the relation $\rho$ is given as it follows:
$(a,b) \rho (c,d) \iff (2 | (a-c) $ or $ 3|(b-d))$
Is $\rho$ equivalence relation and, if not, specify all the ways to expand it to the equivalence relation.

First, I checked if it's equivalence relation.

reflexivity

$(\forall a,b \in \mathbb Z )(a,b) \rho(a,b)$
This is true because $(2|(a-a)$ or $3|(b-b)) \iff(2|0$ or $3|0)$ is true.

symmetry

$(\forall a,b,c,d \in \mathbb Z )(a,b) \rho(c,d) \Rightarrow (c,d)\rho(a,b)$
Also true, $(a,b) \rho(c,d) \Rightarrow (2|(a-c)$ or $3|(b-d))\Rightarrow (2|(c-a)$ or $3|(d-b)) \Rightarrow (c,d)\rho(a,b)$.

transitivity

$(\forall a,b,c,d,e,f \in \mathbb Z )((a,b) \rho(c,d)$ and $(c,d) \rho(e,f)) \Rightarrow (a,b)\rho(e,f)$
Not true, for example while$(5,4) \rho (3,2)$ and $(3,2)\rho(2,-1)$ is true, $(5,4)\rho(2,-1)$ isn't.
Ok so now I have to expand this relation so that it can also be transitive. How do I do that?

I'm not sure what the question means by "all the ways" to make it an equivalence relation, however there is in general a way to make a symmetric, reflexive but not transitive $\rho$ into an equivalence relation.

Regardless of what $\rho$ is, we can define a new equivalence relation $\tilde{\rho}$ such that $x\tilde{\rho} y$ if and only if $x\rho y$ or there exists $z$ such that $x\rho z$ and $y\rho z$.

Note that symmetry and reflexivity of $\tilde{\rho}$ follow from symmetry and reflexivity of $\rho$.

In your case I would recommend writing explicitly what this $\tilde{\rho}$ is.

Hint: for any $a, b, c, d \in \mathbb{Z}$, we have $(a, b) \mathrel{\rho} (c, b)$ and $(c, b) \mathrel{\rho} (c, d)$.


For a graph $G$, why should one expect the ratio $\text{ex} (n;G)/ \binom n2$ to converge? $\text{ex} (n;G)$ is the maximal number of edges of a graph of order $n$ can have without containing $G$ as a subgraph.

There are theorems saying what the limit actually is. But my lecture notes says that it is not hard to see that this ratio converges as $n$ tends to infinity.

Can anyone point out why I should expect this ratio to converge?

It is true that $\operatorname{ex}(n; G)/\binom{n}{2}$ is monotone non-increasing.  As you've observed, if you have a graph on $n$ vertices of density $\theta$, removing just any vertex doesn't give you a subgraph of density $\theta$.  However, it is true that there exists some vertex whose removal gives you a subgraph of density at least $\theta$, as the following "averaging argument" makes precise.

(Throughout the proof, $E(G)$ denotes the edge set of a graph $G$, and $e(G) := \lvert E(G) \rvert$ denotes the number of edges in $G$.)

Proof: Let $G$ be a graph.  Let $F$ be a graph on $n$ vertices.  Define
$$\theta := \frac{e(F)}{\dbinom{n}{2}}$$
to be the density of $F$.

Now choose $n - 1$ vertices of $F$ uniformly at random, and consider the subgraph $H$ induced by these vertices.  There are $n$ choices for $H$, and an edge $uv$ of $F$ is in $H$ as long as both $u$ and $v$ are vertices of $H$.  Hence, for all $e \in E(F)$, we have
$$\mathbb{P}\bigl( e \in E(H)\bigr) = \frac{n-2}{n}.$$
This means that the expected number of edges in $H$ is
$$\mathbb{E}\bigl(e(H)\bigr) = \sum_{e \in E(F)} \mathbb{P}\bigl(e \in E(H)\bigr) = \frac{n-2}{n} e(F) = \frac{n-2}{n} \theta\dbinom{n}{2} = \theta\dbinom{n-1}{2}.$$
Because $H$ has $n - 1$ vertices, this means that the expected density of $H$ is $\theta$.  Since the average density of $H$ is $\theta$, there must exist a choice of $H$ (call it $H_0$) with density at least $\theta$.

Because $H_0$ is a graph on $n - 1$ vertices, if $\theta > \operatorname{ex}(n - 1; G)/\binom{n - 1}{2}$, then by definition $H_0$ must contain a copy of $G$, which means that and $F$ does as well.  We have thus shown that every graph on $n$ vertices of density $\theta > \operatorname{ex}(n - 1; G)/\binom{n - 1}{2}$ must contain a copy of $G$.  This implies that
$$\frac{\operatorname{ex}(n; G)}{\dbinom{n}{2}} \leq \frac{\operatorname{ex}(n - 1; G)}{\dbinom{n-1}{2}},$$
as claimed. $\square$

Note: this proof is based on an argument in this survey paper by Peter Keevash (see Section 2).

If you look at the numerator, there are $\frac{n^2}{2}$ possible edges in an undirected graph, and it seems intuitive that this number $ex(n,G)$ would grow like a constant factor of this.  So we say the numerator is $\Theta(n^2)$.

For the denominator, we see that $\binom{n}{2} = n(n-1) = n^2 - n$, which is $\Theta(n^2)$.

The $\Theta$ notation basically means that both functions grow like $n^2$ does, and you can reason that if two things grow at basically the same rate, then their quotient will converge.  For a reference on this notation, see this wikipedia article.


Subset as a relation for the given set Having an immense amount of trouble trying to figure this problem out, and the more I think and ask about it the more confused I seem to get. I think I have finally figured it out so can someone who truly knows the correct answer justify this?

Problem:Let $A=\{a,b,c\}$,Let $\mathcal{P}(A)=\{S:S\subseteq A\}$

Q1.Is the relation "is a subset of" a relation on $A$?

Q2.Is the relation "is a subset of" a relation on $\mathcal{P}(A)$?

Q1.Attempt:

I would say the relation "is a subset of" is NOT a relation on $A$

Since $A \times A=\{(a,a),(a,b),(a,c),(b,a),(b,b),(b,c),(c,a),(c,b)(c,c)\}$

Knowing $R \subseteq A \times A$

It is clear to me $a,b,c$ are elements, not sets, so an element in the relation on $A$ takes the form of $(a,a)$ for $a \in A$.So a "subset relation" cannot be defined on $A$.

Since $a \subseteq a$ is false for all $a \in A$ because $a$ is an element not a set

Q2.However a subset relation can be defined on $\mathcal{P}(A)$

Since elements of a relation on $\mathcal{P}(A)$ are actually ordered pairs of sets, now it is possible to define a "subset relation" on $\mathcal{P}(A)$.

For example $(\{a\},\{a,b,c\})\in \mathcal{P}(A) \times \mathcal{P}(A)$

and since for a relation on $\mathcal{P}(A)$, $R \subseteq \mathcal{P}(A) \times \mathcal{P}(A)$ and $\{a\} \subseteq \{a,b,c\}$

Now it is possible to define the relation "is a subset" of on $\mathcal{P}(A)$

To summarize my comments above, all that a set needs to be to be called a relation is to be a subset of a cartesian product of sets.  Nothing more, nothing less.  With regards to Q1, after some rewording the question asks if the set of ordered pairs $\{(x,y)\in A\times A~:~x\subseteq y\}$ is a relation.

Regardless of your point of view or how pedantic you want to be about whether the elements of $A$ are able to themselves be sets or not, as soon as you have a decided upon interpretation of how the elements of $A$ behave and are defined you should be able to agree that for each $x,y\in A$ the statement $x\subseteq y$ has a truth value as being clearly true or clearly false with no ambiguity.  We can say that $5\not\subseteq 3$ for instance, whether that is because you treat $5$ and $3$ as sets and you see that $5$ is not a subset of $3$, or whether you choose to treat them as atomic elements who are not sets and you say that $5\not\subseteq 3$ for the reason that these are not sets.

@MatthewLeingang is correct when he poitns out that $(\{a\},\{a\})\not\in A\times A$, however he is incorrect in assuming that this has any relevance to the problem at hand whatsoever.  We never cared about whether $(\{a\},\{a\})$ was an element of our relation or not (and it not being in our relation wouldn't have mattered anyways), we only cared about seeing which elements (if any) of $A\times A$ are included in our relation.

The only reason why the above would not have been a relation is in the event that the condition for an ordered pair to be in the set is ill-defined or undefined.  The set $\{(x,y)\in A\times A~:~x\heartsuit y\}$ is such an example since we don't yet know the meaning of the symbol $\heartsuit$.

Otherwise, just because a relation is empty does not stop it from being a relation.  Again, even if you disagree on whether or not $a,b,c$ should be considered pure "elements" and not sets or not, that does not stop the described relation from being well defined.

That all being said, depending on which style of set theory you are studying, the most common foundations to set theory that people study in the modern age is Zermelo-Fraenkel set theory.  In an attempt to make things as clean and unambiguous as possible, we restrict our attention to pure sets and every mathematical object, number, set, function, operation, statement, etc... are all also considered sets.  (There are things that ZF can't handle, such as proper classes.  Read about other set theories in the wiki link above such as ZFG)

Under this interpretation, even if we didn't intend it, $a,b,c$ are all actually sets too!  And as such, not only is the relation defined, just like before, it is even nonempty as it is guaranteed to contain at least the pairs $(a,a),(b,b)$ and $(c,c)$ and depending on the values of $a,b,c$ possibly more.

With respect to JMoravitz's point of view, I think you understand the problem as intended by the instructor/problem author, and should not extend to the case that elements of $A$ might contain themselves as subsets.

I'm inferring the point of the problem is to understand the difference between something being an element of a set versus a subset.  The notation $A = \{a,b,c\}$ implies (not logically, but idiomatically) that $A$ is a set containing three elements which are not sets.  A representative interpretation might be three integers, or three points in the plane.

Assuming this, then you can say $\{a\} \subset \{a\}$ but $(\{a\},\{a\}) \notin A \times A$ (since $\{a\} \notin A$), so is-a-subset-of is not a relation on $A$.

I'm sure your instructor would be happy to clarify the problem or discuss these counterintuitive edge cases.


Prove that either $r \geq 1$ or $r = \infty.$ Let $\{a_n \}_{n=0}^{\infty}$ be a sequence of real numbers such that $\sum\limits_{n=0}^{\infty} |a_n|^2 < \infty.$ Consider the power series $\sum\limits_{n=0}^{\infty} a_n x^n$ with radius of convergence $r.$ Show that either $r \geq 1$ or $r = \infty.$

How do I prove that? Any suggestion or hint regarding this will be highly appreciated.

Thank you very much for your valuable time.

You need to prove that
$$\sum_{n=0}^\infty a_nz^n$$
converges absolutely when $|z|<1$. I suggest using Cauchy-Schwarz. By hypothesis,
the $\ell^2$-norm of the sequence $(a_n)$ is finite.

For any bounded sequence $(a_n)$ the power series $\sum a_n x^{n}$ converges for $|x| < 1$ (by comparison with a geometric series). Hence the radius of convergence is  a real number $\geq 1$ or $+\infty$. Note that $\sum |a_n|^{2} <\infty$ implies that $a_n \to 0$ and hence $(a_n)$ is bounded.


Proving that a prime ideal $p \subset R$ yields a prime ideal $p[x] \subset R[x]$ I'm curious as to whether I can have my proof critiqued.

Proposition : Let $\mathfrak{p}$ be a prime ideal in a ring $R$. Show that $\mathfrak{p}[x]$ is a prime ideal in $R[x]$.

Proof : Suppose $\mathfrak{p}$ is  a prime ideal in a ring $R$. Then $\mathfrak{p} \subset R$. Therefore, consider that $$R \otimes_R \bigoplus_{i=1}^{\infty} R \cong R[x].$$ Since $\mathfrak{p} \subset R$, we have that $$\mathfrak{p} \otimes_R \bigoplus_{i=1}^{\infty} R \cong \bigoplus_{i=1}^{\infty} \mathfrak{p} \otimes_R R \cong \mathfrak{p}[x] \subset R[x].$$ Given that prime ideal are preserved under any isomorphism, it follows that $\mathfrak{p}[x]$ is a prime ideal in $R[x]$.

I furthermore claim that since isomorphisms do not necessarily preserve maximal ideals, if $\mathfrak{m} \subset R$ is a maximal ideal in $R$, it is not necessarily true that $\mathfrak{m}[x] \subset R[x]$ is a maximal ideal in $R[x]$.

It is infact never true that $\mathfrak{m}[x]\subseteq R[x]$! All of these questions become simple when you consider the following fact.

An ideal $\mathfrak{p}\subseteq R$ is prime if and only if $R/\mathfrak{p}$ is an integral domain. Moreover $\mathfrak{p}$ is maximal if and only if $R/\mathfrak{p}$ is a field.

The first statement is simple, the second might require a little bit of thought. Now note that $R[x]/\mathfrak{p}[x]=R/\mathfrak{p}[x]$, a polynomial ring on an integral domain, and thus an integral domain! But $R/\mathfrak{m}[x]$ is the polynomial ring on a field, and so contains the ideal $(x)$, which proves it is nonmaximal.

Lastly, I'm not really sure about your proof, since your isomorphism is not a ring isomorphism, and appears to be a module isomorphism, which doesn't help with proving an ideal is prime as far as I know.


System of equations modulo primes Is it generally possible to solve a set of linear equations modulo some prime numbers $\{p,q,r\}$. For example if I have the following congruences:

$$
xa_p + yb_p \equiv d \pmod {p}\\
xa_q + yb_q \equiv d \pmod {q}\\
xa_r + yb_r \equiv d \pmod {r}\\
$$

for some known $a_i$ and $b_i$ with $i \in \{p,q,r\}$ can I determine $d$ (assuming a solution exists and $d<p,q,r$)?

Thanks in advance! (and sorry if this is trivial.)

EDIT: Additionally, $a_i$ and $b_i$ are congruences of some unknown (and large) integers $a,b$. (i.e. $a\equiv a_i \pmod{i}$ and $b\equiv b_i \pmod{i}$ )

Any pointers are be welcome! Also knowing that there is no "good" algorithm for this problem would also help me a lot.

Since the problem is finite, a brute force approach will either find a solution or prove none exist. In more detail, if $x,y$ is a solution, then $x+kpqr,y+spqr$ is a solution for all $k,s\in \mathbb Z$ (since $pqr=0$ modulo each of $p,q,r$). It thus follows (by using the division algorithm) that if a solution exists then a solution exists with $x,y\in \{0,1,\cdots ,pqr-1\}$. So, the brute force approach will be to iterate over all $0\le x,y< pqr$ and check for solutions. If one is found then you found one, otherwise none exist.

x+y ≡ d (mod p)

d ≡ (x (mod p)) + (y (mod p)) (mod p)

Example:

a ≡ 1 mod 3,
b ≡ 2 mod 3,
a + b ≡ 3 mod 3 ≡ 0,

Also:

xa ≡ d (mod p),
d = c (mod p),

c = a (mod p) iff (a (mod p) > x (mod p) or a (mod p) = x (mod p) OR a (mod p) is = 0)
c = x (mod p) iff (x (mod p) > a (mod p) or x (mod p) = a (mod p) OR a (mod p) is = 0)


Prove that $lim_{n\rightarrow \infty} a_n/a_{n+1} = z_0.$ I am studying for a complex qual and I am having a little difficulty with the following problem.  Any help that you could give me would be awesome.  Thank you.

Let $f(z) = \sum_{n=0}^\infty a_nz^n$ be the Taylor series around $0$ of a function which is analytic in $\mathbb{C}$ \ ${z_0},$ $z_0\neq 0$ and has only a simple pole at $z_0$. Prove that
$lim_{n\rightarrow \infty} a_n/a_{n+1} = z_0.$

I decided to add another answer rather than delete or change the last since the first answer is nice ill leave it, but here is a complete solution, independent of previous considerations.
$(z-z_0)f(z)$ is an entire function with power expansion
$$\sum (a_n -a_{n+1}z_0)z^{n+1}$$ now as this series converges for all $z$, and the original series has a finite radius of convergence we must have
$$\lim |\frac{a_n -a_{n+1}z_0}{a_{n+1}}|=0$$

And this is exactly what was to be proved.

This is only a partial answer, but a power series converges on the largest radius where the function is defined. The series cannot converge at this point $z_0$. Therefore the radius of convergence is $|z_0|$ now by the ratio test this gives,

$$\lim\limits_{n\to \infty} \frac{|a_n|}{|a_{n+1}|}=|z_0|$$

Of course I have not used that $z_0$ is a simple pole so somehow that must be involved in proving the better limit.


Unsure how to solve equation I have the problem. Solve: $x^3 - 3x^2 + 2x - 5 = 0$. I've done similar, simpler ones like this, but in this case I don't know. Can someone at least tell me what you'd even call this exact type of problem so I can google it and learn how to do this and work on it more?

Here's a way to find all complex $x$ satisfying
$$x^3-3x^2+2x-5=0$$
First we want to remove the $-3x^2$ term, this can be done by doing the substitution $x=y+1$
$$(y+1)^3-3(y+1)^2+2(y+1)-5=0$$
Expanding this yields
$$y^3-y-5=0$$
Substitute $y=\frac\lambda z+z$, this also means that $z = \frac12\left(y+\sqrt{y^2-4\lambda}\right)$

$$-5-z-\frac\lambda z+\left(z+\frac\lambda z\right)^3=0$$

Multiply both sides with $z^3$ and collect in terms of $z$

$$z^6+z^4(3\lambda-1)-5z^3+z^2\left(3\lambda^2-\lambda\right)+\lambda^3=0$$

Substitute $\lambda=\frac13$ and then $u=z^3$

$$u^2-5u+\frac1{27}=0$$

Find the solution where $\pm$ is a $+$

$$u=\frac1{18}\left(45+\sqrt{2013}\right)$$

Substitute back $u=z^3$

$$z^3=\frac1{18}\left(45+\sqrt{2013}\right)$$

The cube root of that is $\frac{\sqrt[3]{45+\sqrt{2013}}}{\sqrt[3]{2}~3^{2/3}}$

Multiply that with the third roots of unity.

$$z=\frac{\sqrt[3]{\frac12\left(45+\sqrt{2013}\right)}}{3^{2/3}}$$
$$z=-\frac{\sqrt[3]{\frac12\left(-45-\sqrt{2013}\right)}}{3^{2/3}}$$
$$z=\frac{(-1)^{2/3}\sqrt[3]{\frac12\left(45+\sqrt{2013}\right)}}{3^{2/3}}$$

Substituting back for $z = \frac y2+\frac12\sqrt{y^2-\frac43}$

$$\frac y2+\frac12\sqrt{y^2-\frac43}=\frac{\sqrt[3]{\frac12\left(45+\sqrt{2013}\right)}}{3^{2/3}}$$
$$\frac y2+\frac12\sqrt{y^2-\frac43}=-\frac{\sqrt[3]{\frac12\left(-45-\sqrt{2013}\right)}}{3^{2/3}}$$
$$\frac y2+\frac12\sqrt{y^2-\frac43}=\frac{(-1)^{2/3}\sqrt[3]{\frac12\left(45+\sqrt{2013}\right)}}{3^{2/3}}$$

Solve the radical equation for $y$

$$y=\sqrt[3]{\frac2{3\left(45+\sqrt{2013}\right)}}+\frac{\sqrt[3]{\frac{45+\sqrt{2013}}{2}}}{3^{2/3}}$$
$$y=(-1)^{2/3}\sqrt[3]{\frac2{3\left(45+\sqrt{2013}\right)}}-\frac{\sqrt[3]{\frac{-45-\sqrt{2013}}{2}}}{3^{2/3}}$$
$$y=\frac{(-1)^{2/3}\sqrt[3]{\frac{45+\sqrt{2013}}{2}}}{3^{2/3}}-\sqrt[3]{\frac{-2}{3\left(45+\sqrt{2013}\right)}}$$

Substitute back $y=x-1$ and add $1$ to both sides

$$x=1+\sqrt[3]{\frac2{3\left(45+\sqrt{2013}\right)}}+\frac{\sqrt[3]{\frac{45+\sqrt{2013}}{2}}}{3^{2/3}}$$
$$x=1-(-1)^{2/3}\sqrt[3]{\frac2{3\left(45+\sqrt{2013}\right)}}-\frac{\sqrt[3]{\frac{-45-\sqrt{2013}}{2}}}{3^{2/3}}\approx$$
$$x=1+\frac{(-1)^{2/3}\sqrt[3]{\frac{45+\sqrt{2013}}{2}}}{3^{2/3}}-\sqrt[3]{\frac{-2}{3\left(45+\sqrt{2013}\right)}}$$

Here's some approximate versions of these numbers

$$
\begin{array}{rclcl}
x&\approx&2.90416085913492\\
x&\approx&0.04791957043254&-&1.311248044077123i\\
x&\approx&0.04791957043254&+&1.311248044077123i\\
\end{array}
$$

This equation has a single real root. The solution has a closed form formula, but you can also find the numerical value by Newton's iterations
$$x_{n+1}=x_n-\frac{x_n^3-3x_n^2+2x_n-5}{3x_n^2-6x_n+2},$$ starting from, say, $x_0=0$. (The denominator is the derivative of the numerator.)
UPDATE:
After finding the single real root, you can show that there is no other by deflation (divide by $x-r$ and show that the quadratic quotient has no real root).


Prove that $1+x+y$ not equal to $xy$, when $x$ and $y$ are positive odd numbers I was engaging in this equation and with the help of desmos.com I noticed it is not possible. So how can we prove? In which way it is easier, should I leave $x$ or $y$ alone, or is there any method different?

First of all we can define odd numbers:

$x=2m-1 \ \  \forall \ m \in \mathbb N \ \backslash \{0\}$ and $y=2n-1 \ \  \forall \ n \in \mathbb N \ \backslash \{0\}$

Then we have $1+x+y=x\cdot y\Rightarrow -1+2m+2n=(2m-1)(2n-1)$

$2m+2n-1=4mn-2m-2n+1$

$4m+4n=4mn+2$

$2(m+n)=2mn+1$

The LHS is always even and the RHS is always odd for all $n,m \in \mathbb N \ \backslash \{0\}$


Any hint for this product of complex numbers I have to calculate the modulus of $z$. I've already tried to find a general formula for $\left(\frac{1+i}{2}\right)^{2^n}$, which seems to be $\frac{1}{2^{2^{n-1}}}, \forall \geq3$, using the trigonometric form and de Moivre's formula.

$z=\left[1+\frac{1+i}{2}\right] \left[1+\left(\frac{1+i}{2}\right)^2\right] \left[1+\left(\frac{1+i}{2}\right)^4\right] \cdots \left[1+\left(\frac{1+i}{2}\right)^{2^n}\right]$

How should I keep solving this?

Thanks!


Implicit function theorem usage Suppose I have the following equation representing level curves of a function:

$$-\left(\frac{1-\alpha}{x}+\frac{\alpha}{y}\right)=k \  \text{where} \ 0<\alpha<1$$

And I want to find $\frac{dy}{dx}$:

Using the implicit function theorem, I get:

$$\frac{dy}{dx}=-\frac{1-\alpha}{\alpha}\left(\frac{y}{x}\right)^2$$

However, I can also put the first equation into a $y$ in function of $x$ form:

$$y=-\frac{\alpha x}{(1-\alpha)+kx}$$

Which gives me:

$$\frac{dy}{dx}=-\frac{\alpha(1-\alpha)}{[(1-\alpha)+kx]^2}$$

I don't understand why the results are different. In the first expression, the constant $k$ disappears, while in the second equation it doesn't. Also, we have $y$ appearing in the first expression but obvisouly not in the second. How can I explain this?

Making use of fredgoodman's comment, we substitute $y$ into the first $\frac{dy}{dx}$ expression and get the second one.


Can you prove that $[(x^2 + 2x + 1)\log(x) – x^2\log(x+1)]/2\log(x)\log(x+1)$ > 1 for all x > 2? Can you prove that $\frac{(x^2 + 2x + 1)\log(x) – x^2\log(x+1)}{2\log(x)\log(x+1)} > 1$ for all $x > 2$?

The target inequality can be rewritten as
\begin{align*}
\frac{(x+1)^2}{2\log(x+1)} > \frac{x^2}{2\log(x)} + 1 \iff f(x+1) > f(x) + 1
\end{align*}
where $f(x) = \frac{x^2}{2\log(x)}$. By Taylor series,
\begin{align*}
f(x+1) = f(x) + \color{blue}{f'(x) + \frac{1}{2}f''(\xi)}
\end{align*}
for some $\xi \in (1, x)$. We strive to prove the blue portion is $> 1$. Note that
\begin{align*}
f'(x) = \frac{x}{\log(x)} - \frac{x}{2\log^2(x)}, \qquad f''(x) = \frac{2(\log(x)-0.75)^2+0.875}{2 \log^3(x)}
\end{align*}
Note that $f''(x) > 0$ for all $x > 2$ (in fact, $x > 1$), so therefore $f'(x)$ is increasing, and
\begin{align*}
f'(x) > f'(2) = \frac{\log(4)-1}{\log^2(2)} > 0.80 \quad \text{for all } x > 2
\end{align*}
Not strong enough, but note that $f'(e) = e/2 = 1.359 > 1$, so if we can prove $\frac{1}{2}f''(\xi) > 0.2$ for $\xi \in [2, e]$, then we are done. Indeed,
\begin{align*}
\frac{1}{2}f''(\xi) \ge \frac{0.875}{4 \log^3(\xi)} \ge \frac{0.875}{4 \log^3(e)} = 0.21875 > 0.2
\end{align*}

Here is my answer, people can correct me if I am wrong.

$$\frac{[(x+1)^2\log{(x)}-x^2\log{(x+1)}]}{2\log{(x)}\log{(x+1)}}$$

$$\frac{[\log(x^{{(x+1)}^{2}})-\log{{({{(x+1)}^x}^2}}{}]}{\log(x^2)\log(x+1)}$$

$$\frac{\log(x^{{(x+1)}^{2}})}{\log(x^2)\log(x+1)} \frac{\log(x^2)\log(x+1)}{\log{{(x+1)^x}^2}}$$

$$\frac{{(x+1)}^{2}\log(x)}{x^2\log(x+1)}$$

I think that the last form is always bigger than 1 if x > 2, just plug-in 2 or any greater number.


$\lim_{n\to\infty}\frac{e^{\frac{1}{n^2}}-\cos \frac{1}{n}}{\frac{1}{n}\log(\frac{n+1}{n})-(\sin\frac{2}{n})^2}$ $$\lim_{n\to\infty}\frac{e^{1/n^2}-\cos \frac{1}{n}}{\frac{1}{n}\log(\frac{n+1}{n})-(\sin\frac{2}{n})^2}, \quad n \in \mathbb{N}$$

I'm stuck trying to evaluate this limit. I have tried applying l'Hôpital's rule to the corresponding function in $\mathbb{R}$, but even after differentiating all that stuff I still end up with a $\frac{0}{0}$ form. There has to be something I'm missing. Any hints?

Hint. Try by using Taylor expansions. As regards the numerator
$$e^{\frac{1}{n^2}}-\cos \frac{1}{n}=1+\frac{1}{n^2}+o(1/n^2)-
\left(1-\frac{1}{2n^2}+o(1/n^2)\right)=\frac{3}{2n^2}+o(1/n^2).$$
Whereas, for the denominator we have that
$$\frac{1}{n}\log(1+\frac{1}{n})-(\sin\frac{2}{n})^2=?$$
Can you take it from here?


How to solve $6x^{2/3}-12-2x=0$? How to solve  $6x^{2/3}-12-2x=0$? I need it to prove that the graphic of $y=6x^{2/3}$ never meets the graphic of function $y=12+2x$.

We have $x+6 = 3x^{2/3}$ i.e $(x+6)^3 = 27x^2$. This has a real solution. You won't be able to prove what you wish.

Let $x^{\frac{1}{3}}=y$.

Thus, $x>0$, $y>0$ and we need to solve
$$y^3-3y^2+6=0.$$
But, by AM-GM
$$y^3+6=2\left(\frac{y^3}{2}\right)+6\geq3\sqrt[3]{6\left(\frac{y^3}{2}\right)^2}>3y^2,$$ which says that our equation has no real roots.


Prove that $\omega$ is not successor ordinal Prove that $\omega$ is not successor ordinal ?

I assumed that $\omega$ is successor ordinal, meaning there is ordinal $\beta$ such that $S(\beta) = \omega$ meaning $\beta \cup \{ \beta \} = \omega$

Resulting that $\beta \in \omega $ so $ \beta$ is finite, but how to proceed ?

Show $\beta$ is a maximum element of $\omega.$
Thus a contradiction.

First prove that $n\notin n$ for every $n\in\omega$.

For that define $b:=\{n\in\omega\mid n\notin n\}\subseteq\omega$ and prove by induction that $b=\omega$.

That comes to proving that $\varnothing\in b$ and $n\in b\implies n\cup\{n\}\in b$, allowing the conclusion that next to $b\subseteq\omega$ also $\omega\subseteq b$.

Now if $\omega=\beta\cup\{\beta\}$ then $\beta\in\omega$ and then also $\omega=\beta\cup\{\beta\}\in\omega$.

This because $\omega$ is closed under the successor operation.

But from $\omega\in\omega$ you can conclude now that $\omega\notin\omega$ and a contradiction is found.


How Many Angles in an Angle? [1)

I'm wondering if there are two angles in an
'angle.'

For example, the image below would typically be
identified as a right angle.

But couldn't you also say that it's a right angle
and a $270^o$ reflex angle? Are there two
angles?

http://tinyurl.com/4fs85kg

[2)

Or-- An equilateral triangle is defined (?) as a
triangle with three $60^o$ angles.

But couldn't you also say that an equilateral
triangle has three $300^o$ reflex angles? Six
angles?

[3)

Would the rays that form an angle be considered
straight angles?

If so: Since a ray consists of infinitely many
points.

And since any two points form a line segment.

And since a line segment can be considered a
straight angle.

Does it follow that an angle consists of infinitely
many angles?

You could also say it's a 90+360 = 450 degree angle, or any number $90+360n$ where n is a natural number. The point is, we say that it's 90 degrees clockwise or 270 mostly by convention, but yes, "an angle consists of infinitely many angles" is true.

Similarly for 2) you called it an "equiangular" triangle and so indeed by definition it must have three equal angles, however you measure them.

A "straight angle" would just be an angle of zero (or 360, or  720...) degrees.

By simply saying an angle, we mean the size of it is between 0 degree and 180 degrees. Its 360-degree counterpart should be marked with a prefix called reflex. Thus, there is no such thing as '2 angles in an angle'.


Find the value of $\lim_{x\to 0} \sum^{2013}_{k=1} \frac{\{\frac{x}{tanx}+2k\}}{2013}$... Problem :

$$\lim_{x\to 0} \sum^{2013}_{k=1} \frac{\{\frac{x}{\tan x}+2k\}}{2013}$$

where $\{.\}$ represent the fraction part function $0\leq \{x\} < 1$

My approach :

We know that $ \lim_{x\to 0}  \frac{\tan x}{x}=1$

and not getting any clue please suggest thanks.

Let $$l=\frac{1}{2013}\lim_{x\rightarrow 0}\sum^{2013}_{k=1}\left\{2k+\frac{x}{\tan x}\right\}$$

Now Using $\displaystyle \{x+n\} = \{x\}\;,$ Where $n\in \mathbb{N}$

So $$l=\frac{1}{2013}\sum^{2013}_{k=1}2k+\frac{1}{2013}\lim_{x\rightarrow 0}\sum^{2013}_{k=1}\left\{\frac{x}{\tan x}\right\}$$

So $$l=\frac{2}{2013}\cdot \frac{2013\cdot 2014}{2}+\lim_{x\rightarrow 0}\left\{\frac{x}{\tan x}\right\}$$

So $$l=2014+\lim_{x\rightarrow 0}\frac{x}{\tan x}-\lim_{x\rightarrow 0}\big\lfloor \frac{x}{\tan x}\big\rfloor$$

So $$l=2014+1-0 = 2015$$

Bcz  Here $\displaystyle \frac{\tan x}{x}$ is an even Function .

So we will calculate limit for $x>0$

Now in $\displaystyle x\in \left(0,\frac{\pi}{2}\right)\;, \tan x>x\Rightarrow \frac{ x}{\tan x }<1$ and $\displaystyle \frac{ x}{\tan x }>0$

So We get $$\lim_{x\rightarrow 0}\big\lfloor \frac{x}{\tan x}\big\rfloor =0$$

See the $2k$ is a natural number so you can directly remove it out from the problem . So now we have $\sum \frac{x/tan(x)}{2013}$ now the sum is indirectly $\sum 1=2013$ so $2013$ cancels out. Now we are left with ${\frac{x}{tan(x)}}$ now we know that $lim {x \to 0} tan(x)/x=1$ also we know ${y}=y-[y]$ so we can write fractional part as $\frac{x}{tan(x)}-[\frac{x}{tan(x)}]$ so the limit becomes $1-1=  0$ hence the limit is $0$.


How should I define symmetric matrices? "A matrix A is a symmetric matrix if its transpose $A^T$ = $A$." This is my definition.

"A square matrix A is a symmetric matrix if its transpose $A^T$ = $A$." This is the textbook definition.

I understand that only square matrices can be symmetric, but I figured it would be all right to cut out the "square", as satisfying the equation $A^T$ = $A$ would ensure it is a square matrix.

So why can't I omit the "square"? (An error I was sternly reminded of.)

Any better definitions would also be greatly appreciated.

I think both definitions are perfectly valid (and mathematically equivalent), however I prefer the textbook definition.

The textbook definition makes a distinction between two cases. Consider the matrices  $$A=\begin{bmatrix}1&1\\0&1\end{bmatrix}, B=\begin{bmatrix}1\\0\end{bmatrix}.$$

In some sense, the textbook definition says different things about them.

For $A$, the textbook definition tells you that $A$ is not a symmetric matrix, because $A^T\neq A$.
For $B$, the textbook definition tells you that $B$ is not a symmetric matrix because the concept of symmetric matrices is incompatible with non-square matrices.

I like this additional clarity in the definition. It makes it clear that $B$ is not symmetric because, in a sense, it doesn't even make sense to ask whether $B$ is symmetric.

Given an $m\times n$ matrix $A=(a_{ij})$ and a $k\times l$ matrix $B=(b_{uv})$ with entries in a commutative ring $R$.

The matrices $A,B$ are equal if the mappings $f:[m]\times [n]\rightarrow R: (i,j)\mapsto a_{ij}$ and $g:[k]\times [l]\rightarrow R: (u,v)\mapsto b_{uv}$ are equal. Equality of maps means equality of the corresponding relations (sets).

Practically, this means that $A,B$ are equal if $A,B$ have the same size $(m=k,n=l)$ and for each pair $(i,j)\in[m]\times [n]$, $a_{ij}=b_{ij}$.

Well, if $A$ is an $m\times n$ matrix, then the transposed $A^T$ is an $n\times m$ matrix.
By the above definition, the equality $A=A^T$ means that $m=n$ and $a_{ij}=a_{ji}$.


Problem simplifying equation using boolean algebra I have this boolean equation:

A'.B'.C'.D' + A'.B.C'.D' + B'.C'.D + B.C'.D

Using a Karnaugh map I find I can simplify the above to:

C'.(A' + D)

I have used Wolfram Alfa website to confirm this simplification is correct.

I would like to try simplifying the original statement using Boolean algebra but get stuck along the way:

A'.B'.C'.D' + A'.B.C'.D' + B'.C'.D + B.C'.D

C'.D'.(A'+ B' + A' + B) + C'.D.(B' + B)

C'.D'.(A' + 1)  + C'.D.1

I would be grateful if someone could point out where I have gone wrong. Thank you.


How do I find $r$ in $2/64 = r^5$? I've never did an equation like this before, so I'm not sure how to remove the $r^5$. Self-explanatory title, any help is much appreciated.

$$2/64 = r^5$$

If $r\in\mathbb R,$ then

$$r^5=\frac{2}{64}\Rightarrow r^5=\left(\frac{1}{2}\right)^5\Rightarrow r=\frac{1}{2}.$$

If $r\in\mathbb C,$ then
$$r=\frac 12e^{2m\pi i/5}\ \ (m=0,1,2,3,4).$$

$r^5=2/64$

then if take put both sides to the power of 1/5 or 0.2

$(r^5)^.2=(2/64)^.2$

then the exponent of $r$ becomes $1$ because the reciprocal of $5$ is $1/5$

$r^1=(2/64)^.2$

plug that in to a calculator and you are good to go

$r=0.5$ or $1/2$

don't forget to click the check mark


New bound for Am-Gm of 2 variables Today I'm interested by the following problem :

  Let $x,y>0$ then we have :
  $$x+y-\sqrt{xy}\leq\exp\Big(\frac{x\ln(x)+y\ln(y)}{x+y}\Big)$$

The equality case comes when $x=y$

My proof uses derivative because for $x\geq y $ the function :
$$f(x)=x+y-\sqrt{xy}-\exp\Big(\frac{x\ln(x)+y\ln(y)}{x+y}\Big)$$

is decreasing and for $y\geq x$ the function is increasing and the maximum occurs when $x=y$

My question is : Have you an alternative proof wich doesn't use derivative ?

Thanks in advance.

This is not an answer to the question, but it's too big to put it to the comment. I will show some connection (which might be interesting) between this inequality and Shannon entropy.

Firstly rewrite this expression as
$$
x + y- \sqrt{xy} \le x^{\frac{x}{x+y}} \cdot y^{\frac{y}{x+y}}.
$$
Then one can use a substitution
$$
a = \frac{x}{x+y}, \; b = \frac{y}{x+y}
$$
and get an equivalelent inequality in terms of $a$ and $b$
$$
(1 - \sqrt{ab}) \le a^a b^b, \;\; a +b = 1.
$$
So, we need to show that given $a+b = 1$, we will have
$$
\sqrt{ab} + a^a b^b \ge 1.
$$
It's equivalent to the following upper bound for Shannon entropy $H(a,b)$
$$
H(a,b) = -a \log a - b\log b \le -\log(1-\sqrt{ab}), \;\; a+b=1.
$$

So, one needs to prove this estimate for Shannon entropy $H(a,b)$. Unfortunately, I have no idea how to do this without calculus. Plots of $H(a,b)$ and its upper bound:

It might happen that this inequality has some meaning in information theory, though I haven't found anything about that.

The alternative forms of the given inequality are
$$x\ln x + y \ln y \ge (x+y)\ln(x+y)+(x+y)\ln\left(1-\dfrac{\sqrt{xy}}{x+y}\right),$$
$$x\ln\,\dfrac x{x+y} + y \ln\, \dfrac y{x+y} \ge (x+y)\ln\left(1-\dfrac{\sqrt{xy}}{x+y}\right),$$
$$\dfrac x{x+y}\,\ln\,\dfrac x{x+y} + \dfrac y{x+y}\, \ln\, \dfrac y{x+y}
\ge \ln\left(1-\sqrt{\dfrac x{x+y}}\;\sqrt{\dfrac y{x+y}}\right).\tag1$$
Let WLOG $\;x\le y,\;$
$$\dfrac yx = \dfrac{1+z}{1-z},\quad \dfrac x{x+y} =\dfrac{1-z}{2}
,\quad \dfrac y{x+y} =\dfrac{1+z}{2},\quad z\in[0,1],\tag2$$
then $(1)$ can be presented in the forms of
$$\dfrac{1-z}{2}\,\ln\dfrac{1-z}{2}+\dfrac{1+z}{2}\,\ln\dfrac{1+z}{2} \ge \ln(1-\,^1\!/_2\sqrt{1-z^2}\,),$$
$$\dfrac{1-z}{2}\,\ln(1-z)+\dfrac{1+z}{2}\,\ln(1+z) \ge \ln(2-\sqrt{1-z^2}),$$
$$-\dfrac{z}{2}\,\ln(1-z)+\dfrac{z}{2}\,\ln(1+z) \ge \ln(2-\sqrt{1-z^2})
-\ln\sqrt{1-z^2},$$
$$f(z) = (1-z)^{-\large\,^z\!/_2}(1+z)^{\large\,^z\!/_2}+1- \dfrac 2{\sqrt{1-z^2}}\ge 0. \tag3$$
Using binomial series in the forms of
$$(1-z)^{-\large\,^z\!/_2} = 1 + \dfrac12 z^2 + \dfrac1{2!}\,\dfrac z2\,\dfrac{2+z}2 z^2
 + \dfrac1{3!}\,\dfrac z2\,\dfrac{2+z}2\dfrac{4+z}2 z^3 +\dots$$
$$= 1+\dfrac12z^2+\dfrac14z^3+\dfrac7{24}z^4+\dfrac14z^5+\dfrac{113}{480}z^6+\dots,$$
$$(1+z)^{\large\,^z\!/_2} = 1 + \dfrac12 z^2 + \dfrac1{2!}\,\dfrac z2\,\dfrac{-2+z}2 z^2
 + \dfrac1{3!}\,\dfrac z2\,\dfrac{-2+z}2\dfrac{-4+z}2 z^3 +\dots$$
$$= 1+\dfrac12z^2-\dfrac14z^3+\dfrac7{24}z^4-\dfrac14z^5+\dfrac{113}{480}z^6+\dots,$$
$$(1-z^2)^{-\large\,^1\!/_2} = 1+\dfrac12z^2+\dfrac38z^4+\dfrac5{16}z^6+\dots,$$
one can get the Maclaurin series in the form of
$$f(z) = \left(1+\dfrac12z^2+\dfrac7{24}z^4+\dfrac{113}{480}z^6+\dots\right)^2
-\left(\dfrac14z^3+\dfrac14z^5+\dots\right)^2$$
$$+1-2\left(1+\dfrac12z^2+\dfrac38z^4+\dfrac5{16}z^6+\dots\right),$$
$$f(z) = \dfrac1{12}z^4+\dfrac3{40}z^6+\dots$$
(see also WA series),
with the all positive coefficients.
Proved!
$\color{brown}{\textbf{About binomial series.}}$

Newton's binomial formulas exist independently of the derivatives associated theory.

If the exponent is the rational, then the coeffitients of the binomial series $(1+x)^{\large\frac pq} = 1+a_1x+a_2x^2+\dots$ are defined via the identity
$$(1+a_1x+a_2x^2+\dots)^q = (1+x)^q,$$
by the comparation of the polynomials in the LHS and RHS.

If the exponent is real, then the binomial series are defined via the limiting transition.

These elementary knowledges point that the binomial series exist independently of the derivatives associated theory.


Characterisation of a Commutative C* Algebra which is an Integral Domain Let $X$ be a compact hausdorff topological space with more than one element.Then prove that the ring $C(X)$ of complex valued continuous functions on $X$ is not an integral domain.
Thanks for any help.
Actually this question arose when I was trying to prove that any commutative C*-algebra which is also an integral domain must be isomorphic to C and I think this statement is correct.
The problem I am having is with the case when X is connected.

Let $X$ be a compact Hausdorff space with more than one point. Then $C(X)$ is a natural, normal uniform algebra on $X$. This means that the character space of $C(X)$ is exactly $X$ and for each closed set $E\subseteq X$ and each closed set $F\subseteq X\setminus E$, there exists a function $f\in C(X)$ such that $f(y)=0$ for all $y\in E$ and $f(y)=1$ for each $y\in F$. Thus $C(X)$ necessarily contains non-zero divisors. This follows from the Tietze extension theorem.

Note that when $X$ is locally compact, we need one of the sets $K$ to be compact and the other closed, but I cannot remember which one needs to be compact and which needs to be closed. I refer you Uniform algebras by Gamelin, Introduction to function algebras by Browder or Banach algebras and automatic continuity (Chapter 4) by Dales for details on these facts.

Let $A$ be a non-commutative C*-algebra that is also an integral domain. If $a$ be a self-adjoint element of $A$, then it is well-known that C*(a), C*-algebra generated by $a$, is a commutative C*-algebra. Since, $A$ is an integral domain, C*(a) is an integral domain, too. If we accept the above conjecture that any commutatve C*-algebra which is also an integral domain must be isomorphic to C, then C*(a) as a commutative C*-algebra which is also an integral domain must be C and it implies that $a$ is a complex number. Since $a$ is an arbitrary self-adjoint element of $A$, $A$ must be isomorphic to C and it is a contradiction beacause of non-commutativity of $A$. Hence, I think the above conjecture in incorrect.

Best regards
A. Hosseini


$e^{-x^4\sin^2x}\in L_2(R)$ proving $\int_{-\infty}^{\infty} e^{-2x^4\sin^2x}dx < \infty$ I am trying to prove that $e^{-x^4\sin^2x}\in L_2(R)$. It means that  $\int_{-\infty}^{\infty} e^{-2x^4\sin^2x}dx < \infty$. So, i tried to use some usual methods, but $\sin(x)$ has no limit of infinity and I can't pick up an integrable upper bound to use Weierstrass theorem. Please give me some hints.

The function $f(x)=2x^4\sin^2x$ is even, so $\int_{-\infty}^{\infty}\exp(-f(x))=2\int_{0}^{\infty}\exp(-f(x))$. Now you can easily bound the integral by noting that $2x^4\sin^2x > 2x^2$ in $(1,\infty)$, so everything is finite.


$1\le p \lt \infty$ and $f_k$ nonnegative increasing. Then $f_k\to f$ in $L_p$ iff $\sup_k||f_k||_p \lt \infty$. Let $1\le p \lt \infty$ and $0\le f_k$ increasing to $f$, and $f_k$ measurable.

Then $f_k\to f$ in $L_p$ if and only if $\sup_k||f_k||_p \lt \infty$.

I was able to show the if part, but I can't show the only if part. How can I show this. I would greatly appreciate any help.

By the monotone convergence theorem, you always will have
$$\int f^p = \lim_{k \rightarrow \infty} \int f_k^p$$
Taking $p$th roots (and using continuity of the function $x \rightarrow x^{1 \over p}$), one therefore has
$$||f||_p = \lim_{k \rightarrow \infty} ||f_k||_p \tag 1$$
Since the $f_k$ increase to $f$, the limit in $(1)$ is an increasing limit.
So one also has
$$||f||_p = \sup_k ||f_k||_p \tag 2$$
If each $f_k \in L^p$ and $\lim_{k \rightarrow \infty} ||f_k - f||_p = 0$, which I'm taking as your assumption, then $f \in L^p$ since by the triangle inequality one has $||f||_p \leq ||f - f_k||_p + ||f_k||_p$. Hence $(2)$ gives that $\sup_k ||f_k||_p = ||f||_p$ is finite.

(You do need each $f_k \in L^p$ for this; otherwise you could just take $f_k = f$ to be equal to some non-$L^p$ function for all $k$ and the problem will be false).

Well, fk is a sequence of increasing functions to f, so with sufficient work you should be able to show that sup(k) ||fk||p = ||f||p.


The minimum number of circles in order to obtain a COVER of a specific square Suppose a unit square $X$, with side length $l=1$ as below, which is COVERed by a set $Y$ of circles with the same constant radius of $r=\dfrac{\sqrt{2}}{10}$, where a cover set $Y$ of another set $X$ means, $x\in Y,\forall x\in X$.

Questions are :

1) how to obtain the minimum number $n^*$ of circles in $Y$?

2) And I am also interested in the relationship between the ratio $\eta=\dfrac{r}{l}$ of circle radius to the side length of the square, and the number of circles $n^*$.

I also guess, if the pattern as above is right, then $n^*=\left\lceil {\dfrac{\sqrt{2}}{2\eta}} \right\rceil^2 $, but don't know how to prove or disprove it. Probably, below is a counter example?

According to Table 1 of following paper by GK Das et al (2006),

$\hspace0.3in$Efficient algorithm for placing a given number of base stations to cover a convex region

the minimum radius for 23 circles to over a unit square is at most $0.14124482238793135951$. Since this is smaller than $\displaystyle\;\frac{\sqrt{2}}{10}$, at most 23 circles is enough.

I have no idea how the configuration for 23 circles look like. However, following is a
page
which has the configuration of best known covering of a square by up to $12$ equal circles.
As your can see, the configuration doesn't seem to follow any obvious pattern. It is highly
unlikely that $n^{*}$ has any simple formula.

Update

The configurations in Table 1 are computed in a research notes (Ref 15 in GK Das' paper)

$\hspace0.3in$Covering a square with up to 30 equal circles by K.J. Nurmela, P.R.J. Ostergard

An online copy (in postscript) is available here. It has pictures for configuration up to 30 circles.

Based on above paper, following is one way to cover the unit square with 23 circles
of radius $\frac{\sqrt{2}}{10}$.
For simplicity of presentation, we will center the unit square at the origin and
only show the $8$ circles on the first quadrant.

In this configuration, the centers of the circles are positioned at:

$$\begin{array}{|c:l|}
\hline
\text{eg.} & (x,y)\\
\hline
A, B, C & (0,0), (\pm 0.25,0), (\pm 0.5,0)\\
D, E & (\pm 0.125,\pm 0.20756513901392), (\pm 0.375,\pm0.20756513901392)\\
F & (0, \pm 0.41513027802785)\\
G & (\pm 0.22172066050239,\pm 0.40940831933773)\\
H & (\pm 0.41515845958082,\pm 0.38685446089527)\\
\hline
\end{array}$$

The centers of the middle 3 layers (i.e those in the same layers as $A,B,C,D,E$) are forming a triangular lattice elongated in the vertical direction.

Well just few thoughts for the problem. The formula for dimension $l=ne^d$ the area that covers a circle is $ E=πρ^2$ and for Square is $E=a^2$ so if we try solve this in order to find the radius, then we take the following result this is if you want a constant radius otherwise if you want multiple circles you have to find the sum $ \sum_{i=1}^n ρ_i =\pm a /\sqrt{\pi} $
for interpretation you can use formulas of Gregory and Leibniz or Abraham Sharp

a = value of square side
ρ = radius
π = 3,14...
Ε = Area
l = length
n = number of shapes
d = dimension
e = the characteristic value of the shape (radius for circle , side for square)


Prove $\lim\limits_{x \to \infty} x \sin^2 x$ does not exist. I need to prove $\lim\limits_{x \to \infty} x \sin^2 x$ does not exist, and request your help verifying my argument is correct. I've got somewhat decent  at making $\epsilon-\delta$ arguments, but proving a limit does not exist has been a bit challenging actually.
We need to prove $\lim\limits_{x \to \infty} x \sin^2 x \neq \ell$, for every $\ell \in \mathbb{R}$, so take $\ell \in \mathbb{R}$. The result follows if we can find some $\epsilon > 0$ such that for every $N \in \mathbb{R}$, there is some real $x$ such that $x > N$ but $\lvert x \sin^2 x - \ell \rvert \geq \epsilon$.
If $\epsilon =\lvert \ell \rvert + 1$, then for every $N \in \mathbb{R}$ we can find $x_0 > \max(N, \quad 2\lvert \ell \rvert + 1)$ such that $x_0 = \cfrac{\pi}{2} + 2\pi k$, for some big enough $k \in \mathbb{Z}$. For this $x_0$ we have that
$$x_0 \sin^2 x_0 = x_0 \cdot 1 = x_0 > 2\lvert \ell \rvert + 1 \geq 1 + \lvert \ell \rvert + \ell$$.
This implies $x_0 \sin^2 x_0 - \ell = x_0 - \ell > 1 + \lvert \ell \rvert = \epsilon$ and the proof is complete.
Any comments or suggestions for improvement are very appreciated.

Your argument is fine, though as mentioned in the comments you don't address the possibility that the limit is infinite.
Here is a slightly simpler idea. Given any $k>0$, then $x_k=\frac\pi2+2k\pi>k$ and $x_k\sin^2x_k=x_k>k$. Also $z_k=2k\pi>k$ and $z_k\sin^2z_k=0$. So
$$
\lim_{k\to\infty}x_k\sin^2x_k=\infty,
$$
while
$$
\lim_{k\to\infty}z_k\sin^2z_k=0.
$$
This precludes the existence of the limit (finite or infinite), since if it existed all sequences that go to infinity would give the same limit.

This is just a variation of Martin Argerami's answer. He constructed a sequence $(x_k)$ such that $\lim_{k\to\infty}x_k\sin^2x_k=\infty$. Let us show that for each $a \in [0,\infty)$ there exists a sequence $(y_k)$ such that
$$\lim_{k\to\infty}y_k\sin^2y_k=a .$$
We have $\sin^2 x = 0$ for $x = m\pi$ and $\sin^2 x = 1$ for $x = \pi/2 + m\pi$ for all $m \in \mathbb Z$. Thus by the IVT the function $f(x) = x \sin^2x$ maps the interval $[m\pi, \pi/2 + m\pi]$ onto the interval $[0,\pi/2 + m\pi]$. Hence for $m \ge m_0 = \max(0,(a-2)/\pi)$ we find a point $y_m \in [m\pi, \pi/2 + m\pi]$ such that $f(y_m) = a$. Let $y_m = m\pi$ for $m < m_0$. Then clearly $y_m \to \infty$ as $m \to  \infty$ and $\lim_{m\to\infty}y_m\sin^2y_m=a$.


Base 10 theorem in number theory The question asks:
Let $n\ \in \mathbb{N}$ in base $10$ as $n=a_ka_{k-1}\cdots a_1a_0$. Let $m=a_k+a_{k-1}+\cdots + a_1+a_0$. Then $9\mid n$ if and only if $9\mid m$
Here is my Proof:
$9\mid n$ implies that $n=9k$ where $k\in \mathbb{Z}$
Next, we can express $n$ as such:
$$n=a_0+10a_1+100a_2+\cdots\\
n=a_0+(9+1)a_1+(99+1)a_2+\cdots\\
n=a_0+9a_1+a_1+99a_2+a_2+\cdots\\
n=(\underbrace{a_0+a_1+a_2+\cdots}_\text{m}) + 9(\underbrace{a_1+11a_2+\cdots}_\text{k})$$
Thus, $n=m+9k$ which implies that $9\mid(n-m)$ which can be written as the congruence, $n\equiv m\pmod 9$.
Now, from here, what steps do I need to take to prove the theorem?

We have $k\in\Bbb Z$. Therefore:
If $9|n$ then $n/9\in\Bbb Z$ so $m/9=(n/9)-k\in\Bbb Z,$ so $9|m.$
If $9\not |\,n$ then $n/9\not\in\Bbb Z$ so   $m/9=(n/9)-k\not\in\Bbb Z,$ so $9\not |\,m.$

The trick is that all of the powers of $10$ are $1\bmod9$.


proof of l'Hôpital's rule that minimizes special-casing A simple form of l'Hôpital's rule looks like this: If $u$ and $v$ are functions with $u(0)=0$ and $v(0)=0$, the derivatives $\dot{v}(0)$ and $\dot{v}(0)$ are defined,
and the derivative $\dot{v}(0)\ne 0$,
then
\begin{align*}
  \lim_{x\rightarrow 0} \frac{u}{v} &= \frac{\dot{u}(0)}{\dot{v}(0)} \qquad .
\end{align*}

To me, the clearest way to arrive at this result uses a little nonstandard analysis: Since $u(0)=0$, and the derivative $d u/d x$ is defined at $0$, $u(d x)=d u$ is infinitesimal, and likewise for $v$. By the definition
of the limit, the limit is the standard part of
\begin{equation*}
  \frac{u}{v} = \frac{d u}{d v} = \frac{d u/d x}{d v/d x} \qquad ,
\end{equation*}
where by assumption the numerator and denominator are both defined (and finite, because the derivative
is defined in terms of the standard part). The standard part of a quotient like $p/q$ equals the quotient of the
standard parts, provided that both $p$ and $q$ are finite (which we've established), and $q \ne 0$ (which is
true by assumption). But the standard part of $d u/d x$ is the definition of the derivative $\dot{u}$, and
likewise for  $d v/d x$, so this establishes the result.

The generalizations to $x\rightarrow a$, where $a\ne 0$, and $x\rightarrow \infty$ are pretty trivial with the changes of variable $x\rightarrow x-a$ and $x\rightarrow 1/x$.

But there are a bunch of other cases of l'Hôpital's that seem to me to involve toxic doses of case-splitting. There are cases where you have to differentiate more than once, and cases where the indeterminate form is $\infty/\infty$ rather than $0/0$.

Is it possible to treat all of this in a unified way, possibly using ideas from projective geometry or inversions with respect to a circle in the complex plane?

Last week I covered L'Hôpital's Rule in my Spivak Calculus course.  I ended up preferring the proof given in Rudin's Principles to Spivak's treatment, because (i) he does compile several different cases in a rather efficient way and (ii) he proves a stronger result in the case where $\lim_{x \rightarrow a} g(x) = \infty$, namely that there is no hypothesis needed on the limiting behavior of the numerator $f$.

I wrote this up in $\S$ 1 of these notes.  However I must confess that I found the proof itself not very interesting and ended up not covering it in class.

I must also confess that I don't really understand the OP's sketch proof.  In particular I do not recognize the hypothesis $g'(a) \neq 0$ that the OP is assuming: who says that $f$ and $g$ are even defined at $a$?  I am also slightly skeptical that NSA really helps here to give a shorter proof, but I would be very interested to be proven wrong about this.

Added: If you are willing to assume that $f$ and $g$ are defined and differentiable at $a$, that $f(a) = g(a) = 0$ and that $g'(a) \neq 0$, the proof becomes almost trivial:

$\lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f(x) - f(a)}{g(x)-g(a)} = \lim_{x \rightarrow a} \frac{ \frac{f(x)-f(a)}{x-a}}{\frac{g(x)-g(a)}{x-a}} = \frac{ \lim_{x \rightarrow a} \frac{f(x) - f(a)}{x-a}}{\lim_{x \rightarrow a} \frac{g(x)-g(a)}{x-a}} = \frac{f'(a)}{g'(a)}$.

(However this version is inadequate for many of the standard applications of freshman calculus.)  So I presume we're talking about a stronger version than this?

Here's a very simple proof. We consider functions $f(x)$ and $g(x)$ which have derivatives $df/dx$ and $dg/dx$ respectively. In general $dy$ means the difference between $y_1$ and $y_0$, and L'Hopital's rule applies to cases where both $f_0$ and $g_0$ are actually $0$, therefore $df$ and $dg$ are the same as $f$ and $g$ respectively. So we have:
$$\frac{f}{g} = \frac{df}{dg} = \frac{df/dx}{dg/dx}$$


What function does $\sum \limits_{n=1}^{\infty}\frac{1}{n3^n}$ represent, evaluated at some number $x$? I need to know what the function $$\sum \limits_{n=1}^{\infty}\frac{1}{n3^n}$$ represents evaluated at a particular point.

For example if the series given was $$\sum \limits_{n=0}^{\infty}\frac{3^n}{n!}$$ the answer would be $e^x$ evaluated at $3$.

Yes, this is homework, but I'm not looking for any handouts, any help would be greatly appreciated.

Let $f(x)=\sum_1^{\infty}{1\over nx^n}$; you've got $f(3)$, so you want to know what $f(x)$ is. Differentiate it.

We have $ \displaystyle -\ln(1-x) = \sum_{n \geq 1 } \frac{x^n}{n} $ by course for $|x|<1$
So this is simply $f(x)=-ln(1-x) $ evaluated at $x=\frac{1}{3}$
 Which gives the value :  $-\ln(1-\frac{1}{3})=\ln(\frac{3}{2})$


Which of the following defines norm on V? Let $V$ denote the vector space of all polynomials over $\mathbb R$ of degree less than or equal to $n$. Which of the following defines a norm on $V$.

$\|p \|^2 = |p(1)|^2 + \cdots + |p(n+1)|^2, p \in V $.
$\| p \| = \sup_{t \in [0,1]} |p(t) |, \ \  p\in V$.
$\| P \|  =  \int_0^1 |p(t)|\, dt , \ \  p \in V$
$\| p \| = \sup_{t \in [0,1]} |p'(t) |, \ \  p\in V$.

I  think (1), (2), (3) and (4) satisfy all  the properties of norm. I have done this problem, I want only to confirm that my answer is right or not.

1. $\|p\|=0$ implies $p(1)=p(2)=\cdots =p(n+1)=0$. That is the polynomial $p$ has $n+1$ zeros ; which is possible only when $p \equiv 0$ , because $p$ is of degree at most $n$. Also , if $p\equiv 0$ then clearly  $\|p\|=0$.

$\|\alpha p\|=|\alpha|.\|p\|$ ,very clear ! Triangle inequality follows from Minkowski's inequality . Hence , it defines a norm.

2. is $\sup$ norm and 3. is integral-norm , they are standard norms !

4. If you take $p$ as any non-zero constant then $\|p\|=0$. So it does not satisfy the property of norm.

The answer is not right. (I do not elaborate, because you only asked for confirmation.)


Finding the expected value of a nonnegative integer valued random variable If $X$ is a nonnegative integer valued random variable, show that

$$\mathbb{E}(X) = \sum_{i=1}^\infty\mathbb{P}(X\geq i) = \sum_{i=0}^\infty \mathbb{P}(X\geq i).
$$

I'm not sure how to do this. I only know the definition that
$$\mathbb{E}(X):= \sum_{i=-\infty}^\infty i \mathbb{P}(X=i)\mathbf{1}_{\{X\geq 0\}} = \sum_{i=1}^\infty i \mathbb{P}(X=i).
$$

Assert that most surely $X\in \Bbb N^+$ , then use that for all $k\in\Bbb N^+$, by definition of counting: $k=\sum_{j=0}^{k-1} 1$ .

$$\mathsf E(X) {= \sum_{k=1}^\infty k~\mathsf P(X=k) \\ = \sum_{k=1}^\infty\mathsf P(X=k)\sum_{j=0}^{k-1}1\\ \vdots \\ = \sum_{j=0}^\infty \mathsf P(X>j)}$$

If $E[X] < \infty$, then see here:

Expectation of an non-negative integer-valued random variable
Find the Mean for Non-Negative Integer-Valued Random Variable

If $E[X] = \infty$, I guess you have to show

$$\lim_{i \to \infty} P(X \ge i) \ne 0 \leftarrow \lim_{i \to \infty} iP(X = i) \ne 0$$


Prove that $\lim_{s \to \infty} \sum_{x=1}^{2s} (-1)^x\sum_{n=1}^{x}\frac{1}{n!}=\cosh (1) -1$ How can we prove that $$\lim_{s \to \infty} \sum_{x=1}^{2s} (-1)^x\sum_{n=1}^{x}\frac{1}{n!}=\cosh (1) -1$$

It seems like this is some kind of telescopic series, but I don't know how to find the limit of this sum. Any help would be greatly appreciated.

$$\cosh(x)=\frac{e^x+e^{-x}}{2}\Rightarrow \cosh(1)=\frac{e+e^{-1}}{2}$$

$$S=\sum_{x=1}^{2s}(-1)^x\sum_{n=1}^x\frac{1}{n!}$$
$$e^x=\sum_{n=0}^\infty\frac{x^n}{n!}\Rightarrow \cosh(1)=\frac 12\sum_{n=0}^\infty\frac{1^n+(-1)^n}{n!}=\frac{1}{2}\sum_{n=0}^\infty\frac{\left[1+(-1)^n\right]1^n}{n!}$$
now since for even odd n the terms will equal zero, we want:
$$\cosh(1)=\sum_{n=0}^\infty\frac{1^{2n}}{(2n)!}=\sum_{n=0}^\infty\frac{1}{(2n)!}$$
going back to $S$ now we have:
$$S=\sum_{x=1}^{2s}(-1)^x\sum_{n=1}^x\frac{1}{n!}=-\left(\frac 1{1!}\right)+\left(\frac 1{1!}+\frac 1{2!}\right)-\left(\frac 1{1!}+\frac 1{2!}+\frac 1{3!}\right)+...$$
and if you look at which terms you will cancel you notice that we are left with:
$$S=\sum_{x=1}^s\frac 1{(2x)!}$$
Now we are simply left with:
$$\lim_{s\to\infty}S_s=\cosh(1)$$


rational number solutions to $\frac{a}{a^2+1} + \frac{b}{b^2+1} = \frac{c}{c^2+1}$ with $abc\ne 0$ This question concerns the equation
$$\frac{a}{a^2+1} + \frac{b}{b^2+1} = \frac{c}{c^2+1}$$
and the possibility of rational number solutions with $abc \ne 0$.

In comments arising from: Using Modularity Theorem and Ribet's Theorem to disprove existence of rational solutions
It is mentioned that this equation has an infinite number of such solutions.

How can we determine that there are an infinite number of such solutions?

And is there a way to generate these solutions (or at least find a few of them)?

Let $C$ be the curve defined by the equation
$$3(a^2 + 1)(b^2 + 1) = 10( a(b^2 + 1) + b(a^2 + 1)),$$
so rational points on $C$ correspond to solutions of your original equation with $c = 3$. Let $E$ be the elliptic curve with equation
$$y^2 = x^3 + x^2 + 65x + 3458.$$

The point is that $E$ and $C$ are "birationally equivalent" over the rational numbers $\mathbf{Q}$ -- there are maps given by ratios of polynomials with $\mathbf{Q}$-coefficients from $E$ to $C$ and back again, which are well-defined away from a finite set of bad points. (I won't write down the polynomials explicitly; I got the computer program Magma to compute them for me).

Now, elliptic curves over $\mathbf{Q}$ are very well-studied objects, and we can do (or, rather, ask Magma to do) a "descent" computation to find that $E$ has infinitely many rational points and enumerate the first few. Then we just feed them into the map from $E$ to $C$ to get rational points on $C$. This gives the following solutions:
$$ (a, b) = (-627/119, 160/123), (-1287933/7579360, 20097649/13679883), (176228877641067/9078545489689, 14356927982089/53923856957760), \dots $$

If $(a, b)$ is any such solution then clearly $(a, b, 3)$ is a solution to your original problem. So your equation has infinitely many rational solutions.

(I got to this by a little trial and error: I spotted that for a fixed $c$ the equation in $a, b$ gave an elliptic curve, and I tried a few small values of $c$ until I found an elliptic curve with nontrivial rational points. Here is the Magma session:

Magma V2.21-4     Thu Oct  8 2015 10:14:42 on fermat   [Seed = 3701259969]
Type ? for help.  Type <Ctrl>-D to quit.
> A<x, y> := AffineSpace(Rationals(), 2);
> S := Curve(A, 3/(3^2 + 1)*(x^2 + 1)*(y^2 + 1) - (x*(y^2 + 1) + y*(x^2 + 1)));
> Sp := ProjectiveClosure(S);
> E, phi := EllipticCurve(Sp, Sp ! [3, 0, 1]);
> Rank(E);
1 true
> _, nu := IsInvertible(phi);
> nu(2*Generators(E)[2]);
> (-627/119 : 123/160 : 1)
> nu(3*Generators(E)[2]);
(-1287933/7579360 : 13679883/20097649 : 1)

Try it yourself at the online Magma calculator.
)

PS. One can even write down infinitely many parametric families of solutions: for instance, for any rational $c$ we can take
$$ a = \frac{-c^6 + 3c^4 + c^2 + 1}{c^7 + c^5 + 3c^3 - c}, b = \frac{c^6 - c^4 - c^2 + 1}{2(c^5 + c)}. $$

-b/(bb+1)=a/(aa+1)-c/(cc+1). Put a=0. Then b/(bb+1)=c/(cc+1). Then b=c. Infinity number of rational solutions. We can replace the problem with -f(b)/f(b)f(b)+1=f(a)/(f(a)f(a)+1)-f(c)/(f(c)f(c)+1) such that f(0)=0 and f is 1:1.


Proof of Nesbitt's Inequality: $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$? I just thought of this proof but I can't seem to get it to work.

Let $a,b,c>0$, prove that $$\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge \frac{3}{2}$$

Proof: Since the inequality is homogeneous, WLOG $a+b+c=1$. So $b+c=1-a$, and similarly for $b,c$. Hence it suffices to prove that $$\frac{a}{1-a}+\frac{b}{1-b}+\frac{c}{1-c}\ge \frac{3}{2}$$. From $a+b+c=1$ and $a,b,c>0$ we have $0<a,b,c<1$, so we have $$\frac{a}{1-a}=a+a^2+a^3+...$$ and similarly for $b,c$, so it suffices to prove that $$\sum_{cyc} a+\sum_{cyc} a^2+\sum_{cyc} a^3+...\ge \frac{3}{2}$$, or equivalently (by $a+b+c=1$) $$\sum_{cyc} a^2+\sum_{cyc} a^3+...\ge \frac{1}{2}$$, where $\sum_{cyc} a=a+b+c$ similarly for $\sum_{cyc}a^n=a^n+b^n+c^n$. Here I get stuck. For example, doing the stuff below yields a weak inequality, because of too many applications of the $a^2+b^2+c^2\ge (a+b+c)^2/3$ inequality.

"stuff below": Now, from $0<a<1$ we have $a^3>a^4$, $a^5>a^8$, $a^7>a^8$, $a^9>a^{16}$, and so on, so it suffices to prove that $$\sum_{cyc} a^2+2\sum_{cyc} a^4+4\sum_{cyc} a^8+8\sum_{cyc} a^{16}+...\ge \frac{1}{2}$$, or, multiplying by 2, $$2\sum_{cyc} a^2+4\sum_{cyc} a^4+8\sum_{cyc} a^8+...\ge 1$$, which by a simple inequality (i.e. recursively using $a^{2^n}+b^{2^n}+c^{2^n}\ge \frac{(a^{2^{n-1}}+b^{2^{n-1}}+c^{2^{n-1}})^2}{3}$) is equivalent to $$2(1/3)+4(1/3)^3+8(1/3)^7+...+2^n(1/3)^{2^n-1}+...\ge 1$$. But then http://www.wolframalpha.com/input/?i=%5Csum_%7Bi%3D1%7D%5E%7B5859879%7D+2%5Ei+*%281%2F3%29%5E%282%5Ei-1%29 and so we're screwed.

I'm sorry but AM-GM Inequality is all that I know,
So, By using AM-GM Inequality for $ (a+b) , (b+c) , (c+a)$ and
$\frac{1}{b+c},\frac{1}{c+a},\frac{1}{a+b}$
We get
$$\frac{(a+b)+(b+c)+(c+a)}{3}
\ge\sqrt[3]{(a+b)(b+c)(c+a)}\tag{1}$$
$$\frac{\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}}{3}
\ge\sqrt[3]{\frac{1}{(b+c)(c+a)(a+b)}}\tag{2}$$
Multiplying $(1)$ and $(2)$ We get.
$$\frac{1}{9}\left[(a+b)+(b+c)+(c+a)\right]\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\ge\sqrt[3]{\frac{(a+b)(b+c)(c+a)}{(b+c)(c+a)(a+b)}}$$
$$\left[(a+b)+(b+c)+(c+a)\right]\left(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b}\right)\ge9$$
After Expanding
$$3+\frac{a+b}{b+c}+\frac{c+a}{b+c}+
\frac{a+b}{c+a}+\frac{b+c}{c+a}+
\frac{b+c}{a+b}+\frac{c+a}{a+b}
\ge\ 9$$
By rearrangement
$$\left(\frac{a+b}{b+c}+\frac{c+a}{b+c}\right)+
\left(\frac{a+b}{c+a}+\frac{b+c}{c+a}\right)+
\left(\frac{b+c}{a+b}+\frac{c+a}{a+b}\right)
\ge\ 6$$
$$\frac{2a+b+c}{b+c}+
\frac{a+2b+c}{c+a}+
\frac{b+2c+a}{a+b}
\ge\ 6$$
$$\frac{2a}{b+c}+\frac{b+c}{b+c}+
\frac{2b}{c+a}+\frac{a+c}{c+a}+
\frac{2c}{a+b}+\frac{b+a}{a+b}
\ge\ 6$$
$$3+\frac{2a}{b+c}+
\frac{2b}{c+a}+
\frac{2c}{a+b}
\ge\ 6$$
$$\frac{2a}{b+c}+
\frac{2b}{c+a}+
\frac{2c}{a+b}
\ge\ 3$$
$$\frac{a}{b+c}+
\frac{b}{c+a}+
\frac{c}{a+b}
\ge\ \frac{3}{2}$$
As Simple as that


Two circle intersection: help on understanding a specific explanation As someone with basic algebra knowledge, I am having trouble understanding Paul Bourke's explanation on "Intersection of two circles" on this page.

The specific part that I don't understand is where the last three equations come from:

1) $P_2 = P_0 + a \cdot \dfrac{ P_1 - P_0 } {d}$

2) $x_3 = x_2 \pm h \cdot\dfrac{ y_1 - y_0 }{d}$

3) $y_3 = y_2 \mp h\cdot  \dfrac{ x_1 - x_0} {d}$

Can anyone help shed some light on how they are derived? Or at the very least point me in the right direction?

All segment lengths are considered as positive. You have to add in the negative signs somewhere when necessary.

$x_3 = x_2 + k$

$= x_2 + h \sin \theta$ ... [from the smallest triangle]

$= x_2 + h \cdot \dfrac {y_1 – y_0}{d}$ ... [from the largest triangle]

It is part of the statement #2. Other statements can be proved in a similar fashion.

In eq. 1) the second term is a vector distance (P1-P0) which is then factorized by a/d so that P2 lies in the line connecting P0 and P1 at a distance a/d starting at P0. This equation is not used for calculations anyway.

To better understand equation 2 you could draw the axis x and y in anyway you want and you'll find out that (y1-y0)/d corresponds to sin(θ) where θ is the angle between the line P0P1 and the x axis.

h⋅(y1−y0)/d = h.sin(θ)

The angle θ happens to be also the same angle between the line P2P3 and the y axis which you use to derive the x component of P3 by computing h.sin(θ).

Eq. 3 can be explained in the same way as eq. 2, but in this case (x1−x0)/d corresponds to cos(θ).

It would be clearer by looking at a drawing but hopefully you'll be able to draw it yourself after this explanation.


Prove that there is no probability measure on $\mathbb{N}$ Prove that there is no probability measure on $\mathbb{N}$ such that for any $n\geq 1$, the probability of the set of multiples of $n$ is $1/n$.

I'm trying to proof this by contradiction, but actually I have no clue so far. Much appreciated if someone can help.

Suppose there were such a measure, $\mu$. $\newcommand\of[1]{\left({#1}\right)}\newcommand\N{\Bbb{N}}$

Let $X_n=\{m\in\Bbb{N} : n\mid m\}$ be the set of multiples of $n$. Certainly these sets must be measurable.

First let's deal with a simple case, computing
$\mu(X_{p_1}\cup \cdots \cup X_{p_i})$, where the $p_i$ are all distinct primes. Since the primes are all distinct, we get that finite intersections of sets have measure the product of the measures of the individual sets.
By PIE, we then get that
$$
\mu(X_{p_1}\cup\cdots \cup X_{p_i})
= \sum_{j=1}^i \frac{1}{p_j} - \sum_{j_1<j_2} \frac{1}{p_{j_1}p_{j_2}} + \cdots + (-1)^{i+1} \frac{1}{p_1p_2\cdots p_i}.
$$
Now, if we subtract both sides from 1, then the right hand side factors to give
$$
1-\mu(X_{p_1}\cup\cdots \cup X_{p_i})
= \prod_{j=1}^i \of{1-\frac{1}{p_j}},
$$
so
$$\mu(X_{p_1}\cup\cdots \cup X_{p_i}) = 1- \prod_{j=1}^i \of{1-\frac{1}{p_j}}
=1-\frac{\phi(p_1\cdots p_i)}{p_1\cdots p_i}.$$

  Now we make an observation: For any sequence $n_1,\ldots,n_k,\ldots$ containing all but finitely many prime numbers,
  $$\mu\of{\bigcup_{k\in\N} X_{n_k}} = 1.$$

Proof.

We can assume that the $n_i$ are all prime, since adding extra sets only increases the measure. The result then follows from the previous result combined with the following fact:
$$\prod_{p\text{ prime}} \of{1-\frac{1}{p}} = 0.$$

To prove this fact, take $-\log$ of a partial product across primes below some upper bound, $p<N$, which gives
$$\sum_{p < N} -\log(1-1/p) = \sum_{p<N}\sum_{n=1}^\infty \frac{1}{p^nn} \ge \sum_{p<N} \frac{1}{p}.$$
The reciprocals of the primes are well known to be a divergent series.
(Most proofs of this actually go the other direction and prove that this product diverges, but I'm just trying to get a quick reduction to a well known fact, since this is not the point of my answer).
Thus $-\log$ of the partial products go to $\infty$, which implies that the partial products converge to $0$.

This completes the proof of this fact. $\blacksquare$

Now we can show that if all the $X_n$ are measurable and there is a measure with value $\mu(X_n)=1/n$, then for all naturals $m$, $\{m\}$ is measurable and has measure $0$, which contradicts the claim.

Proof.

Induct on $m$ by divisibility.
First $\{1\}$ is the complement of
$$\bigcup_{p\text{ prime}}X_p,$$ which we just showed has measure $1$, so our base case holds.

For $m>1$, assume the result for all divisors of $m$. Then
$$\mu(\{ d : d\mid m\}) = \mu\of{\Bbb{N} \setminus \bigcup_{d\nmid m} X_d}=1-1=0.$$
By the inductive hypothesis, $\mu(\{d:d\mid m\}) = \mu(\{m\})$, so $\mu(\{m\})=0$, as claimed.

Finally, for $m=0$, since $\{0\}=\bigcap_{n\in\N} X_n$, we have $\{0\}$ is measurable and $\mu(\{0\}) < 1/n$ for all $n$, which implies $\mu(\{0\})=0$.

Thus the claim is proved. $\blacksquare$

I believe such a probability measure would need to be uniform, so we have two cases: either $P(X=n)=0$ for all $n \in \mathbb{N}$, or $P(X=n)=\epsilon>0$ for all $n \in \mathbb{N}$.

If $P(X=n)=0$, then $1=P(\mathbb{N})=\sum_{n=1}^{\infty} P(X=n)=0$, a contradiction.

If $P(X=n)=\epsilon>0$, then $1=P(\mathbb{N})=\sum_{n=1}^{\infty} P(X=n)=\sum_{n=1}^{\infty} \epsilon=+\infty$, another contradiction.

Therefore, $\mathbb{N}$ cannot have a uniform probability measure.


Applying Stokes Theorem Im struggling to solve this problem :

    Consider F$(x,y,z) = (y,z,x)$ and the surface $\phi(u,v) = (u,v,sin(u^2 + v^2))$ where $(u,v)$ belongs to some disc centred at $(0,0)$ and $r=3$. Find the surface integral using Stoke's Theorem.

So far I have parametrised the curve so that
$$ \gamma(t) = (3cos(t),3sin(t), 0)$$

I'm slightly unsure of the method used to solve this integral using Stoke's Theorem. What should i be taking the curl of?

The correct answer should be zero which is not what im getting.

Hint: You can write
$$F=\nabla\times G$$
where
$$G(x,y,z)=\left (\frac{z^2}{2},\frac{x^2}{2},\frac{y^2}{2}\right )$$

in order to use Stokes' theorem to express
$$\int_S F\cdot \mathrm d\Sigma=\int_S \left (\nabla \times G\right ) \cdot \mathrm d \Sigma=\oint_{\partial S}G\cdot \mathrm d s$$


How to evaluate this partial derivative? If we have some function:

$$ f(x,y)= \begin{cases} \frac{xy(x^2-y^2)}{x^2+y^2} &\text {if }(x,y) \neq (0,0) \\
0 & \text{otherwise}\end{cases}$$

then find $$\frac{d^2f}{dxdy}(0,0).$$

This is my question I need to solve. Now I was thinking that it would be $0$ since $f(x,y)=0$ at $(0,0)$ but according to the workings it is not, now I have no idea how to evaluate this derivative and would like some help?

Also the question is posed asking for this  $$\frac{d^2f}{dxdy}(0,0)$$.
 but shouldn't it be a partial derivative or something? I don't know I'm really confused. Any help?

In order to find $\frac{\partial ^2f}{\partial x\partial y}(0,0)$ you first need to find $\frac{\partial f}{\partial y}(t,0)$ for $t$ in a neighborhood of $0$.

$\frac{\partial f}{\partial y}(0,0)$ is easily seen to be $0$ because $f(0,y)=0$ no matter whether $y$ is zero or nonzero.

For nonzero $t$ you only need to look at the first case of the definition of $f$, because that case governs $f(t,y)$ for all possible $y$. Holding $t$ constant, you would use the quotient rule to differentiate with respect to $y$, but since the derivative of the denominator is $0$ one of the terms vanish and we get just
$$ \frac{\partial}{\partial y}\frac{ty(t^2-y^2)}{t^2+y^2} \bigg|_{y=0}
= \frac{\frac{\partial}{\partial y}ty(t^2-y^2)}{t^2+y^2} \bigg|_{y=0} $$
When the dust clears you get a very simple expression for this which agrees with what we just found for $t=0$ and which is trivial to differentiate with respect to $t$.

Use the quotient rule and differentiate partially for x (treating y as a constant) and then partially for y (treating x as a constant)


Proving little o by first principles I'm trying to prove that $15n+7$ is $o(n\log n)$ (by first principles, i.e. no limits).

My idea is to solve for n to determine $n_0$ and then work backwards from there. But I can't seem to find a way.

$15n+7 < cn\log n$

$7 < n(c\log n-15)$

I got stuck here.

The problem is to prove that
$$
\bigg( \frac{15n + 7}{n\log n} \bigg) \to 0
$$
as $n \to \infty$;
for then we have by definition $15n + 7$ is $o(n\log n)$.

If $n > 1$, then
$$
\frac{15n + 7}{n\log n} = \frac{15}{\log n} + \frac{7}{n\log n} < \frac{15}{\log n} + \frac{7}{(\log n)^{2}}.
$$
Let $\varepsilon > 0$. Note that we have
$$
\frac{15}{\log n} < \frac{\varepsilon}{2}
$$
if $n > e^{30/\varepsilon} =: n_{1}$ and that we have
$$
\frac{7}{(\log n)^{2}} < \frac{\varepsilon}{2}
$$
if $n > e^{\sqrt{14/\varepsilon}} =: n_{2}$.
So taking $n_{0} := \max \{ 1, n_{1}, n_{2} \}$ suffices.

If $15n+7$ is $o(n \log n)$ then

$$15n+7 < cn \log_2n$$

for every $n \geq n_0$ for some positive real $c$ and $n_0$.

$$7 < n(c\log_2n - 15)$$

If we chose $n_0 = 1$ then this would change to $7 < -15$ which is false, so we need a larger $n$. Let's do $n_0 \geq 2$.

$$7 < 2(c - 15) = 2c - 30$$

And $7 < 2c -30$ is true for $c > 18.5$, so the proof is complete.


What sorts of problems can fractals solve? After doing a bit of research on fractals, I was wondering what sort of real-life applications do fractal have and in what way would they be used to help solve a problem. I already know people use fractals to map coastlines and borders, but is there any other 'smaller' problems that could be solved by using the principle of fractals and what would be some examples of these and how do they work?

Also, is there a way to generate a problem of a certain situation that could be solved by a fractal or its principles?

Fractals are used in many computer games to render realistic graphics for mountains, landscapes & 3D terrains, especially for flight simulations, computer games, digital artworks & animations. Rather than storing a huge amount of detailed height data in the computers memory, fractal-based algorithms generate the data 'on-the-fly' to render realistic landscapes in real-time. Algorithms for generating fractal-based landscapes take advantage of the 'self-similarity' which exist in the natural world to render visually rich and complex images with attention to fine detail at all scales to appear smooth and not suffer from 'pixelation' even when zooming in close.

These algorithms use methods such as recursive subdivision and fractional Brownian motion to generate a 3D landscape which is then smoothed using a variety of techniques such as image filtering and polynomial interpolation (splines & Bezier curves) to generate photorealistic images of rolling hills & other natural scenes.

Pioneers in this field include Ken Musgrave, digital artist and CEO of Pandromeda (http://www.pandromeda.com/) and  Loren Carpenter who went on to form Pixar Studios.

Not clear of what you mean by 'smaller' problems but anyway I suppose you should take a glance of the journals 'Fractals' and 'Chaos, Solitons & Fractals'.


Baby Rudin examples 2.21: openness and closedness of subsets of $\mathbb{R}^2$ On page 33 of Principles of Mathematical Analysis by Walter Rudin(3rd Ed.), there are the examples:

2.21 Examples Let us consider the following subsets of $ \mathbb{R}^2 $:
(excluding the irrelevant parts)
(g) The segment $ (a,b) $.

I can't understand why Rudin says (g) can be regarded as a subset of $ \mathbb{R}^1 $. I know $ ((1,0),(2,0)) $ can be a subset of $ \mathbb{R}^1 $,  but what about $ ((1,2),(3,4)) $? Rudin also says the segment is not open if we regard it as a subset of $ \mathbb{R}^2 $, but is an open subset of $ \mathbb{R}^1 $. I understand such a segment is open in the real line, but how can it be close in $ \mathbb{R}^2 $?

This may be an old thread, but I am currently reading the book, so to speak I will give my $2$ cents:

If $(a,b)$ is a subset of $\mathbb{R}^1$, then the elements of $(a,b)$ are those $x$ for which we have
$$a < x < b , x\in \mathbb{R}^1$$

My idea for $\mathbb{R}^2$(probably not what Rudin meant):
If $(a,b)$ is a subset of $\mathbb{R}^2$, we can think of $(a,b)$ as an interval, just not being closed, that is:
then $a=(q,w), b=(z,r)$, that is $a$ and $b$ are from $\mathbb{R^2}$,
then if $x=(o,p)$ is an element of $(a,b)$,
it follows that $q < o < z$ and also $w < p < r$.
That is we have a $2$-cell but we have removed it's borders.

From my understanding of the answers I've read so far, it seems Rudin thinks of something else.
Most likely the meaning is that we think of $(a,b)$ in $\mathbb{R}^2$ as of points of type $(x,0)$ for which $a < x < b$, that is $a,x,b$ are from $\mathbb{R}^1$, hence $(a,b)$ is a line in the plane and it follows that we can't have a neighbourhood in just a line, cause in $\mathbb{R}^2$ it's a circle instead.

You can think of the segment $ ((1,2),(3,4)) $ as a subset of the (infinite) line that passes through the points $ (1,2)$ and $ (3,4) $. Every such line is an $ \mathbb{R}^1 $.

You are right, it is not closed in $ \mathbb{R}^2 $. Remember that a set can be both not closed and not open.


When $cov(Z_s,Z_u)=\sigma^2_{\max(s,u)}$ for normal random variables? Let $Z_1, Z_2, \dots, Z_n$ be normal random variables with mean zero and $Var(Z_i)=\sigma_i^2$.
Under what condition $cov(Z_s,Z_u)=\sigma^2_{\max(s,u)}$?

I will only consider the $n=2$ case for a counter-example, as my linear algebra is not strong. But a covariance matrix M must be positive semi-definite, i.e. $z^T M z \geq 0$ for all column vectors $z$ of length $n$. As you have defined it, the covariance matrix would be of the form $M = (a,b;b,b)$, for some $a,b > 0$. Let $z = (x;y)$. Then to be positive semi-definite, $ax^2 + 2bxy + by^2 = ax^2 + b(2xy + y^2)\geq 0$ for all $x,y$ (this expression can be obtained by standard matrix operations). Only $b(2xy + y^2)$ can be negative, so we need $ax^2 \geq -b(2xy + y^2)$. For any $y, a, b$ there should be a suitable $x$ which breaks this inequality, so we have shown that this matrix cannot be a covariance matrix.


Show that $1/T$ has the same distribution as $T$ Let T be the ratio X/Y of two i.i.d. N (0, 1) r.v.s. X, Y . This is the Cauchy distribution and, as shown  it has PDF $f_T(t)= \frac{1}{\pi (1+t^2)}$
(a) Show that $1/T$ has the same distribution as $T$ using calculus, after first finding the CDF of $1/T$.
(b) Show that $1/T$ has the same distribution as $T$ without using calculus, in 140 characters or fewer.

a) $Y= \frac 1T$, $P(Y \le y)=P(\frac1T \le y)=P(T \le 1/y)= \int_{-\infty}^{1/y} \frac{1}{\pi} \frac {1}{(1+t^2)}dt=\frac {tan^{-1}(1/y)}{\pi}$; $y \in (-\infty;\infty)$
$f_Y(y)= \frac {-1}{\pi x^2+\pi}$
b) Can you help me with $b$ and check $a$?

$$P(\frac{1}{T} \le y) = P(T \ge \frac{1}{y}) =  \int_{\frac{1}{y}}^{\infty} \frac{1}{\pi(1+t^2)}dt$$

$$=\left[\frac{1}{2}-\frac{cot^{-1}(y)}{\pi} \right]$$

$$ F_{Y}(y)= \left[\frac{1}{2}-\frac{cot^{-1}(y)}{\pi} \right]$$

$$f_{Y}(y) = \frac{d}{dy}F_{Y}(y)$$

$$f_{Y}(y) = \frac{1}{\pi(1+y^2)}$$

For Part B)

As @Hw Chu said, 1/T is the ratio Y/X, but X and Y are i.i.d. so this is really the same as X/Y and hence the same density.


find the general solution(ode) Isn't the general solution yh + yp? What am I doing wrong? Got everything else correct, but stuck on question d.

HINT

To start with, rearrange the given ODE in the following manner

\begin{align*}
y^{\prime\prime} - 2y^{\prime} + y = (y^{\prime} - y)^{\prime} - (y^{\prime} - y)
\end{align*}

Then make the substitution $u = y^{\prime} - y$, which reduces the problem to solving a first order ODE twice.

The given answer is not correct.

Your $$y_h = c_1e^t+c_2te^t$$ is correct but the $y_p$ is not .
To find $y_p$ they used the method of variation of parameters as explained in the solution but they have their $y_p$  incorrect. The correct $y_p=-t^2e^t$ does not need the $c_1$ and $c_2$ but it uses $u_1e^t+u_2te^t$ where $u_1$  and $u_2$ are  found by the integrals.

The correct answer is $$y=  c_1e^t+c_2te^t -t^2e^t $$


Using continuity of the measure in a proof... I'm trying to understand the following proof:

I don't understand how the conclusion came from the equation in the green box, did they use continuity of the measure?

The existing answers are confused about what the relevant part of the proof is trying to conclude. Let's recap: The green box and the conclusion that followed at the bottom are all part of the $\Rightarrow$ direction, which is

  If equality $\mu^*(E) = \mu^+(E)$ holds, then there exists an $A \in \mathcal M^*$ such that $E \subset A$ and $\mu^*(E) = \mu^*(A)$.

Since the asker appears to accept that $B \in \mathcal M^*$ and $E \subset B$, what we need to prove is that $\mu^*(E) = \mu^*(B)$ (not $\mu^*(E) = \mu^*(E)$ as was written in the proof – that would be a complete triviality).

We use the monotonicity of outer measures. Note that $E \subset B$ implies

$$
\mu^*(E)
\leq \mu^*(B). \tag{1}
$$

On the other hand, by the definition of $B$, we have $B \subset A_{1/n}$ for all $n$. So monotonicity and the green box tells us that

$$
\mu^*(B)
\leq \mu^*(A_{1/n})
\leq \mu^*(E) + \frac{1}{n}
\;\text{for all $n \in \mathbb Z_{>0}$}. \tag{2}
$$
The statements $(1)$ and $(2)$ combine to force $\mu^*(E) = \mu^*(B)$, as desired.

It looks like there is a typo in the green box. The far right should probably be $$\mu^+(E) + \frac{1}{n}.$$ And so we then have that for all $n$, $$\mu^*(E) \le \mu^+(E) + \frac{1}{n}.$$ This implies that the same inequality holds without the $\frac{1}{n}$. But from the middle of the large box, we have the reverse inequality. Therefore, equality holds as desired. (Of course, I'm explaining only the last sentence of the proof.)


What is the importance of the Collatz conjecture? I have been fascinated by the Collatz problem since I first heard about it in high school.

  Take any natural number $n$. If $n$ is even, divide it by $2$ to get $n / 2$, if $n$ is odd multiply it by $3$ and add $1$ to obtain $3n + 1$. Repeat the process indefinitely. The conjecture is that no matter what number you start with, you will always eventually reach $1$. [...]

  Paul Erdős said about the Collatz conjecture: "Mathematics is not yet ready for such problems." He offered $500 USD for its solution.

How important do you consider the answer to this question to be? Why?

Would you speculate on what might have possessed Paul Erdős to make such an offer?

EDIT: Is there any reason to think that a proof of the Collatz Conjecture would be complex (like the FLT) rather than simple (like PRIMES is in P)?  And can this characterization of FLT vs. PRIMES is in P be made more specific than a bit-length comparison?

Most of the answers so far have been along the general lines of 'Why hard problems are important', rather than 'Why the Collatz conjecture is important'; I will try to address the latter.
I think the basic question being touched on is:

In what ways does the prime factorization of $a$ affect the prime factorization of $a+1$?

Of course, one can always multiply out the prime factorization, add one, and then factor again, but this throws away the information of the prime factorization of $a$.  Note that this question is also meaningful in other UFDs, like $\mathbb{C}[x]$.
It seems very hard to come up with answers to this question that don't fall under the heading of 'immediate', such as distinct primes in each factorization.  This seems to be in part because a small change in the prime factorization for $a$ (multiplication by a prime, say) can have a huge change in the prime factorization for $a+1$ (totally distinct prime support perhaps).  Therefore, it is tempting to regard the act of adding 1 as an essentially-random shuffling of the prime factorization.
The most striking thing about the Collatz conjecture is that it seems to be making a deep statement about a subtle relation between the prime factorizations of $a$ and $a+1$.  Note that the Collatz iteration consists of three steps; two of which are 'small' in terms of the prime factorization, and the other of which is adding one:

multiplying by 3 has a small effect on the factorization.
adding 1 has a (possibly) huge effect on the factorization.
factoring out a power of 2 has a small effect on the factorization (in that it doesn't change the other prime powers in the factorization).

So, the Collatz conjecture seems to say that there is some sort of abstract quantity like 'energy' which cannot be arbitrarily increased by adding 1.  That is, no matter where you start, and no matter where this weird prime-shuffling action of adding 1 takes you, eventually the act of pulling out 2s takes enough energy out of the system so that you reach 1.  I think it is for reasons like this that mathematicians suspect that a solution of the Collatz conjecture will open new horizons and develop new and important techniques in number theory.

Aside from the mathematical answers provided by others, the computational verification of the Collatz problem is a good exercise for programmers. There are many optimization opportunities (e.g., time-space trade-off using lookup tables, parallelism), many pitfalls (e.g., integer type overflow), possibilities of exploiting various implementation tricks (e.g., count trailing zeros instructions available in modern hardware), etc. It is a simple task where you can practice many basic programming constructions (branching the program, do-while loops, recursion). And for these reasons, this is arguably the most common task you can find in many online or university courses (e.g., Harvard University's CS50 course).


Symmetry of trig integrals = $0$? $\int_{-\pi}^{\pi}\frac{\cos(x)}{\sqrt{4+3\sin(x)}}$ $$\int_{-\pi}^{\pi}\frac{\cos(x)}{\sqrt{4+3\sin(x)}}dx$$

Because the integral limits go from $-\pi$ to $\pi$ ...can one immediately conclude this integral equals zero? The area under the curve should cancel itself out, across the $y$-axis, right?

You can certainly use symmetry, since if
$$
f(x) = \frac{\cos(x)}{\sqrt{4+3\sin(x)}},$$
then
$$
f(x-\pi/2) = \frac{\sin (x)}{\sqrt{4-3 \cos (x)}},$$
which is an odd function. Thus,
$$
\int_{-\pi}^{\pi} f(x-\pi/2) = 0.
$$
Geometrically, this means that $f$ has odd symmetry about the line $x=\pi/2$, thus
$$
\int_{-\pi/2}^{3\pi/2} f(x) dx = 0,
$$
as you can see:

Of course, we then get that the integral is zero over any interval of length $2\pi$ by periodicity.

You can also use periodicity directly since, with substitution $u=\sin(x)$, we get
$$
\int_{-\pi}^{\pi} \frac{\cos(x)}{\sqrt{4+3\sin(x)}} \, dx =
 \int_0^0 \frac{1}{\sqrt{4+3u}} \, du = 0.
$$
In fact, this shows that
$$
\int_I g(\sin(x))\cos(x) \, dx = 0,
$$
for any integrable function $g$ over any interval $I$ of length $2\pi$.

If $f(x)$ is an odd integrable  function and $\alpha $ is a real number then we have $$ \int _{-\alpha} ^ \alpha  f(x) dx =0$$

Unfortunately in the case of your function,  $$ f(x) = \frac {\cos x}{\sqrt {4+3\sin x}}$$ is not an odd function.

The integral may or may not be zero, but you can not use the idea of an odd function here.


Non linear ODE Problems I encountered with this problem very recently  here it is

$\frac{dx}{dt}=-x$

$\frac{dy}{dt}=2x^2 y^2$

Find the critical points

find the different equation of paths

solve equation ti find the paths

I really dont know how to solve this equations, if it is linear format i can work out easily .

The first equation is linear, yielding $x(t)=x_0e^{-t}$.
You can put this into the second equation, giving
$$\frac{dy}{dt}=2x_0^2e^{-2t}y^2$$
which is separable.  Can you take it from there?

afther solved we have:
$$y(t)=\frac1{x_0^2 e^{-2t} +y_0 }$$


Expected value is a linear operator? Under what conditions is median also a linear operator? I have always taken for granted that expected value is a linear operator. For any random variables $X$ and $Y$: $E(aX + bY) = aE(X) + bE(Y)$. Can anyone point me to a rigorous proof of this?

Also, I know that generally median $Med()$ is not a linear operator, meaning $Med(aX + bY)$ might not be equal to $a Med(X) + b Med(Y)$. Are there absolute criteria / rules when $Med$ is a linear operator, and when it is not?

First question: By definition, on a probability space ($\Omega,\mathcal{F},P$), the expected value of a random variable $X:\Omega\to \mathbb{R}$ is defined as
$$E(X)=\int_\Omega X(\omega) dP(\omega).$$ Note that it is only well-defined if the integral converges absolutely, i.e. $$\int_\Omega |X(\omega)| dP(\omega)<\infty$$

(The integrals above are Lebesgue integrals. If $X$ is discrete, $P$ is the point measure and this integral turns into a sum.)

Therefore, if the expected values of both $X$ and $Y$ exist, then $E(X+Y)$ exists (triangle inequality) and, for $a,b\in \mathbb{R}$, we can use the linearity of the Lebesgue integral to conclude
$$E(aX+bY)=\int_\Omega aX +bYdP=a\int_\Omega X dP+b\int_\Omega YdP=aE(X)+bE(Y).$$

I would venture that expectation being a linear operator is somewhat in the nature of an axiom—that each generalization to more complex sorts of distributions would have to satisfy this axiom to be considered sensible.


Prove that at a party of $25$ people there is one person knows at least twelve people. So, the full problem goes like this:

  There are $25$ people at a party. Assuming that among any three people, at least two of them know each other, prove that there exists one person who must know at least twelve people.

I've been stuck on this problem for a while and haven't really figured out how to proceed. I'm pretty sure that there is an answer that can be found via the pigeonhole principle or some graph theory, but I'm not really sure how to get started. Any help would be appreciated.

If everyone knows everyone, then you are done.

Otherwise choose two people, A and B say, who don't know each other.  These two people are part of $23$ triples.  In each of these triples, either A knows the third person, or B knows the third person.

Thus one of A or B knows (at least) $12$ people.

Let A is the person who knows 12 people. Now, if there are 12 pairs (12*2) and the A is the third person then A knows at least one of them (pairs) So A knows 12 persons. :)


Prove there exists $\xi \in (a,b)$ such that $f(\xi)=f^{(n+1)}(\xi)$. Problem
Let $f(x)$ be $n$-times differentiable over $[a,b]$ and $n+1$-times differentiable over $(a,b)$. $f^{(k)}(a)=f^{(k)}(b)=0$, where $k=0,1,2,\cdots,n$. Prove there exists $\xi \in (a,b)$ such that $f(\xi)=f^{(n+1)}(\xi)$.
Attempt
Consider applying Taylor's formula expanding at $x=a,b$. We have
\begin{align*}
f(x)&=f(a)+f'(a)(x-a)+\cdots+\frac{f^{(n+1)}(\xi_1)}{(n+1)!}(x-a)^{n+1}
=\frac{f^{(n+1)}(\xi_1)}{(n+1)!}(x-a)^{n+1}.
\end{align*}
and
\begin{align*}
f(x)&=f(b)+f'(b)(x-b)+\cdots+\frac{f^{(n+1)}(\xi_2)}{(n+1)!}(x-b)^{n+1}
=\frac{f^{(n+1)}(\xi_2)}{(n+1)!}(x-b)^{n+1}.
\end{align*}
Can we go on from these?
Edit
Later I consulted some reference books and find a similar problem in the book named Problems In Real Analysis: Advanced Calculus On The Real Axis.

Thus, I wonder whether the conclusion holds or not ,if we are only given that   $f(x)$ is $n$-times differentiable over $[a,b]$ and $n+1$-times differentiable over $(a,b)$.

Failed attempt:

Consider the function
$$g(x)=(f(x)+f'(x)+\ldots+f^{(n)}(x))e^{-x} $$
that is continuous(!) over $[a,b]$ and differentiable over $(a,b)$ with derivative
$$\begin{align}g'(x)&=(f'(x)+f''(x)+\ldots+f^{(n+1)}(x))e^{-x}-(f(x)+f'(x)+\ldots+f^{(n)}(x))e^{-x}\\&=(f^{n+1}(x)-f(x))e^{-x}.\end{align}$$
As $g(a)=g(b)=0$, the Rolle theorem tells us that here exists $\xi\in(a,b)$ with $g'(\xi)=0$.

Correction:

Unfortunately, $g$ need not be continuous at $a$ or $b$, so the Rolle theorem cannot be applied. $g$ will only be continuous (and differentiable) over $(a,b)$. In other words, the Rolle theorem holds under somewhat weaker assumptions, but it is usually not presented that way - for the simple reason that Rolle is one of the first theorems about derivatives one encounters.
Here's how to mend the mistake above:

Assume $g$ is injective over $(a,b)$.
Then $g$ must be strictly monotonic over $(a,b)$ (and in particular not constant). Pick $c\in(a,b)$ with $g(c)\ne 0$, wlog. $g(c)>0$ (or else replace $f$ with $-f$). Then either $g(x)>g(c)$ for all $x\in(a,c)$ or for all $x\in (c,b)$; wlog the former (or else replace $f$ with $x\mapsto f(a+b-x)$).
As $e^x$ is bounded over $[a,c]$, it follows that there exist $\delta>0,\epsilon>0$ such that $$\tag1h(x):=e^xg(x)>\epsilon$$ for all $a<x<a+\delta$. But
$ h(x)=f(x)+f'(x)+\ldots+f^{(n)}(x)$ is the derivative of $x\mapsto \int_a^xf(t)\,\mathrm dt+f(x)+\ldots +f^{(n-1)}(x)$ and hence has the Darboux (aka. Inermediate Value) property. In particular, as $h(a)=0$, there must exist $x\in(a,a+\delta)$ with $h(x)=\frac12\epsilon$, contradicting $(1)$.

We conclude that $g$ cannot be injective over $(a,b)$. Hence, there are $u,v$ with $a<u<v<b$ and $g(u)=g(v)$. By Rolle, there exixts $\xi\in(u,v)$ with $g(\xi)=0$. This implies that $f^{(n+1)}(\xi)=f(\xi)$.

First, let's prove the case of $n=1$

You have two contnuous functions. $f(x)$ and $g = f'(x)$. Let's look at the interval $[a, a+ s]$ such that for the first time $f(s)=0$ That means that the derivative has different signs at $a$ and $a+s$. (The case of the function touching the line should be treated carefully, but nothing special there.) So, we have $f-g$ that is positive on the one side of $[a, a+s]$ and negative on the another. By continuousity, there is $0$ of $f-g$ somewhere.

I don't know how to make an indction step, but looks like that's possible.


Fermat Little Theorem The problem I am trying to solve is how to use Fermat Little theorem to prove that the number 66013 is not prime. I found this problem on another website (Question Cove). The student who had to solve this problem didn't know how to proceed. I offered to help. I read the following lecture notes Princeton lecture notes on cryptography and thought I would follow the same procedure. I wrote
$66013 = 257^2 + 36 = 256^2 + 513 + 36 = 2^{16} + 2^9 + 37$ but I am stuck after that. What is the correct approach?

Fermat's little theorem is an incredibly inefficient method for proving primality. In this case the standard test for divisibility by 3 (by adding the digits and checking whether the sum is a multiple of 3) shows that $66213$ is a multiple of $3$.

Pick $a:=3$.  We have
$$a^{66212}=3^{66212}\not\equiv1\pmod{66213}$$
because $3\mid 66213$.  Well, there, I used Fermat's Little Theorem!

Edit:  Originally, the OP's question was to disprove the primality of $66213$.

For the corrected question (with number $66013$), choose $a:=15781$.  Then,
$$a^2=249039961\equiv 38925\pmod{66013}\,.$$
That is,
$$a^4\equiv 38925^2=1515155625\equiv25249\pmod{66013}\,.$$
Finally,
$$a^5=a\cdot a^4\equiv 15781\cdot 25249=398454469\equiv 1\pmod{66013}\,.$$
That is,
$$a^{66013-1}=a^2\,\left(a^5\right)^{13202} \equiv a^2\cdot 1^{13202}\equiv a^2\equiv 38925\pmod{66013}\,.$$
Thus, $66013$ is not prime.


$\sum_{n=1}^{\infty}(a_{n}-a_{n-1})x^n$ and $\sum_{n=0}^{\infty}a_nx^n$ has the same convergence radius Proof:
Suppose power series $\sum_{n=0}^{\infty}a_n x^n $, if series ${a_n}$ is divergent. Then $\sum_{n=1}^{\infty}(a_{n}-a_{n-1})x^n$ and $\sum_{n=0}^{\infty}a_nx^n$ has the same convergence radius.

I only can see that when $x=1$ these are both divergent. Thank you.

EDIT: Solution is WRONG. I would delete if I could.
Let $R$ be the radius of convergence of $\sum_{n=0}^\infty a_n x^n$. That is, $\sum_{n=0}^\infty a_n x^n$ converges when $|x| < R$, but diverges when $|x| > R$. We need to show $\sum_{n=1}^\infty (a_n - a_{n-1}) x^n$ converges when $|x| < R$, and diverges when $|x| > R$.
If $|x| < R$, then
$$\sum_{n=1}^\infty (a_n - a_{n-1}) x^n = \sum_{n=1}^\infty a_n x^n - x\sum_{n=0}^\infty a_n x^n,$$
i.e. a combination of two convergent series, making the series convergent.
Suppose $|x| > R$. Without loss of generality, we may assume $x \neq 1$; if $x = 1$ is possible, then $R < 1$, and the $x = 1$ case is covered by $x = \frac{1 + R}{2}$.
For the sake of contradiction, suppose $\sum_{n=1}^\infty (a_n - a_{n-1}) x^n = L$. We can similarly assume that this sum is absolutely convergent, since we can replace $x$ with $\frac{R + x}{2}$, taking $x$ off the boundary of the interval of convergence. Then
\begin{align*}
\frac{L}{1 - x} &= \sum_{m=0}^\infty x^mL \\
&= \sum_{m=0}^\infty x^m\sum_{n=1}^\infty (a_n - a_{n-1}) x^n \\
&= x\sum_{n=0}^\infty \sum_{m=0}^\infty (a_{n+1} - a_n) x^{m+n}.
\end{align*}
I wish to express $\sum_{n=0}^\infty \sum_{m=0}^\infty (a_{n+1} - a_n) x^{m+n}$ as a simple power series, collecting each power of $x$ as a single term. Note that the $x^k$ term occurs when $n + m = k$, giving us:
\begin{align*}
\frac{L}{1 - x} &= x\sum_{k=0}^\infty\sum_{l=0}^k (a_{l+1} - a_l) x^k \\
&= x\sum_{k=0}^\infty(a_{k+1} - a_0)x^k \\
&= \sum_{k=0}^\infty a_{k+1} x^{k+1} - a_0 x \sum_{k=0}^\infty x^k \\
&= \sum_{k=0}^\infty a_k x^k - a_0 - a_0 x \sum_{k=0}^\infty x^k \\
&= \sum_{k=0}^\infty a_k x^k - a_0 \sum_{k=0}^\infty x^k \\
&= \sum_{k=0}^\infty a_k x^k - \frac{a_0}{1 - x} \\
\frac{L + a_0}{1 - x} &= \sum_{k=0}^\infty a_k x^k.
\end{align*}
That is, $\sum_{k=0}^\infty a_k x^k$ is convergent too, contradicting $|x| > R$.

I think maybe I found the solution:
When $x=1$, the two are both divergent, so since they are power series, when $x\ge 1$, it is divergent.
When $x< 1$, by summation by part (Abel transformation),
$$\sum_{n=1}^{N}(a_n-a_{n-1})x^{n}\\=(x^na_n-xa_0)-(1-\frac{1}{x})\sum_{n=1}^{N-1}a_nx^n,$$ since $x^n$ is convergence when $x<1$, the two series has the same convergence and divergence.


Could G. H. Hardy make a product of two primes so big he couldn't find out which? This question of course began as a slightly irreverent play on the riddle, "Can God make a stone so big He can't lift it?" Then I wondered about the answer.

Is it possible to exhibit a number that is provably a product of exactly two primes without knowing which primes? More to the point, without having the theoretical and technological means to compute which primes? If so, how is that done, and how old are the results that are needed in order to do it? And if not, does recent theory and technology make it impossible?

The magic phrase for this problem is 'Zero-Knowledge Proof', and a bit of hunting around suggests that there is in fact a zero-knowledge scheme for the product of two primes - not an exact proof, but a statistical one (so that it can be verified to arbitrarily high probability even with an adversarial prover).  From the excerpt for the paper An efficient non-interactive statistical zero-knowledge proof system for quasi-safe prime products by Gennaro, Micciancio and Rabin:

  [...] we present the first simple and efficient zero-knowledge proof that an alleged RSA modulus is of the correct form, i.e. the product of two primes.

It looks like the proof goes in stages; first, a zero-knowledge proof that a number $n$ is squarefree by exhibiting the ability to extract $n$th roots of numbers (numbers $n$ that aren't square-free have $\mathrm{GCD}(n,\phi(n))>1$ and thus only a small fraction of numbers will have $n$th roots mod $n$), then a zero-knowledge proof that a number is a product of two prime powers (both of which are inequal to each other and inequal to 1 mod 8), where the verifier supplies a number $x$ and the prover provides a square root for one of $\pm x, \pm 2x$; by quadratic reciprocity, this is always possible for numbers satisfying the condition, whereas products $m$ of three or more prime factors have at most one-eighth of the numbers less than $m$ as quadratic residues, so a randomly-chosen $x$ will fail to have any of $\{\pm x, \pm 2x\}$ be a square at least half the time.  Combining the two zero-knowledge proofs gives a zero-knowledge proof for (a subset of) the property 'product of at most two primes', and then combining this with the standard primality tests gives a proof for (a subset of) 'product of exactly two primes'.

Note that this is a statistical proof, though, not an exact one; you can be convinced of the result to an arbitrary degree of certainty but it's not a proof in the classical sense.  For more details on zero-knowledge proofs in general, the Wikipedia page is a decent starting point.

Depends on what you mean by 'exhibit a number'. How about $14$ iff for all $N&lt{10^{10^{99999}}}$, $2N$ is a sum of two primes, else $15$?

This is a well-defined, computable and unique number.

In what sense is the string $56566633746456456$ a better representation of a number then "$x=14$ iff Goldbach is true for all $N&lt{10^{10^{99999}}}$, else $x=15$", other then the artificial human base-10 convention ?