I just invented this last week

Copy the following into Mathematica, select all, right-click on Convert To, choose StandardForm.

N[Sum[2/10^((n*(Floor[c/n] + 1) - c)s), {n, 1, Floor[Sqrt[ c]]}] + Sum[ 2/(10^((n(Floor[c/n] + 1) - c)s) (10^(ns) - 1)), {n, 1, Floor[Sqrt[ c]]}] + Sum[2/(10^((o^2 - o - c) s)(10^(os) - 1)), {o, Floor[Sqrt[c]] + 1, max}], ((max + 1)^2 - c)*s]; AbsoluteTiming[Flatten[ Position[Partition[ RealDigits[%][[1]], s, s, -1], {(0) .., 2}]]] + c

-end

c=crossover s=spacing

The crossover is the number which you want to surpass. The spacing is difficult to determine, so I'll say that in general, if you use a number which is equal to the number of digits of the max^2, you should be safe. So for all primes up to 100^2 = 10000, that is five digits, so use a 5 for the spacing. Follow this rule and you should be safe.

For those who are wondering, the code is based on a bigger structure of numbers formed by the Sieve of Eratosthenes.

I was successful in working out the bugs, now I can specify a range of numbers that I want to test for primality. I'll provide an example, here I test for primes between 10 and 100

N[Sum[2/(10^(Mod[-(10 - n^2), n]*4)*
          (10^(n*4) - 1)),
     {n, 1, Sqrt[10000]}], (10000 - 10)*4]
str = StringDrop[ToString[%], 2]
(1/4)*Last /@ StringPosition[str,
         "0002"] + 10

Out[175]= \
0.02040204060602080210060402120404060802120208040406120204041202100206\
0804021204080406021004100404021402040808040802060410021402040606040802\
1206040212040404080214040604040412020606100

Out[176]= \
"020402040606020802100604021204040608021202080404061202040412021002060\
8040212040804060210041004040214020408080408020604100214020406060408021\
206040212040404080214040604040412020606100"

Out[177]= {11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, \
67, 71, 73, 79, 83, 89, 97}

It's really easy, just follow the formula:

[\!\(
\*UnderoverscriptBox[\(\[Sum]\), \(n = 1\),
RadicalBox[\(max\), \(2\)]]\(2/
   10^\((\((Mod[\(-\((\((min)\))\)\) - \((n^2)\),
        n])\) spacing)\) \((\((10^\((n*spacing)\))\) -
    1)\)\)\), (max - min)*spacing]

For those of you who are following, you will need to know how to find the spacing. The spacing is determined using a couple of formulas, in these examples below, I am solving for the LCM of 1-1000. The First Formula, I would keep adding the log10 of primes until I arrive at the max number I want to test for. Once I find that prime, I input that into the second formula to find the bonus range. The output from the first formula and the second are added together properly by converting the logs back into numbers first before adding them together, otherwise just adding the logs is like multiplying those numbers together. This yields the overall max that could be used to test with. The third formula yields the number of divisors of an LCM of 1-n and taking the log10 of it determines the spacing needed to pad the decimal expansion with. Using that prime number again from the first formula, it is inputted into the third formula to yield the spacing. It all looks like this:

AbsoluteTiming[N[Sum[Log[Prime[n]]/Log[10], {n, 1, 1000}], 13]]

{0.031222, 3392.831632804}

N[10^3392.83163280386907722938003534866124707481`13.]

6.78629608*10^3392

N[Sum[(1/Log[10])*(-2 + x)*Log[Prime[p + PrimePi[1000^(1/x)]]],
     {x, 3, 1 + Floor[Log[1000]/Log[2]]},
     {p, 1, PrimePi[1000^(1/(-1 + x))] - PrimePi[1000^(1/x)]}], 13]

17.56097841949

N[10^17.56097841949371816986880298169279418524`13.]

3.63897*10^17

6.786296084197555149532660048969925646446`9.107222002718924*^3392 + \
3.63896953291872`*^17

6.786296084197555*10^3392
is the max

N[1 + Floor[Log10[Product[x^(PrimePi[Prime[1000]^(1/(-1 + x))] -
                PrimePi[Prime[1000]^(1/x)]),
           {x, 2, 1 + Floor[Log[Prime[1000]]/Log[2]]}]]]]

Out[178]= 308.

308 is the spacing required to test for primes under the max, 6.786296084197555*10^3392. That alot of primes for such little spacing. Pretty neat, right? Don't forget to vote, I could use some reputation improvement.
Oh, I just got an answer on how to get more zeros in front of the two so that you will have something to search with. Use:

PaddedForm[2, spacing, NumberPadding -> {"0", "0"}]

And I should point out that my computer is pretty tops, and it easily does a range of 10000 at a time. I wouldn't try for 6.786296084197555*10^3392. It was just an example, sorry I didn't use one that is practical or usable.
A spacing of 4 will get you pretty far, up to 6.46969359`*^9 at least.

One more thing, a bug I haven't worked out yet but is easy to work around, when you do get the display of primes and there is a number that did not divide by the spacing, it will display as a fraction and can be ignored as a false positive. I've seen a lot of false positives when the computer comes across a 0022 instead of a 0002. I've compared lists and they do happen, it's not a perfect system yet. That will be my next question...


Function never equal to 0 Let $p,q,x,y:(a,b)\to\mathbb{R}$ be $C^1 ((a,b))$ functions. Knowing that for  $(u,v), (\tilde{u},\tilde{v})\in (a,b)\times (0,1)$ we have that:

$$\begin{cases} p(u)+vx(u)=p(\tilde{u})+\tilde{v}x(\tilde{u}) \\ q(u)+vy(u)=q(\tilde{u})+\tilde{v}y(\tilde{u}) \end{cases}\ \Longleftrightarrow\ (u,v)=(\tilde{u},\tilde{v}),$$

is it true that the function:

$f:(a,b)\times (0,1)\to\mathbb{R}$ given by:

$f(u,v)=p'(u)y(u)-q'(u)x(u)+v(x'(u)y(u)-x(u)y'(u))$, is NEVER equal to $0$?

Leg $r=(p,q)$ and $z=(x,y)$. Define $h(u,v):=r(u)+vz(u)$.  From the given equivalence we know that $h$ is injective.  Now verify that $f$ is the determinant of the Jacobian of $h$ which is doomed not to be zero because $h$ is injective.


Closed form for an infinite sum over Gamma functions? I am having quite a bit of trouble trying to find a closed form (or a really fast way to compute) for the infinite sum
$$\sum_{n=1}^{\infty} a^n \dfrac{\gamma(n+1,b)}{\Gamma(n+1)\Gamma(n)}$$
where $\gamma(n,b) = \int_0^bz^{n-1}e^{-z} dz$ is the lower incomplete Gamma function and $a,b > 0$.

Any ideas how to solve this?


Some doubts about conditional statemenr 1.) First of all if P is false and Q is true then result is false. Why this?

2.) If Juan has a smartphone, then 2+3=5.

If Juan has a smartphone then both propositions true and answer is true.

If Juan does not have a smartphone then first proposition is false and second is true and answer is false.

How false does not implies truth?

I want to explain this to person for which this is new topic. So very basic explanation can help.

There are 4 possibilities:  TT, TF, FT and FF, for P and Q, respectively. ..  Perhaps a little counter-intuitively, 3 of the 4 give a true conditional (as $P\rightarrow Q$, or if P then Q, is called ).  The combination  that's considered false is TF.

1) I assume that you are talking about a statement of the form -If P the Q . If P is TRUE AND Q IS FALSE THEN THE STATEMENT IS FALSE . In all other cases the statement is TRUE . So you are in error here .
2) No , If Juan does not have a smart phone then any Q  makes the conditional true even 2+3=7


I would like to ask about 3R Planes. [Hi, I have a question in my homework and I attached my solution. By any chance somebody check is it a good solution and correct answer? Thank you.][1]
[1]: https://i.stack.imgur.com/XcHWC.jpg![enter image description here](https://i.stack.imgur.com/X9fgl.jpg)

Your solutions are correct. A tip on how to check these answers for yourself: dot your solution line vector with the normals of the planes. If you get zero, it means they are perpendicular, so the line is in the plane (if at least one point on the line is in the plane).
Details: Your solution has vector $(-1,1,1)$, while your plane $\pi_1$ has normal vector $(1,2,-1)$. And note that $(-2,1,0)$ is on your line, but obviously that satisfies $x+2y-1z=0$.


Linear Combination of Roots of Unity Let $\omega_n$ be a primitive $n$the root of unity and $\lambda_k$ be natural numbers.

  Does $\sum_{k=1}^{n} \lambda_k w_n^k =0$ imply $\lambda_1 = \lambda_2 = ... = \lambda_n $?

I am aware of this generalised question, but I'm unable to follow the solutions

If $n=ab$ is $a,b>1$, then write:

$$p(x)=\frac{x^{n}-1}{x^a-1}=1+x^a+x^{2a}+\cdots + x^{(b-1)a}$$

Now, $w_n$ is a primitive $n$th root of unity, so is not a root of $x^a-1$, so it is a root of $p(x)$.

But that means that, given any positive integers: $\lambda_0,\lambda_1,\dots,\lambda_{a-1}$ we get that that $p(x)(\lambda_0 + \lambda_1 x+\cdots+\lambda_{a-1}x^{a-1})$ is of degree $n-1$ and has root $w_n$ and positive integer cooefficients.

More generally, look at the Cyclotomic poylomials, $\Phi_n$. They are the minimal polynomials for each $w_n$, and $\Phi_n$ has degree $\phi(n)$. They have integer coefficients, but you can add to them to get natural numbers.

For example:

$$\Phi_{6}(x)=1-x+x^2$$

has root $w_6$. Then add $2$ to every coefficient for $0$ to $5$, and you get:

$$\Phi_6(x)+2\frac{x^6-1}{x-1} = 3+x+3x^2+2x^3+2x^4+2x^5$$

has $w_6$ as a root.

Your rule is only true when $n$ is prime. This is because $\Phi_n(x)$ is the minimal polynomial for $w_n$ and, when $n$ is prime, it is not hard to show that $\Phi_n(x)=1+x+\cdots+x^{n-1}$, which means that any polynomial which has $w_n$ as a root must be divisible by $\Phi_n(x)$. You (essentially) have a polynomial of degree $n-1$ with root $w_n$, so your polynomial must be a scalar thans $\Phi_n(x)$.

Hint $i+i^2+i^3+i^4=0$ summation of four consecutive powers of iota is $0$


Show that $c = \max(a, b)$ on $\mathbb{Z}_2$ is not a binary operation Let $*: \mathbb{Z}_2\times\mathbb{Z}_2 \to \mathbb{Z}_2$, be defined as $[a] * [b] = [c]$, where $c = \max\{a, b\}$, for all $[a], [b] \in \mathbb{Z}_2$. Prove that $*$ is not a binary operation on $\mathbb{Z}_2$. Hint: is it well defined?

--

I don't want someone to solve the problem for me, but just help clear up my misunderstanding or point me in the right direction.

I don't see how $*$ is not a binary operation, nor how it isn't well defined.

A binary operation is a function on a set $S$ that associates every ordered pair of elements in $S$ to another element of $S$.

Here, $*$ associated ordered pairs of elements $\mathbb{Z}_2$ to another element $\mathbb{Z}_2$, through the max function. It has four possible inputs:

$$
\begin{align}
(0, 0) & \mapsto 0 \\
(0, 1) & \mapsto 1 \\
(1, 0) & \mapsto 1 \\
(1, 1) & \mapsto 1
\end{align}
$$

So it associates every unique input to one output, and meets the definition of a function. Every possible input returns an output that is in $S$ ($\mathbb{Z}_2$).

How is it not well-defined?

The definition of $[a]\ast[b]$ is defined using representatives of the congruence classes $[a],[b]$. In order for it to be well-defined, it must work the same on all pairs of representatives of $[a]$ and $[b]$. For instance, it must satisfy $[0]\ast[1]=[2]\ast[1]$ since $[0]=[2]$.

Note that $[1]=[3]$. But whe have $[2]*[1]=[\max\{2,1\}]=[2]=[0]$, but $[2]*[3]=[\max\{2,3\}]=[3]=[1]$, so if you operate $[2]$ with $[1]$, and $[2]$ with $[3]$, you get two different elements, so this operation is not well defined.


Is there a 3-dimensional "matrix" by "matrix" product? Is it possible to multiply A[m,n,k] by B[p,q,r]? Does the regular matrix product have generalized form?

I would appreciate it if you could help me to find out some tutorials online or mathematical 'word' which means N-dimensional matrix product.

Upd.
I'm writing a program that can perform matrix calculations. I created a class called matrix
and made it independent from the storage using object oriented features of C++. But when I started to write this program I thought that it was some general operation to multiply for all kinds of arrays(matrices). And my plan was to implement this multiplication (and other operators) and get generalized class of objects. Since this site is not concerned with programming I didn't post too much technical details earlier. Now I'm not quite sure if that one general procedure exists. Thanks for all comments.

The general procedure is called tensor contraction. Concretely it's given by summing over various indices. For example, just as ordinary matrix multiplication $C = AB$ is given by

$$c_{ij} = \sum_k a_{ik} b_{kj}$$

we can contract by summing across any index. For example, we can write

$$c_{ijlm} = \sum_k a_{ijk} b_{klm}$$

which gives a $4$-tensor ("$4$-dimensional matrix") rather than a $3$-tensor. One can also contract twice, for example

$$c_{il} = \sum_{j,k} a_{ijk} b_{kjl}$$

which gives a $2$-tensor.

The abstract details shouldn't matter terribly unless you explicitly want to implement mixed variance, which as far as I know nobody who writes algorithms for manipulating matrices does.

In multiplication of two $n$ dimensional matrices starting $n-2$ numbers should be same and last two numbers should follow the rule of $2d$ multiplication i.e., column of left should be equal to row of right.

$$(a\times b\times c\times d\times e)\times(a\times b\times c\times e\times f) = a\times b\times c\times d\times f$$


A question on calculus The question is:

In the accompanying figure the curve represented by $\rm C$ is given by $y= \rm f (x)$ where $\rm f$ is one-to-one and continuous. ${\rm C1}$ and ${\rm C2}$ are respectively $y=2x^2$ and $y=x^2$. At each point $P$ on $\rm {C1}$, assume that the areas $OAP$ and $OPB$ are equal. Here $PA, PB$ are the horizontal and vertical segments. Determine the function $f$.

My progress:
I took any arbitrary point $x > 0$ and calculated the areas and found out that $$\int_{0}^{x}{\rm f}(t)dt=x{\rm f}(x)-(\frac{{\rm f}(x)}{2})^{\frac {3}{2}}.$$Now the problem is that we do not know that $f$ is differentiable or not. If it was, then we could differentiate both sides of the equation(since $f$ continuous is given) w.r.t $x$ to obtain $$f^{'}(x)(x-\frac {3}{4\sqrt {2}}\sqrt {f(x)})=0.$$ Now since $f$ is one-to-one and continuous, we can easily show that $f$ is monotonic and hence $f$ cannot attain a relative extremum anywhere and hence $f^{'}(x)=0$ for no $x$, i.e. for all $x$, $$\sqrt {f(x)}=\frac {4\sqrt {2}x}{3}$$ wherefrom we conclude $f(x)=\frac{32x^{2}}{9}.$  How do we prove that $f$ is differentiable? Also I think we can solve the problem another way round, i.e we first show that $\frac {32x^2}{9}$ indeed is a solution and then that there can be at most one solution of $f$. Can we do it this way, i.e show the uniqueness? Thank you for your help in advance.
(P.S.  It would also be helpful if someone tells me how to draw good diagrams here. I have drawn this in paint and the graphs look horrible.)

Extend the segments $PA$ and $PB$ until they meet the axes at $A'$ and $B\,'$, respectively.

Since arc $OP$ is a graph of $y = 2x^2$ and arc $OB$ is a graph of $y = x^2$, we know that the area enclosed by the figure $OBB\,'$ is half the area enclosed by $OPB\,'$; and the area enclosed by $OPB$ is the difference of those two areas, so it also is half the area of $OPB\,'$.

But we also know that the area of $OPB\,'$, that is, under the parabolic arc $OP$, is one third the area of the rectangle $OB\,'PA'$, and the area of $OPA'$ (between the arc $OP$ and the $y$-axis) is two-thirds of the area of the rectangle.  The area of $OPA$, which is equal to the area of $OPB$, is therefore one quarter of the area of $OPA'$, and $OAA'$ has three-fourths of the area of $OPA'$.

Now, since $f$ is one-to-one, $f^{-1}$ exists on the range of $f$,
and the arcs $OP$ and $OA$ can be described by the equations
$x = \sqrt{y/2}$ and $x = f^{-1}(y)$, respectively.  From the preceding facts,
the areas between those arcs and the $y$-axis satisfy the equation

$$ \frac{3}{4} \int_0^Y \sqrt{\frac{y}{2}} dy = \int_0^Y f^{-1}(y) \, dy $$

where $Y$ is the $y$-coordinate of $P$.

Taking the integral on each side of the equation above as a function of $Y$, by the fundamental theorem of calculus the derivatives of these integrals exist and are equal to  $\sqrt{Y/2}$ and $f^{-1}(Y)$, respectively.  Therefore,

$$ \frac{3}{4} \sqrt{\frac{y}{2}} = f^{-1}(y). $$

If $x = \frac{3}{4} \sqrt{\frac{y}{2}}$, then $y = \frac{32}{9} x^2$, so

$$ x = f^{-1} \left( \frac{32}{9} x^2 \right), $$

and applying $f$ to both sides, we find that

$$  f(x) = \frac{32}{9} x^2. $$

First, show that the function you derived (using stronger assumption) solves the problem. Then show that the problem has unique solution on the class of monotone continuous functions.

I also sketch a proof of uniqueness. Let $f_1$ and $f_2$ be two different monotone continuous functions solving the integral equation. Necessarily $f_1(0)=f_2(0)=0$, hence there exist (using continuity) a points $0\leq t_0<t_1$ such that $f_1(t)=f_2(t)$ for all $t\in [0,t_0]$ and (WLOG) $f_2(t)<f_1(t)$ for $t\in(t_0,t_1]$. Now there exists $t_2>t_1$ such that $f_1(t_1)=f_2(t_2)$. When you sketch this situation, you should realize contradiction.

A regularity result for integral equatlity holds and can be obtained by the Implicit function theorem.


How to evaluate $\lim\limits_{x \rightarrow +\infty}{e^x (e - (1+\frac{1}{x} )^x)}$ without L'Hospital? Using several times L'Hospital Rule I got
$$\lim_{x \rightarrow +\infty}{e^x \left (e - \left(1+\dfrac{1}{x}\right )^x\right)} = +\infty.$$ Is it possible find this limit without L'Hospital?

The natural thing to do is to look at the logarithm of $\left(1+\frac{1}{x}\right)^x$, that is, at $x\log\left(1+\frac{1}{x}\right)$. Use the series
$$\log(1+t)=t-\frac{t^2}{2}+\frac{t^3}{3}-\frac{t^4}{4}+\cdots.$$
From this we can obtain good estimates of the difference between $e$ and $(1+1/x)^x$ when $x$ is large. For the calculation, the series expansion of $e^t$ is useful.

We need to prove that,
$$\lim_{x \rightarrow +\infty}{e^x \left (e - \left(1+\dfrac{1}{x}\right )^x\right)} = +\infty.$$

consider
$$\lim_{x \rightarrow +\infty}{e^x \left (e - M\right)} $$

where, $$M = \left(1+\dfrac{1}{x}\right )^x$$

if we prove that $M$ has a finite limit, we are done.

Note that,

1. M is increasing function of x
2. M is bounded above

first one you can prove as an exercise, for second
$$M = \left(1+\dfrac{1}{x}\right )^x = \left(\left(1+\dfrac{1}{x}\right )^{x/k} \right)^k < \left(\frac{1}{1-\frac{1}{x} \cdot\frac{x}{k}}\right)^k = \left ( \frac{1}{\left(1-\frac{1}{k}\right)^k}\right)$$

so that, $$M< \frac{1}{\left(1-\frac{1}{k}\right)^k}$$ for any whole k.

$$\lim_{x \rightarrow +\infty}{e^x \left (e - M\right)} =  \lim_{x \rightarrow +\infty}{e^x L} = +\infty $$


Prove that a number is even, given the cube is even If it is known that $x^3$ is even, can we say that $x$ is even? It seems to be the case because an odd*odd*odd=odd (if we are dealing with natural numbers). But is there a proof?

It sounds like you've more or less given a proof without realizing it.  Proof by contrapositive

On top of the mathematical proofs, you can try this yourself with very high numbers with a bit of code.

from __future__ import division

for i in range(0,1000):
    if type((i*i*i)/2)==int:
        print(i)
else:
    print("Done")

With any number you input, you can see it will never find any number that this statement does not hold true for.

http://labs.codecademy.com/
You can try the python code out here if you like.


Proof that $\log_23 +\log_52$ is irrational number Problem is to prove that
$$\log_23 +\log_52$$
is irrational number.
My attempt:
I try to write number like $$\log_23 +\frac{1}{\log_25}$$ but I didn't get anything(proof by contradiction). I also try to find polynomial such that given number is zero point but also without success.

Refer to Qiaochu Yuan's answer from (Question 986227), which states that, contingent on the currently unproven Schanuel's conjecture, the logarithms of the primes, $\ln 2,\, \ln3\, \,\ln 5\,\ldots$, are algebraically (not just linearly) independent over $\mathbb{Q}$. The stronger statement would follow that $\log_2 3+\log_5 2$ is transcendental and consequentially irrational.

  Proof. By unique prime factorization, the logarithms of the primes are linearly independent over $\mathbb{Q}$. If $p_1, p_2, \dots$ is an enumeration of the primes, then by Schanuel's conjecture it follows that $\mathbb{Q}(\log p_1, \log p_2, \dots \log p_k)$ has transcendence degree at least $k$, hence exactly $k$, for all $k$. $\Box$

Your problem will likely remain open until Schanuel's conjecture is resolved. To quote GH from MO's answer to (Math Overflow: Question 185540),

  ...[that $\log_35+\log_25$ is irrational] is probably true, but proving it might be out of reach at the moment.

$2^{log_23}.2^{log_52}= 2^{\frac {p}{q}}$ (where p and q are natural numbers as the number is graeter than 0).

(and $a^{log_bc}=c^{log_ba}$=>$2^{log_23}=3$)

therefore  $3^q . 2^{qlog_52}=2^p$ (we know that $3^q$ and $2^p$ are natural number so we now just need to show $2^{qlog_52}$ is always irrational for any natural 'q').

$2^{qlog_52}=\frac{m}{n}$ (we know $\frac{m}{n}$ is not pf the form $2^k$ where $k$ is a natural number becaise $qlog_52$ cannot be natural)

therefore $q=log_25.log_2(\frac{m}{n})$=c (where q is a natural number,and if $log_2\frac{m}{n}$ becomes rational then $log_25$ being irrational $q$ becomes irrational which is not possible.)

$log_25=a,log_2\frac{m}{n}=b$ ,where a and b are both irrational.I dont know how to proceed further  , irrational $\times$ irrational can be rational.


Sum of unit and nilpotent element in a noncommutative ring. Something similar is asked here but it is not exactly the same thing.

An element $a$ of a ring $R$ is called nilpotent if $a^n=0$ for an $n \in \mathbb{N}$.  Show that if $u$ is a unit and $a$ is nilpotent in $R$, then $u+a$ is a unit.

My proof:

Let $a^n=0$.  The equality $$(u+a)^na^{n-1}=u^na^{n-1} \tag{1}$$  results from expanding $(u+a)^n$ together with the property that $a$ is nilpotent.

Let $u_1$ be the left inverse of $u$.  Then by multiplying both sides of (1) on the left by $u_1^n$, the resulting equality is $u_1^n(u+a)^na^{n-1}=a^{n-1}$.  This is where I am nervous.  I thought that you could then write $u_1^n(u+a)^n=1$ by the property of the multiplicative identity in an associative ring with unity and setting the coefficents equal to each other, but I feel like this looks like applying a cancellation to the $a^{n-1}$ on both sides, which I do not mean to be doing.

The last part is to rewrite the resulting equality as $(u_1^n(u+a)^{n-1})(u+a)=1$ to show the existence of a left inverse of $u+a$ and then do a similar thing starting from (1) to show the existence of a right inverse.

Is there an error with this?

This fails, for instance,
$$\begin{pmatrix}0&-1\\1&0\end{pmatrix}+\begin{pmatrix}0&1\\0&0\end{pmatrix}=\begin{pmatrix}0&0\\1&0\end{pmatrix}$$
is singular.

In a ring $R$ the nilpotent elements constitutes an ideal which is called nilradical, $N(R)$ .

$N(R)$ is the intersection af all prime ideals; so if $u$ is a unit and $a$ is nilpotent $a+u$ must be a  unit.

Otherwise there would be a maximal ideal $M$ with $a+u \in M$. But a maximal ideal is prime, and so $$a \in N(R) \subseteq M$$ and so $$u = (u+a) - a  \in M$$ which is impossible because $u$ is a unit.


How do I prove when $2\sin^2x$ is greater than $\sin x $? I need to show when

$2\sin^2x \geq \sin x $ for $x$ between $0$ and $2\pi$.

I see that for values $\pi/6$ and $5\pi/6$ and after $\pi$. How do I prove or reason out these answers?

If $x\geqslant\pi$, then $2\sin^2x\geqslant0\geqslant\sin x$. If $x=0$ this is again true.

If $0<x<\pi$, then $\sin x>0$ and therefore$$2\sin^2x\geqslant\sin x\iff 2\sin x\geqslant 1\iff \sin x\geqslant\frac12$$and this takes place if and only of $\frac\pi6\leqslant x\leqslant\frac{5\pi}6$.

$2\sin^2x \geq \sin x $ can simplify to $2\sin x \geq 1$ after factoring out a $\sin x$.

$2\sin x \geq 1$ is true only when $\sin x \geq 1/2$.

Therefore $30^\circ \leq x \leq 150^\circ$. $(x \neq 0,360)$


How To Determine If A Large Number is Prime? For a very large number n, how many divisibility tests are required to establish if its prime?

I know this has something to do with the Golden Number, but I can't figure out what. I did try searching for an answer but not much luck.

!!EDIT!!
(It wont let me answer my own question for upto 8hours)

I found something posted by someone else on the primality test by golden ratio, although just like Fermat's probability test, it also fails at times.

  There is a primality test by Golden ratio that is used in conjunction with the Lucas N+1 primality test. It is based on the relation between Lucas numbers and Fibonacci numbers. Primality test by Golden ratio states that if

  $g^p+(1-g)^p \equiv 1\mod p$ , where g is golden ration, is true then p is prime. In other words, if

  $\frac{g^p+(1-g)^p-1}{p} $  divides wholly then p is prime. The expression

  $g^p+(1-g)^p$   is a formula for the p-th Lucas number, i.e.

  $g^p+(1-g)^p = L_p$.   As a result, we can say that if p-th Lucas number minus 1 divides by p wholly then p is prime, i.e.
  $ \forall p \in \mathbb{N}, \frac{L_p-1}{p}=a$  where  a $\in \mathbb{N} \Rightarrow $  p is prime.

  Aaaand it is not true. If you check a composite number 705 which is equal to 3*5*47:

  $ \frac{L_{705}-1}{705} = \frac{g^{705} +(1-g)^{705}}{705} = 3.031556 * 10^{144}$

  $3.031556 *10^{144}$ is a whole number and the test fails. Fermat's primality test suffers from a similar problem.

To test if some $x$ is prime, we generally have to do divisibility tests only up to and including $\sqrt{x}$.

That's because if some $y > \sqrt{x}$ were a factor of $x$, then there would have to be some $z$ such that $zy = x$. And $z < \sqrt{x}$ because if $z > \sqrt{x}$, then clearly $zy > x$ (as both $z$ and $y$ would be greater than $\sqrt{x}$). But if $z < \sqrt{x}$, then we've already tested $z$ in going up to $\sqrt{x}$!

And we don't have to do trial division for every integer up to $\sqrt{x}$. Using the Sieve of Eratosthenes, for example, after testing some $n$, we 'cross out' the integers that are multiples of $n$, because if some multiple of $n$ divides $x$, then $n$ has to divide $x$. So we only need to check the case for $n$.

In that spirit, there are lots of heuristics we can use to make prime-testing more efficient. However, finding large primes remains computationally very difficult.

Finally, I cannot immediately find an application of the golden ratio to this. You may be misremembering. I could dig up Prime Number Spirals, but I don't see how they'd be related.

Why would you not just check any given number to see if it is a prime number by simply attempting to divide it by preceding prime numbers? No matter how big the number is there can not be that many you would need to check it against.
Other than that unless you know a formula that can tell you what any given prime number may be then there can be no way to test to see if a given number is a prime other that trial and error.


Intuition Behind Power Law Distribution I know that the pdf of a power law distribution is $$ p(x) = \frac{\alpha-1}{x_{\text{min}}} \left(\frac{x}{x_{\text{min}}} \right)^{-\alpha}$$

But what does it intuitively mean if, for example, stock prices follow a power law distribution? Does this mean that losses can be very high but infrequently?

This means that a stylized fact about stock returns distributions, high kurtosis or "fat tails", is better taken into account with a "Pareto-type distribution" than with a normal distribution.  The latter has a finite variance, whereas the former is such that: "all moments $m \geq \alpha - 1$ diverge: when α < 2, the average and all higher-order moments are infinite; when 2 < α < 3, the mean exists, but the variance and higher-order moments are infinite, etc. For finite-size samples drawn from such distribution, this behavior implies that the central moment estimators (like the mean and the variance) for diverging moments will never converge - as more data is accumulated, they continue to grow."(Source: Wikipedia)

A power law distribution just means that y = ax^n for some a,n. In practice, and for large values of x, a polynomial can be approximated as a power law where n = the degree of the polynomial. (Ie ignore terms x^(n-1), x^(n-2) etc as these will be much smaller than ax^n for large x).

So y = 5x^1000 is a power law. y = 2x^67 + x^2 can be approximated as a power law y = 2x^67 for large x. y = e^x and y = sin(x)^10 are not power series.


Rademacher functions independent random Variables You call $r_n$=sign($sin^{2n}(\pi t))$ Rademacher functions.

I read " the $r_n's$ are independant random variables"

Why is this true ? I don't get it...

Read the Wikipedia Rademacher system article. I think you want
$\;r_n = \textrm{sign}(\sin(2^n 2\pi t))\;$ instead. The  $r_n$ are orthogonal as functions on $\;t\in[0,1]\;$ and thus the random variables associated with them are independant.

because the probability of its values are independent such that $p ( r_1=e_1 , r_2=e_2 , ... , r_n=e_n ) = p ( r_1=e_1 )  p ( r_2=e_2 ) .... p ( r_n=e_n )$ where $p$ is the probability measure and $e$ are borel subsets of the space , if this satisfied then $r_n$'s are independent random variables , you can take some specific values of n it would be much easier to prove it , GOOD LUCK.


Proof About Division of Integers Here is a problem I just finished working on:

  Prove that if $n$ is composite then there are integers $a$ and $b$ such that $n$ divides $ab$ but not $n$ does not divide either $a$ or $b$.

One thing I noticed while proving this is, that we are showing that there exist a pair of integers such that the conditional relationship is true; is it possible to find a pair such that it is not true, that $n$ divides both $a$ and $b$ individually?

The words "there are" tell you that the problem is asking about existence.

Words that mean existence ($\exists$) include "there is", "there exists", "one can find", etc.

Words that mean the statement is true for all ($\forall$) members of the specified set include "all", "every", etc.

"Any" is a tricky word that is best avoided.

EDIT: For a specific answer to your specific question, the statement is not true for all integers. Let $n = a = b = 4$. Then $n$ is composite and it divides $a$ and $b$ and $ab$.

$60 = 15\times 4$.  So $60$ divides $15\times4$ but $60$ does not divide $15$ and $60$ does not divide $4$.

You can do the same with any composite number.


What is linear approximation? What is the purpose of linear approximation questions?
For example, one question reads:
(a) Use the Linear Approximation for f(x) = ln(x) at a = 1 to estimate ln(0.84)...
(b) That Linear Approximation has error...

Every differentiable function $f:\Bbb R\to \Bbb R$ (by definition) can be written as $$f(x+h)=f(x) + f'(x)h + E(h)$$ in some interval of $x$, where $E(h)$ is an error term such that $E\to 0$ as $h\to 0$ faster than $h$ (meaning for somewhat small $h$'s the error term, $E(h)$, will be a very small number).

We call $$L(h)=f(x)+f'(x)h$$ the linear approximation of $f$ at $x+h$ (note: this is a misnomer as this is really an affine function, not a linear one).

So when we only move a little ways away from $x$ (by an small number $h$) the function $L(h)$ should give a good approximation for $f(x+h)$.  The function $L$ has the benefit of not only being pretty simple to evaluate in most instances, but also linear/affine functions have very nice properties.  So this is an easy approximation in practice that necessarily has a small error for small enough $h$.

Note that if $f$ is $k$-times differentiable (with $k\gt 1$), we can extend this approximation method to polynomials of higher degree.  For instance the quadratic approximation to $f$ at $x+h$ is $$Q(h) = f(x) + f'(x)h + \frac{f''(x)}{2}h^2$$  You can see that this is just a quadratic polynomial in $h$.  In general, this will evaluate to a number even closer to the value of $f$ at $x+h$ and will have a small error for even larger values of $h$ than the linear approximation.  You'll learn more about this when you cover Taylor polynomials.

Just to give you a little something to look forward to, this animation shows a sequence of Taylor polynomials for the sine function.  Note that $N=1$ is the linear approximation.

In understood it in this way: sse the Taylor series for $\ln(x+a) = \ln(a)  + \frac{x}{a} + \mathsf{O}(x^2)$

with $a = 1$ and $a = 0.84$, namely:

$$\ln(0.84 + 1) \approx \ln(1) + 0.84 = 0.84$$
whereas
$$\ln(0.84) = 0.609$$

The error is then $E = 0.23 \equiv 137.93\%$

OTHERWISE

Use the Taylor Series in its general first order expression

$$F(x) \approx F(a) + F'(a)\cdot(x-a)$$

obtaining

$$\ln(0.84) \approx \ln(1) + \frac{0.84 - 1}{1} = -0.16$$

Knowing that $\ln(0.84) \approx -0.07$ you got an error of about 228\%$


Maximize this bonus. I'd like to understand if my strategy to maximize this welcome bonus is mathematically correct.

A bookmaker is offering a bonus of €50 to all its new users after they deposit for the first time on its website.

There are some rules:

Before you are granted the €50 bonus you have to bet at least €50. This could be done cumulatively, e.g. you could make ten €50 bets or just one €50 bet.
After you get those €50 bonus you have to bet other €50 before you can cash out.
Every bet you make has to be at least 1.5 decimal odds (EU), e.g. if you bet €50 you get €75.

My strategy is based on a few thoughts I've made:

Since the bookmakers get a commission on each bet it's better to make just one €50 bet to get the bonus and just another €50 bet to be able to cash out.

At the same time I would choose an event with the highest allowed probability to happen (1.5 odds).

Do you think this strategy correctly maximizes our expected value?

Theoretically speaking,

a) The amount you bet does not matter, as long as the money you can win and it's probability are all in proportion. For example, there is no difference in betting \$1 or \$100 on a dice roll, as long as the winning amount is \$6 or \$600 respectively.

b) Each win or loss has no effect on the next bet.

c) You should never bet if the opponent can win more easily than you. So if the bet is fair, but an additional commission is being taken, you are at a losing side and should not bet.

If these rules are followed, you should make only 2 bets of any amount (it doesn't matter). Then again, if these rules are followed, anybody betting in a casino is at a slight losing edge due to the commission paid.


placing 15 balls into 6 urns There are $6$ urns indexed from $1$ to $6$. We place $15$ balls one by one randomly into the urns.

What is the probability that the first urn stays empty?

I found $\frac{19\choose4}{20\choose5}$ because there are ${15+6-1\choose 6-1}$ way to distribute the balls in total and ${19\choose4}$ ways to distribute the $15$ balls into $5$ urns ($2$ through $6$).

I have trouble with the following questions:

1)  What is the probability that the urns $1,2,3,4$ contain $3$ balls, the $5$th urn contains $2$ balls and the $6$th urn contains $1$ balls?

2) What is the probability that $4$ urns contain $3$ balls each, one urn contains $2$ balls and $1$ contains 1 ball?

For 1) I thought I could use the multinomial coefficient to obtain $\frac{15\choose3,3,3,3,2,1}{20\choose5}$ but this is not a probability (by a huge margin).

For 2) it should be whatever we got for 1) multiplied by $6!$ because that's the number of permutations of the $6$ urns

Edit Maybe this is correct for 1):

${15\choose 3,3,3,3,2,1}({1\over6})^3({1\over6})^3({1\over6})^3({1\over6})^3({1\over6})^2({1\over6})^1={15\choose 3,3,3,3,2,1}({1\over6})^{15}$

Your first calculation is incorrect because those ways of populating the urns are not equi-probable.   The probability that any specified  ball misses the first urn is $\frac 56$ so the probability that the first urn is empty is $\left(\frac 56\right)^{15}$.

To see the difference between the two probability distributions, just consider the $2-$ball, $2-$ urn case.  There is a $\frac 14$ chance that both balls go in the first urn, and a $\frac 14$ chance they both go in the second.  But there is a $\frac 12$ chance that they go into different urns.

To do the first question, say, there are $\binom {15}3$ ways to pick the balls that go into the first urn, then $\binom {12}3$ ways to pick the balls that go into the second urn, and so on.  We come to the result $$\binom {15}3\times \binom {12}3\times \binom {9}3\times \binom {6}3\times \binom 32\times \left(\frac 16\right)^{15}\approx .0011$$

There's a 1:6 chance for a single ball to drop into 1st urn.

That means there's a 5:6 chance of it getting in any of the others, which is the desired result.

Being 15 balls, that means the probability for all of them to land on different than 1st one is (5:6)^15 or 0.0649%.

Note that the not integer distribution is irrelevant because ball #2 has no interdependency or information on how ball#1 landed and so forth. Each drop is an independent action with a statistical chance result.


If $G$ is an abelian group and $H$ be any subgroup, then why is $G/H$ also a group? If $G$ is an abelian group and $H$ be any subgroup, then why is $G/H$ also a group?

I get that every subgroup of an abelian group is normal, but how can I use that to prove that $G/H$ is a group?

Because it is a quotient group.

A less glib answer is that the group operation $\cdot$ is $$\begin{align}gH\cdot g'H&=gHg'H \\ &\stackrel{(1)}{=}(gg')H\end{align}$$ for each $g,g'\in G$. Checking that $(G/H, \cdot)$ is a group is routine and $(1)$ holds because $G$ is abelian (and, in particular, $H$ is thus normal in $G$). Here is a proof that it is indeed a group.

The quotient group G/H is the set of all left cosets gH, where g is an element of G, and with the binary operation gH * g'H = (gg')H.
 It is closed under * ; for a given g and g' in G, gg' is in G since it is a group. So, for two given cosets gH and g'H in G/H, gH * g'H = (gg')H is in G/H. It has an identity element; eH *gH = (eg)H = gH for all g in G. So, eH is our identity in G/H. Every element has an inverse; let g' be the inverse of g in G. Then, gH * g'H = (gg')H = eH. So, for a given element gH, g'H is its inverse in G/H. * is Associative; let g, g', and g'' be in G. Then, (gH * g'H) * g''H = (gg'')H * g''H = ((gg')g'')H = (g(g'g''))H =
= gH * (g'g'')H = gH * (g'H * g''H).
So, it satisfies the group axioms and is therefore a group.


Need help with taylor series. Evaluate the limit
  $$\lim\limits_{x \to 1} \frac{1-x + \ln x}{1+ \cos πx}$$

The limit im trying to get is $-\frac{1}{π^2}$ as I've solved from l'Hopitals rule.

Now I need to solve the limit by using Taylor Series and this is what i did so far

$$\begin{align*}
f(x) &= 1-x + \ln x = 1 -x + (x-1) + \frac{1}{2} (x-1)^2 + \frac{1}{3} (x-1)^3 - \frac{1}{4} (x-1)^4 + \ldots \\
g(x) &= 1+\cos πx = 1+\left[ 1+\frac{1}{2!} (πx)^2 + \frac{1}{4!} (πx)^4 - \frac{1}{6!} (πx)^6  +\ldots \right] \\
\frac {f(x)}{g(x)} & = \frac{\frac{1}{2} (x-1)^2 + \frac{1}{3} (x-1)^3 - \frac{1}{4}(x-1)^4 + \ldots} {2-\frac{1}{2!} (πx)^2 + \frac{1}{4!} (πx)^4 - \frac{1}{6!} (πx)^6+\ldots}
\end{align*}$$

I have no idea where to go to solve for $-\frac{1}{π^2}$ now. Please help

You do not need the Tailor expansion but only the first derivative, the rule of L'Hôpital gives us:
$$
\lim\limits_{x \to 1} \frac{1-x + \ln x}{1+ \cos πx}=\lim\limits_{x \to 1} \frac{\frac{1}{x}-1}{-\pi\,\sin\pi x}=\lim\limits_{x \to 1} \dfrac{\dfrac{1}{x^2}}{-\pi^2\cos\pi x} = -\dfrac{1}{\pi^2}
$$


Take 2: When/Why are these equal? This didn't go right the first time, so I'm going to drastically rephrase the query.

As per this previous question, I am wondering if the two series $$\frac{f(a)+f(b)}{2}\frac{(b-a)}{1!}+\frac{f'(a)-f'(b)}{2}\frac{(b-a)^2}{2!}+\frac{f''(a)+f''(b)}{2}\frac{(b-a)^3}{3!}+\cdots$$

and $$\frac{f(a)+f(b)}{2}\frac{(b-a)}{1!}+\frac{f'(a)-f'(b)}{2^2}\frac{(b-a)^2}{2!}+\frac{f''(a)+f''(b)}{2^3}\frac{(b-a)^3}{3!}+\cdots$$ could possibly be equal. The only difference is the powers of $2$ in the denominator.

NOTE: the numerator is $f^{(k)}(a)+(-1)^kf^{(k)}(b)$ in general.

The identity you give is quite pretty! Yes they are equal.

Let
$$\phi(t) = \sum_{n \ge 0}\frac{(b-a)^{n+1}}{(n+1)!}
\left( f^{(n)}(a) \cdot t^{n+1} + f^{(n)}(b) \cdot (-1)^n (1-t)^{n+1}\right).$$
The first series in question is $\frac12(\phi(1)+\phi(0))$ and the second is $\phi(\frac12)$. The "why" is that these are equal because, in fact, $\phi$ is constant:
$$\phi(t) = \int_a^bf(x) \, dx$$
for all $0 \le t \le 1$. The proof is by splitting the integral into two as
$$\int_a^bf(x) \, dx = \int_a^{a(1-t)+bt}f(x) \, dx + \int_{a(1-t)+bt}^bf(x) \, dx$$
and evaluating the first by expanding $f$ as a power series about $a$ and the second by expanding $f$ as a power series about $b$. You will get $\phi(t)$ as the result.

Specifically, the "when" is the condition that the power series expansions
$$f(x) = f(a)+f'(a)(x-a)+\frac{1}{2!}f''(a)(x-a)^2+\cdots$$
and
$$f(x) = f(b)+f'(b)(x-b)+\frac{1}{2!}f''(b)(x-b)^2+\cdots$$
are valid for all $x \in [a,b]$.

At the risk of sounding totally banal, I've basically just briefly rehashed (and slightly generalized) the derivation you linked to $-$ that post in itself already gives the proof.

Denote a primitive of $f$ by $F$ and add ${1\over2}\bigl(F(a)-F(b)\bigr)$ to both series. Then the first series becomes
$$\eqalign{{1\over2}\sum_{k=0}^\infty \bigl(F^{(k)}(a)-(-1)^kF^{(k)}(b)\bigr){(b-a)^k\over k!}&={1\over2}\sum_{k=0}^\infty F^{(k)}(a){(b-a)^k\over k!}
-{1\over2}\sum_{k=0}^\infty F^{(k)}(b){(a-b)^k\over k!}\cr
&={1\over2}\bigl(F(b)-F(a)\bigr)\ ,\cr}$$
and similarly the second series becomes
$$-{1\over2}\bigl(F(a)-F(b)\bigr)+\sum_{k=0}^\infty \bigl(F^{(k)}(a)-(-1)^k F^{(k)}(b)\bigr)\>{\bigl({b-a\over2}\bigr)^k\over k!}$$
$$={1\over2}\bigl(F(b)-F(a)\bigr)+F\left({a+b\over2}\right)-F\left({a+b\over2}\right)={1\over2}\bigl(F(b)-F(a)\bigr)\ .$$


How can I numerically implement $\delta(f(x,y))$ I would like to numerically implement a Dirac Delta function whose argument is another 2 variable function. I know that I can model a Dirac Delta numerically using a Gaussian. What can I do if I want to implement $\delta(f(x,y))$. Will using a function like $e^{-f(x,y)^2/\sigma^2}$ work?
Edit: $\sigma$ is a parameter I can control to make the peak of the Gaussian sharper. By 'implement', I mean generate numerically, that is to generate a 2 variable Delta function over a 2D grid.

The Dirac delta function $\delta$ is a linear functional which maps your functions into your scalars. To be precise, suppose $V$ is a vector space over $\mathbb{R}$ consisting of continuous functions $f : \mathbb{R}^n \rightarrow \mathbb{R}$ and $x_0 \in \mathbb{R}^n$, then the functional $T : V \rightarrow \mathbb{R}$ given by $$T(f) = f(x_0)$$ is the Dirac delta function with respect to the point $x_0$. The term functional rather than function is used to emphasize that the codomain of $T$ is your set of scalars. It is clear that $T$ is linear, because $$T(af + bg) = a f(x_0) + b g(x_0) = a T(f) + b T(g)$$ for all $a,b \in \mathbb{R}$ and $f, g \in V$.

Many, many functionals $S : V \rightarrow \mathbb{R}$ can be written exactly as
$$ S(f) = \int_{\mathbb{R}^n} f(x) s(x)dx$$
for a suitable choice of $s$. Dirac's delta function is an exception. It can be approximated using such functionals or you can simply implement it using the definition.

Yes, exactly: if the delta function $\delta$ can be well-approximated by a gaussian function $N$, then $\delta(f(x,y)) \approx N(f(x,y)) = \frac{1}{Z_0}\exp{(-f(x,y)^2/\sigma^2)}$, with $Z_0$ a normalizing factor.

The only difficulty I can see is that depending on the possible values of $f$, it may be important for your application to have a sharper gaussian function $N$.


Finding the probability density function of $Z = X + Y$ The problem
Let there be two independent random variables given. Let's name them $X$ and $Y$. Suppose the variables have both normal distribution $N(\mu, \lambda^2).$
Let's define third random variable $Z = X + Y$. The task is to find the density function of $Z$.
My attempt
I know that the density function can be easily calculated using the distribution function (I would need to differentiate it). That's why I started with finding the distribution.
$$F_Z(t) = Pr(Z \le t) = Pr(Y + X  \le t) = Pr(Y \le t - X) = \int \limits_{-\infty}^{\infty}Pr(Y \le t-x)f_X(x) dx$$
The equation above leads us to the following one:
$$ \int \limits_{-\infty}^{\infty} \int \limits_{-\infty}^{t-x} f_Y(y)f_X(x) \mbox{d} y\mbox{d}x \tag{1}.$$
Using the knowledge about $f_X$ and $f_Y$ we can write $(1)$ in this form:
$$\theta\int \limits_{-\infty}^{\infty} \int \limits_{-\infty}^{t-x}e^{-\frac{(y- \mu)^2}{\lambda^2}} e^{-\frac{(x- \mu)^2}{\lambda^2}} \mbox{d} y\mbox{d}x,$$
where $\theta = \frac{1}{\sqrt{2 \lambda^2}}.$
I wonder if my attempt was correct because the integral is quite horrible. Is it possible to calculate it in a "nicer" way?

Is $(1)$ equivalent to this expression $(f_X*f_Y)(x)$?
If yes I would get the following integral:
$$\int \limits_{\mathbb{R}} e^{-\frac{(y- \mu)^2}{\lambda^2}} e^{-\frac{(x- y - \mu)^2}{\lambda^2}} \mbox{d}y.$$
Is there a method of calculating such things?

From your expression, $F_Z(t) = \int_\mathbb{R}P(Y \leq t-x)f_X(x)dx$, differentiate to obtain the density (you may need to justify why you can differentiate under the integral),

$$ f_Z(t) = \int_\mathbb{R}f_Y(t-x)f_X(x)dx = \int_\mathbb{R}\frac{1}{\sqrt{2\pi}}e^{-(t-x-\mu)^2/2\lambda^2}\frac{1}{\sqrt{2\pi}}e^{-(x-\mu)^2/2\lambda^2}dx$$

which we recognize as $(f_Y*f_X) (x)$. To actually evaluate the integral, expand the exponent terms inside the integrand,

$$-\frac{1}{2\lambda^2}\left[(t-x-\mu)^2 + (x-\mu)^2\right] = -\frac{1}{2\lambda^2}(t^2-2\mu t + 2\mu^2 + \underbrace{2x^2-2tx}_{(*)})$$

Complete the square on the $(*)$ term, $2x^2 - 2tx = 2\left(x-\frac{t}{2}\right)^2-\frac{t^2}{2}$. Putting things together,

$$-\frac{1}{2\lambda^2}(t^2-2\mu t + 2\mu^2 + 2x^2-2tx) = -\frac{1}{4\lambda^2}(t^2-4\mu t + 4\mu^2)-\frac{1}{\lambda^2}(x-t/2)^2$$

So, we can pull out the $t$ terms from the integrand to be left with,

$$ f_Z(t) = \frac{1}{\sqrt{4\pi\lambda^2}}e^{-(t-2\mu)^2/4\lambda^2}\underbrace{\int_\mathbb{R}\frac{1}{\sqrt{\pi\lambda^2}}e^{-(x-t/2)^2/\lambda^2}dx}_{=1} = \frac{1}{\sqrt{4\pi\lambda^2}}e^{-(t-2\mu)^2/4\lambda^2}$$

The integral above equals $1$ because the integrand is the density function of a $N(t/2,\lambda^2/2)$ distribution.

The problem can also be solved by using the following property:
If $X\sim N(\mu_1,\sigma_1^2) $ and $Y \sim N(\mu_2,\sigma_2^2)$ and are independent, then $aX + bY \sim N(a\mu_1 + b\mu_2,a^2\sigma_1^2 +b^2\sigma_2^2)$ and thus in your case $Z \sim N(2\mu,2\lambda^2)$.

Edit: I just noticed the comment by @saulspatz above who already pointed this out.


Number of integer solutions to $3i^2 + 2j^2 = 77 \cdot 6^{2012}$ here is another problem I did not manage to solve in the contest I mentioned in my previous question.

  Determine the number of integer solutions $(i, j)$ of the equation:
  $3i^2 + 2j^2 = 77 \cdot 6^{2012}$.

Applying logarithms is not useful, since on the left hand side we have a sum; I also tried some algebraic manipulations that led me to nothing useful. Is there a simple solution to the problem?

Thank you,
rubik

Observe that $i$ has to be even, so do $j$ because $3i^2$ and the right side can be divided by 4.
Let $i'=2i$, $j'=2j$; the equation becomes $3i'^2+2j'2=77 \cdot 3^{2012} \cdot 2^{2010}$. The same argument applies 1006 times, and we get something like $3i^2+2j^2=77 \cdot 3^{2012}$.
Now do the same looking the divisibility by 3. It works. Finaly check $3i^2+2j^2=77$

Edit: It's my first posting here, so hello to all from France !


Axiomatic definition of groups To define natural numbers one can either:

use the Peano axioms in second-order logic;

encode them in set theory as von Neumann ordinals.

The relation between those two definitions of natural numbers is that
(2) satisfies (1).
Now, groups are usually defined as sets equipped with operations and axioms. This is akin to (2) above since it is an encoding of groups in set theory.
What would be a definition of groups akin to (1)?

This is too long for a comment. Given the negative reception of the current answers, and the confusion on the original question, I thought I would add what I think is being asked, and why the current answers don't work. (For what it's worth, I did not downvote either answer.)
It appears to me that the OP is trying to compare the class of all groups to the set of natural numbers. In other words, a single group is compared to a single natural number. This is why it is claimed that the definition of groups via axiomatization is akin to (2), rather than (1). Specifically:

Given a model $M$ of ZF, one can define which sets in $M$ are  natural numbers using the von Neumann definition: a set is a natural number if and only if blah blah...

Given a model $M$ of ZF, one can define which pairs of sets in $M$ are groups (where the pair consists of the underlying set of the group and the graph of the binary operation) using straightforward axioms: a pair of sets is a group if and only if blah blah...

There are some problems with this, as pointed out in Ned's comment. For example the natural numbers don't change depending on the model of ZF, while on the other hand different models of ZF could disagree about which groups they contain. But even still, the analogy does make sense.
On the other hand, the current answers make a comparison between the set of natural numbers and a particular group (rather than the class of all groups). So this doesn't seem to be what the OP is after in something akin to (1).
In my estimation, something akin to (1) would be an axiomatic theory $T$ which fits the analogy
$$
\text{ $T$ : group :: Peano axioms : natural number}
$$
So the class of groups would somehow be the "standard model" of $T$ (whatever that might mean) in the same way the natural numbers are the standard model of PA. But already there are issues with how to make this precise. At the very least, I think one would want to represent groups as isomorphism classes or presentations, rather than sets. Then one could make some "meta-axiomatizations" of special sets of group presentations. For example, one could "axiomatize" sets of group presentations that contain the finite cyclic groups and are closed under direct product. Then the finite abelian groups would be something like the "standard model". But this is not nearly rigorous enough.
So I don't know what the right answer is, or if there is a good answer at all. Hopefully what I've said isn't total nonsense, and I haven't misconstrued the original intent of the question.

Side note: The collection of finite von Neumann ordinals is a set, but the collection of all groups (up to isomorphism) is a proper class, unless you want to restrict yourself to finite groups.
The finite von Neumann ordinals are natural numbers in the sense that $\langle \omega, \varnothing, x \mapsto x \cup \{x\} \rangle \models \langle \mathbb{N}, 0, S \rangle$. How do you expect elements of your "class of groups" to relate to each other?
For groups $G$ and $H$, let $G \sim H$ if they are isomorphic. Let
$$\mathrm{Grp} = \{\langle S, \bullet \rangle \mid S \in V, \bullet \in S \times S \rightarrow S, \text{$\bullet$ satisfies the group axioms}\}$$
be the class of all groups, where $V$ is the class of all sets. Is the quotient $\mathrm{Grp} / {\sim}$ what you're looking for? See Scott's trick for equivalence classes on a proper class.
Or are you perhaps looking for something like a "canonical encoding" of the finite groups, like the finite von Neumann ordinals are a "canonical encoding" of the natural numbers?


Birthday $pmf$ and joint $pmf$ for $n$ people Let a discrete r.v be denoted $X_1,..,X_n$ denote the birthdays of $n$ people in a room. Assume that $X_1,..,X_n$ are mutually independent and that $X_i$ is a distribution such that $X_i \sim U(\left \{ 1,2,...,365 \right \})$ for all $i \in {(\left \{ 1,...,n\right \}}$.

Find and write out the pmf $p_i: \mathbb{R} \rightarrow \mathbb{R}$ for the r.v $X_i$.
Find the joint pmf $p:\mathbb{R}^n \rightarrow \mathbb{R}$ for the vector $(X_1,..,X_n)$ and show that it is a valid joint pmf.

My try
First: The pmf would just be the pmf of a uniformal distribution. That would mean that $p_i(x)={x_i}^{-1}$ for the $x=1,...,n$. (Now the notation is weird for me as it is for the $i$'th $X_i$). (I know that $p_i: \mathbb{R} \rightarrow \mathbb{R}$)
Second: I know that for it to be a valid pmf the total probability has to sum up to $1$. The joint pmf should be denoted as $p_{X_1,...,X_n}(x_1,...,X_n):=P(X_1=x_1,...,X_n=x_n)$ but no idea on how to write this out from my results. Wouldnt I just have to replace $x_i$ with $1/365$? (I know that $p_i: \mathbb{R}^n \rightarrow \mathbb{R}$)

The first question is easier than you think: each day is equally likely and there are $365$ possibilities so $p_i(x)=\frac1{365}$ for $x \in \{1,2,3,\ldots,365\}$ and $0$ otherwise
Since the birthdays are independent you have $p(\mathbf{x}) $ $=p(x_1,x_2,\ldots,x_n) $ $= \mathbb P(X_1=x_1,X_2=x_2,\ldots,X_n=x_n) $ $= \mathbb P(X_1=x_1)\mathbb P(X_2=x_2)\cdots\mathbb P(X_n=x_n) $ $= p_1(x_1)p_1(x_2)\cdots p_n(x_n)$.  This will be $\frac{1}{365^n}$ when all of the $x_i \in \{1,2,3,\ldots,365\}$ i.e. when $\mathbf{x}\in \{1,2,3,\ldots,365\}^n$, and $0$ otherwise.  You can easily check that it adds up to $1$.

It was claimed that $X_i\sim\mathcal U\{1..365\}$ for all $i\in\{1..n\}$.

The probability mass function $p(u)$ for a uniform discrete distribution of $\mathcal U\{a{.}{.}b\}$, is $$p(u)=(b-a+1)^{-1}\mathbf 1_{u\in\{a..b\}}$$

So for any $i\in\{1..n\}$, the marginal probability mass function $p_i(x)$ is...

For the second, the keywords are "mutually independent".    There's something about the joint probability mass function of mutually independent random variables ...


Problem with the solution of differential equation My try:

$\phantom{}\phantom{}\phantom{}$

Start by noticing that $P$ doesn't appear in the end result.  So,  we must eliminate it , instead of $Q$. Multiply by $y_2$  to the first and $y_1$ to the second and subtract to get
$$y_2 \frac{dy_1}{dx}  - y_1 \frac{dy_2}{dx}  = Q(y_2-y_1) $$
Then substitute $y_2 = y_1*z$ and integrate it to get the answer.  Remember P and Q are functions of $x$. Your attempt shows that you forgot it while solving (integration part).

Comment if you any more help.

Plugging $y_2$ in the equation,

$$y_1z'+y_1'z+Py_1z=Q$$ implies

$$y_1z'+Qz=Q$$

and

$$\frac{z'}{z-1}=-\frac{Q}{y_1},$$

$$\log(z-1)=\log a-\int\frac{Q}{y_1}dx.$$


If $f$ is diagonalisable then its minimal polynomial is the product of distinct linear factors I'm trying to prove that if a linear operator $f$ is diagonalisable then its minimal polynomial is the product of distinct linear factors. This is what I have so far:

Let $f$ be diagonalisable. So there exists a basis relative to which $f$ has a diagonal matrix, say $D$. So the characteristic polynomial of $f$ is given by $p_f(x)=(x-\lambda_1)(x-\lambda_2)\ldots(x-\lambda_s),$ where $\lambda_i$ are the diagonal entries of $D$.

I know that the minimal polynomial must divide the characteristic polynomial and have the same linear factors. Without loss of generality let the first $i$ linear factors be distinct. So I claim that the minimal polynomial $m_f(x)=\pm (x-\lambda_1)(x-\lambda_2)\ldots(x-\lambda_i).$

However, how can I now verify that $m_f(D)=0$ ?

This is very basic, and you do not need to use the characteristic polynomial or the fact that the minimal polynomial divides it at all. You just need to realise that "diagonalisable" means that the sum of the eigenspaces fills the whole space, so a linear operator is zero if (and obviously only if) it is zero on each of the eigenspaces.

Now on the eigenspace for an eigenvalue$~\lambda$, our $f$ acts by scalar multiplication by$~\lambda$. It easily follows that on this eigenspace any polynomial $P[f]$ acts by scalar multiplication by$~P[\lambda]$ (just check that $f^k$ acts by multiplication by $\lambda^k$, and then combine the monomials of the polynomial $P$ linearly). So by the above, $P[f]=0$ iff $P[\lambda]=0$ for every eigenvalue$~\lambda$. The minimal monic polynomial$~P$ with that property is the product of (just) one factor $X-\lambda$ for each distinct eigenvalue$~\lambda$ of$~f$; there are distinct linear factors.

Without loss of generality, let the diagonal form $D$ have all the $i$ distinct eigenvalues in the top rows.  For every eigenvalue $\lambda$, one of the top $i$ rows of the matrix $(D - \lambda I)$ will consist of zero entries.  Hence, the determinant of the top left $i$-by-$i$ submatrix of $(D - I \lambda)$ is zero.  But, this determinant is $m_{f}(\lambda)$.


Why do we invert then multiply when dividing fractions? I have looked up various sites online. They only explain it in a very basic arithmetic way. Would someone explain to me why is $\displaystyle\frac{x}{\frac{1}{2}}$ is $\displaystyle\frac{2x}{1}$?

Division can be defined as multiplication by the inverse. Explicitly, for any associative division algebra (including real numbers, rational numbers, complex numbers, quaternions, etc.) we can define division by

$$a \div b = \frac{a}{b} := ab^{-1}$$

for any $b \neq 0$. If the algebra is commutative (that is $ab = ba$ for any $a, b$), which the real numbers, rationals, and complex numbers are, then we must have

$$\frac{a}{b} \div \frac{c}{d}= ab^{-1}\div (cd^{-1})= ab^{-1}(cd^{-1})^{-1}=ab^{-1}(dc^{-1})=adb^{-1}c^{-1}=(ad)(bc)^{-1}= (ad) \div (bc) =\frac{ad}{bc}= \frac{a}{b} \cdot \frac{d}{c}.$$

Thus, when you divide fractions you "flip and multiply".


Please find the radius of the circle. Hello,
I want to find the radius of red circle. I tried it with several ways like trigonometric. But there is a special value is given with this. I can not understand why it is given. It is the blue colored value "2", the distance between two arcs of circles. Please help me to find this.
Thank you.

Strictly speaking, you do not need the information shown in blue,
because you can deduce it from the other information shown on the diagram.
Hence there is really no reason why that particular piece of
information is given.

The important thing in this problem is that you can determine exactly
where the red circle intersects the $45$-degree diagonal line
as shown in the diagram.
You know that the green circle has radius $20$ (because two of the radii
are explicitly labeled $20$), so you know the portion of the diagonal line
between the circumference of the green circle and the center of the green circle (the green dot) is $20$.
To determine exactly where the red circle intersects that line, you can
use the information that the distance from that point to the green dot
is $18$ (as shown on the diagram), or you can use the information
that the distance from that point to the circumference of the green
circle is $2$, which means the distance to the green dot is $18$.

The reason it is important to know where the red circle intersects the
diagonal line is that three points determine a circle.
You have two points on the red circle (where it intersects the green circle),
so this third point determines the center and radius of the red circle.

That is the abstract reason why the information in the diagram should be
sufficient to answer the question.
But as to exactly how to use that information, here are some hints:

Use the two perpendicular black lines as axes of a Cartesian coordinate system
such that the green dot is at $(0,0)$
and the two intersections of the two circles are at $(0,20)$ and $(-20,0)$.
In that case, we can see that the red circle also passes through the point
$(-18\cos\frac{\pi}{4},18\sin\frac{\pi}{4}) = (-9\sqrt{2},9\sqrt{2})$;
that's where the red circle intersects the diagonal line in the upper left quadrant.

Let the coordinates of the red dot be $(x_0,y_0)$. Then $y_0 = -x_0$,
so we can write the coordinates as $(x_0,-x_0)$.
The distance from the red dot to $(0,20)$ is
$\sqrt{x_0^2 + (-x_0 - 20)^2}$
and the distance from the red dot to $(-9\sqrt{2},9\sqrt{2})$ is
$\sqrt{(x_0 + 9\sqrt{2})^2 + (-x_0 - 9\sqrt{2})^2}$.
These two distances must be equal (because each is a radius of the red circle), that is,

$$\sqrt{x_0^2 + (-x_0 - 20)^2} = \sqrt{(x_0 + 9\sqrt{2})^2 + (-x_0 - 9\sqrt{2})^2}.$$

Solve for $x_0$.
(Hint: simplify the last equation so that you can apply the quadratic formula.)
With that information, it is easy to find the radius of the circle.


Formal proof of limit By using $\epsilon-\delta$ definition, prove that $$\lim_{x\rightarrow1}\frac{3}{x^2}=3$$

Here is what I have tried.

$$|\frac{3}{x^2}-3|=3|\frac{1}{x^2}-1|=3|\frac{1-x^2}{x^2}|=3\frac{|x-1||x+1|}{x^2}$$

So I have problem with how to elimate the denominator.

Since $x$ is close to $1$ then $1/x^2 < 4$ and $|x+1|<3$.


Let $ X $ be compact and let $f: X \rightarrow Y$ be a local homeomorphism. Show that for any point $y \in Y$, $f^{-1}(y)$ is a finite set. PROBLEM: Let $ X $ be compact and let $f: X \rightarrow Y$ be a local homeomorphism. Show that for any point $y \in Y$, $f^{-1}(y)$ is a finite set.

I was trying out this problem from Massey, but got stuck.

MY ATTEMPT: My thought was to pick an open neighborhood $U_i$ of arbitrary but fixed $y_i \in Y$, then show that $U_i$ is evenly covered by $f$, so $f^{-1}(U_i)$ is a disjoint union of open sets in $X$,

$f^{-1}(U_i) = \underset{\alpha \in X} \amalg V_{{y_i}_{\alpha}} $ such that for each $\alpha$, $f|_{V_{{y_i}_{\alpha}}}: V_{{y_i}_{\alpha}} \rightarrow U$ is a surjective homeomorphism. This also implies $f$ is a covering map.

Then $X = \underset{\alpha \in X}\amalg V_{{y_1}_{\alpha}} \cup \underset{\alpha \in X}\amalg V_{{y_2}_{\alpha}} ...$, which is an open covering. Since $X$ is compact, that means there is a finite subcover. Specifically, this means only finitely many $f^{−1}(U_i) = \underset{\alpha \in X} \amalg V_{{y_i}_{\alpha}}$ cover $f^{-1}(y_i)$ for each $y_i$.

Does this outline seem correct? I have a problem at the beginning, not knowing how to use the local homeomorphism to show that $f$ evenly covers $U_i$, and I'm not sure about if my conclusion appropriately proves the statement.

Without going into evenly covered maps etc., it might be simpler: Suppose for some $y$, $F_y := f^{-1}[\{y\}]$ is infinite. As $X$ is compact (in fact we need only countably compact) there is a limit point $p$ of $F_y$. But if $U$ is a neighbourhood of $p$ such that $f \restriction U: U \to f[U]$ is a homeomorphism, then $U$ contains at least 2 (in fact infinitely many; here I'm assuming $X$ is $T_1$ as well) points of $F_y$, which makes $f$ not even injective on $U$, as both map to $y$ under $f$.
So in fact, for a locally injective map $f$ on a countably compact $T_1$ space, all fibres are finite.
So if some $F_y$ were infinite, by compactness it would have an $\omega$-accumulation point $x \in X$. If $O$ were the witnessing neighbourhood of $x$ wrt local homeomorphism, a contradiction to injectivity ensues; no sep. axioms needed.

I don't have a full answer for you but here are some thoughts.  If f-1(Ui) is a disjoint union of open sets in X, that does not mean that they are a cover for X (and we don't care).  They are a subset of X and by definition none of them are compact (being open).  So I think you should close them, forming a compact set V ⊂ X (V is compact because X is).  Now you can conclude as you say that a finite set of the f-1Ui) cover V.

Since you obviously can form all the f-1(Ui) you want, the problem comes down to whether you can make them disjoint.  For this I think you need to pay attention both to the fact that f is a local homeomorphism, and  that it is (therefore) continuous.  The former rules out cases such as X = [0,1] on the real line and f(x) ≡ 1.  The latter allows you to choose
each Ui so that yi is unique.

You need to do this so that f-1(yi) is unique in f-1(Ui), which I think it will be because f is 1-1; otherwise having a finite cover doesn't necessarily imply a finite number of f-1(yi).

I still don't quite see how to get from there to disjoint, but I think that plus the compactness of X should be the right approach.

I'll keep thinking about it.  If you solve it, please post the solution so I can see it.

Further thought: First just to be sure we are clear, I am using the set {yi} to be all the y's which have the same value.  Following up on the idea of there being a Ui for each yi which contains only one yi, it must be that there is a there is a constant a > 0 such that the "radius" of each Ui can be a while still preserving just one yi for each Ui.   Otherwise there is a convergent sequence of yi which goes to say yc.  I believe that f-1 would not be continuous at yc, so that f would not be a homeomorphism. (I haven't quite shown this, but maybe you can do that).

Having gotten that far, maybe the disjointness of f-1(Ui) will follow.


Evaluate $\lim_{x \to 0+} (\sin x)^x$ I have tried to evaluate this by :--

$$(\sin x)^x = e^{(x \cdot \log (\sin x))} = (e^x)^{\log (\sin x)}$$

$$\lim_{x \to 0+} (e^x)^{\log (\sin x)} = \lim_{ x \to 0+} e^{\log (\sin x)}(e^x-1)$$

But after this I couldn't proceed.

Let $L=\lim\limits_{x\to0^+}(\sin x)^x$. Then $\ln L=\lim\limits_{x\to0^+}x\ln(\sin x)=\lim\limits_{x\to0^+}\frac{\ln(\sin x)}{1/x}$. This puts the limit in a form to use L'Hopital's rule:
$$\lim\limits_{x\to0^+}\frac{\ln(\sin x)}{1/x}=\lim\limits_{x\to0^+}\frac{\cos x/\sin x}{-1/x^2}=\lim\limits_{x\to0^+}-\frac{x^2\cos x}{\sin x}=-\lim\limits_{x\to0^+}\frac{x}{\sin x}\cdot\lim\limits_{x\to0^+}x\cos x$$
Using the fact that $\lim\limits_{x\to0^+}\frac{x}{\sin x}=1$ (this can be verified using the squeeze theorem), we get that $$-\lim\limits_{x\to0^+}\frac{x}{\sin x}\cdot\lim\limits_{x\to0^+}x\cos x=-1\cdot0\cdot1=0\implies \ln L=0$$
Now exponentiating:
$$\ln L=0\implies L=1$$
Therefore, $\lim\limits_{x\to0^+}(\sin x)^x=1$.

I like binomial expansions, so why not try one?

$$
\begin{align}
\lim_{x\rightarrow 0} (\sin(x))^x & = \lim_{x\rightarrow 0}(\tfrac{e^{ix}-e^{-ix}}{2i})^x \\
& = \lim_{x\rightarrow 0}\tfrac{1}{(2i)^x}\sum_{k=0}^{x} {x\choose k} e^{ix(x-k)}(-e)^{-ix(k)} \\
& = \sum_{k=0}^{0} {0\choose k} e^{i0(0-k)}(-e)^{-i0(k)}\\
& = {0\choose 0} e^{0}(-1)^0e^{0}\\
& = 1
\end{align}
$$


Proof of $\frac{d}{dx}e^x = e^x$ I'm working through the proof of $\frac{d}{dx}e^x = e^x$, and trying to understand it, but my mind has gotten stuck at the last step.

Starting with the definition of a derivative, we can formulate it like so:

$$\frac{d}{dx} e^x = \lim_{h \to 0} \frac{e^{x+h}-e^x}{h}$$

After some algebra, we arrive at:

$$\frac{d}{dx} e^x = e^x \lim_{h \to 0} \frac{e^h-1}{h}$$

As $h\to0$, the expression approaches $\frac{0}{0}$, which makes it indeterminate. And, this is where my understanding ends. I've tried looking at wikipedia and other descriptions of the proof, but couldn't understand those explanations. It has usually been something along the lines of, "plot $\frac{e^x - 1}{x}$ and see the function's behavior at $0$," which ends up approaching $1$, which can substitute the limit to give the result of the derivative:

$$\frac{d}{dx} e^x = e^x \cdot 1 = e^x$$

I vaguely understand the concept of indeterminate forms, and why it is difficult to know what is happening with the function. But is there a better explanation of how the result of $1$ is obtained?

Everything in the "proof" will depend on your definition of the function $e^x$. I will choose the definition
$$
e^h \overset{\text{def}}{=} \sum_{k=0}^\infty \frac{h^k}{k!}.
$$
Using this, one sees that
$$
\frac{e^h - 1}{h} = \frac{\sum_{k=0}^\infty \frac{h^k}{k!} - 1}{h} = \sum_{k=1}^\infty \frac{h^{k-1}}{k!} = 1 + h \sum_{k=0}^\infty \frac{h^k}{(k+2)!}.
$$
If you have studied convergence tests, you know that the last series on the right converges for all $h \in \mathbb R$, hence taking the limit when $h \to 0$, the RHS goes to $1$ because the series will converge to something and the $h$ factored out will make the product go to zero.

Another way to do this would be to show that this series is an analytic function and is its own Taylor expansion around zero (this is not hard to do using convergence tests), so to differentiate it you can go term by term and readily see that its derivative is itself.

A third approach, which will sound a little stupid and meaningless but is nonetheless funny, is choosing another definition for $e^x$ : consider the differential equation
$$
f'(x) = f(x), \quad f(0) = 1.
$$
Using differential equation theory it is really not hard at all to show that the solutions to the equation (without the initial condition) is a one-dimensional vector space and there exists an unique element of this vector space which satisfies the initial condition $f(0) = 1$, because the solutions are of the form $Cg(x)$ for some solution $g(x)$. Let $exp(x)$ be defined as a solution to this differential equation satisfying the initial condition. Then clearly $exp'(x) = exp(x)$. Then you can easily see that $exp'(x)$ is a differentiable function, so by induction $exp(x)$ is an infinitely differentiable function with Taylor expansion
$$
\exp(x) = \sum_{k=0}^\infty \frac{x^k}{k!}.
$$
I mentioned it to show the importance of which definition one decides to choose ; it can change the whole structure of an argument.

Hope that helps,

l'Hôpital's rule applies to indeterminate form of the limit you have in
$$
\frac{d}{dx} e^x = e^x \lim_{h \to 0} \frac{e^h-1}{h}
$$
So you get:
$$
\lim_{h \to 0} \frac{e^h-1}{h} = \lim_{h \to 0} \frac{\frac{d}{dh}(e^h-1)}{\frac{dh}{dh}}=\lim_{h \to 0}\frac{e^h}{1}=1
$$


Solving a system of Pell and Pell-like equations Solve (Find all the solutions, if there are, or prove there are not)  the system of 2 Pell or Pell-like simultanous equations over the positive integers :

$2b^{2}= a^{2} +1 = 3k^{2} + 2 $ with the 3 variables (a,b,k).

Note that the 2 equations are :
 $a^{2} -2b^{2} = -1$ and  $a^{2} -3k^{2} = 1$
The third one: $ 2b^{2} -3k^{2} =2$ does not need to be solved if the other two are.

László Szalay, On the resolution of simultaneous Pell equations, Annales Mathematicae et Informaticae 34 (2007) pp. 77–87, is available here. I don't know whether the precise pair of equations you give is discussed in the paper, but it does give you the tools and references you need to approach this kind of problem.

EDIT: Having looked a bit more closely at the paper, I believe a solution to your problem is in fact given in the paper and attributed to Riele, reference 24.

$a^2 - 3*k^2 = 1$ (1) , general solution:

$a(n+1) = 2a(n) + 3k(n)$, $k(n+1) = a(n) + 2k(n)$. The first 6 solutions, without counting the  trivial solution: (1,0) are:
$(a,k) = (2,1); (7,4); (26,15); (97,56); (362,209); (1351,780);\dots$

$a^2 - 2*b^2 = -1$ (2), general solution:

$a(n+1) = 3a(n) + 4b(n)$, $b(n+1) = 2a(n) + 3b(n)$. The first 6 solutions are:
$(a,b) = (1,1); (7,5); (41,29); (239,169); (1393,985); (8119,5741);\dots$

Once we match the same a value in both equations: a=7 asociated to k=4 and b=5; we do not need to look any  more for solutions, since both a and b grow faster in one equation than a and k in the other ---> the only solution non trivial is (a,b,k) = (7,5,4). The problem would be if a(n) grew faster and b(n) slower in one equation than in the other, or the contrary. In that case we would have to match another coincident value of "a" among the infinite values in both equations. But no problem here.

The general solution of the third equation $2b^{2} -3k^{2} = 2$ is $b(n+1) = 5b(n)+ 6k(n) ; k(n+1) = 4b(n) + 5k(n)$ with solutions: $(1,0);(5,4);(49,40);(485,396);(4801,3920), ...$


Get numbers that have only 2,3 and 5 as prime factors I am given an integer N. I have to find first N elements that are divisible by 2,3 or 5, but not by any other prime number.

N = 3
Results: 2,3,5
N = 5
Results: 2,3,5,6,8

Mistake number = 55.. 55/5 = 11.. 11 is prime number.. so means that it divides by any other prime number and doesn't counts in..

I guess there is need of recursive function,but I cant imagine what would algorithm look like

I was sent here from stackoverflow.

They gave me advice.. 2^i*3^j*5^k , for i,j,k = 0,1,2... , but it won't give an ordered array, because 2*2 < 2*3 .. and so on

I am trying to get this work in C++

The algorithm in this answer, in the end, is the same as the one in Bill Dubuque's answer, but I hope to be more elaborate and describe how one could arrive at the algorithm. Let's frame the problem as that of building the list of all such numbers in ascending order, i.e., each time adding to the list the smallest number that is not already in it.

First, note that all the numbers you want are of the form $2^i3^j5^k$, where $i, j, k$ are nonnegative integers. This means that after the first number $1$ (corresponding to $i = j = k = 0$), each "next" number can be obtained by multiplying some number already generated in the list, by either $2$, or $3$ or $5$.

So you start the list with the first number, $1$. To find the next number, you have three choices: multiply the number $1$ by either $2$, or $3$, or $5$. Since you want the smallest result, you pick $2$, and your list becomes $[1, 2]$.

Now you can no longer multiply $1$ by $2$ to get a new number, it is "used up". Your three choices now are: multiplying $1$ by $3$, multiplying $1$ by $5$, or multiplying $\color{red}{2}$ by $2$. The smallest result is from multiplying $1$ by $3$, so you add the result to the list, which becomes $[1, 2, 3]$.

Now you can no longer multiply $1$ by $3$, it is "used up" for multiplication by $3$. The smallest number you can multiply by $3$ is the next number, $2$. Your three choices now are: multiplying $\color{red}{2}$ by $2$, multiplying $\color{red}2$ by $3$, and multiplying $1$ by $5$. You pick the smallest result, which corresponds to multiplying $2$ by $2$ and gives you $4$ which you add to the list, and now $2$ is also used up for multiplication by $2$ and for multiplication by $2$ you have to go to the next number in the list, namely $3$.

In general, you always have three choices: one number that you can multiply by $2$, another that you can multiply by $3$, and another that you can multiply by $5$. You pick whichever of these three choices gives the smallest result, and update the choice ("advance the pointer") for 2 or 3 or 5, whichever you chose.

The first few steps worked out explicitly:

List is $[1]$. Choice for 2 is $1$, choice for 3 is $1$, choice for 5 is $1$. (Choose multiplication by 2.)

List is $[1, 2]$. Choice for 2 is $2$, choice for 3 is $1$, choice for 5 is $1$. (Choose multiplication by 3.)

List is $[1, 2, 3]$. Choice for 2 is $2$, choice for 3 is $2$, choice for 5 is $1$. (Choose multiplication by 2.)

List $[1, 2, 3, 4]$. Choice for 2 is $3$, for 3 is $2$, choice for 5 is $1$. (Choose multiplication by 5.)

List is $[1, 2, 3, 4, 5]$. Choice for 2 is $3$, for 3 is $2$, for 5 is $2$. (Choose multiplication by 2.)

List is $[1, 2, 3, 4, 5, 6]$. Choice for 2 is $4$, for 3 is $3$ (note that we updated this also), for 5 is $2$. (Choose multiplication by 2.)

List is $[1, 2, 3, 4, 5, 6, 8]$. Choice for 2 is $5$, for 3 is $3$, for 5 is $2$. (Choose multiplication by $3$.)

List is $[1, 2, 3, 4, 5, 6, 8, 9]$. Choice for 2 is $5$, for 3 is $4$, for 5 is $2$.

List is $[1, 2, 3, 4, 5, 6, 8, 9, 10]$. Choice for 2 is $6$, for 3 is $4$, for 5 is $3$.

List is $[1, 2, 3, 4, 5, 6, 8, 9, 10, 12]$. Choice for 2 is $8$, for 3 is $5$, for $5$ is $3$.

List is $[1, 2, 3, 4, 5, 6, 8, 9, 10, 12, 15]$. Choice for 2 is $8$, for 3 is $6$, for $5$ is $4$.

In code, you could implement it as a growing array, with three array indices pointing to the choices for 2, 3, and 5. Here is a complete working Python program (time.sleep is to keep it slow enough to see the output):

import time
num = [1]
c2 = c3 = c5 = 0

while True:
  next = min(2*num[c2], 3*num[c3], 5*num[c5])
  num.append(next)
  print num
  if next == 2*num[c2]: c2 += 1
  if next == 3*num[c3]: c3 += 1
  if next == 5*num[c5]: c5 += 1
  time.sleep(0.5)

Make a loop over numbers i.  Check the numbers one at a time, it doesn't need to be recursive.

Added:
If you want, you can use a nice recursive algorithm f(i) which is true or false depending on whether the number i should be allowed into the list.
f(i):
if i == 1: return true
else if i % 2 == 0: return f(i/2)
else if i % 3 == 0: return f(i/3)
else if i % 5 == 0: return f(i/5)
else: return false

Added:
Start the loop at 2, otherwise this function will wrongly allow 1 into the list.


How can I find the entries of the continued fraction of $2^{1/3}$ more efficiently? I wanted to detect large entries in the continued fraction of $$2^{1/3}$$

I found two ways to find the continued fraction of an irrational number like $$2^{1/3}$$

in PARI/GP. The most obvious is the contfrac-command which requires a very large precision, if I want to calculate, lets say , $10^7$ entries.

If the continued fraction expansion of $2^{1/3}$ is $[a_1,a_2,a_3,\cdots]$, then define $x_k:=[a_k,a_{k+1},a_{k+2},\cdots]$

I had the idea that I can calculate the minimal polynomials of the numbers $x_k$ (See the above definition). I wrote the program

? maxi=0;j=0;f=x^3-2;while(j<10^7,j=j+1;s=0;while(subst(f,x,s)<0,s=s+1);s=s-1;if
(s>maxi,maxi=s;print(j,"  ",s));f=subst(f,x,x+s);f=subst(f,x,1/x)*x^3;if(pollead
(f)<0,f=-f))
1  1
2  3
4  5
10  8
12  14
32  15
36  534
572  7451
1991  12737

which works very well. The output shows the first few successive record entries in the continued fraction expansion. The program does not need floating-point calculations. It only calculates the new minimal polynomial and determines the truncated part of the (unique) positive root.

But the coefficients of the minimal polynomial get soon extremely large, hence the program needs already much longer for the first $2\cdot 10^4$ entries than the cont-frac-command.

  Does anyone know a faster method with which I can calculate $10^7$ or $10^8$ entries in a reasonable time ? The contfrac-command requires much memory , so a way to avoid large arrays would be very appreciated.

I've computed the first 9.6 million coefficients and get the following maximum values. There is only one additional large value added to your list.

       n      a_n
       1        1
       2        3
       4        5
      10        8
      12       14
      32       15
      36      534
     572     7451
    1991    12737
   20857    22466
   27432    68346
   28763   148017
  155122   217441
  190271   320408
  288108   533679
  484709  4156269
 1395499  4886972
 9370521 10253793

Here is my Ruby code.

The big improvement over your code is that I do a binary search, first finding the smallest $n$ such that $p(2^n)<0<p(2^{n+1})$. This should reduce the number of times you have to evaluate $p$, and $p(2^n)$ can be computed faster with bit-shifts rather than multiplications. We are dealing with numbers with millions of digits. so the cost of multiplication can be high.

Once we have the values $2^n$ and $2^{n+1}$, this algorithm performs a binary search between those values to find $m$ so that $p(m)<0<p(m+1)$.

Even this could be optimized, since the binary search is pure, so we are always doing $p(N+2^k)$ for some $N$ where $p(N)$ is already known. We get $$p(N+2^k)=p(N)+p'(N)2^k + p''(N)2^{2k}/2 + p'''(N)2^{3k},$$ so we might be able to get away with speeding up those calculations. The second code example implements that. It is a bit more opaque.

# Evaluate the polynomial at a value
def evaluate(v,poly)
    total = 0
    poly.each do |c|
       total = total*v + c
    end
    total
end

# Performance hack - evaluate the polynomial at 2**n
# Uses bit-shifting to improve performance in this cases
def evaluate_bits(n,coefficients)
    total = 0
    coefficients.each do |c|
       total = (total << n) + c
    end
    total
end

# Given the current polynomial, and a value, compute
# y^3p(v+1/y)
#
# p(x)=ax^3+bx^2+cx+d
# y^3p(v+1/y)=a(yv+1)^3 + by(yv+1)^2+cy^2(yv+1)+dy^3
#            = p(v)y^3+(3av^2+2bv+c)y^2+(3av+b)y +a
def nextPoly(v,poly)
   a,b,c,d = poly
   v2= v*v
   v3 = v2*v
   a1 = a*v3+b*v2+c*v+d
   b1 = 3*a*v2 + 2*b*v + c
   c1 = 3*a*v + b
   d1 = a
   # Take the negative, so the new polynomial is negative at 1
   [-a1,-b1,-c1,-d1]
end

# Find 2^n so that p(2^n)<0<p(2^{n+1})
def next_value_logarithmic(poly)
   bits = 0
   while evaluate_bits(bits+1,poly)<0 do
      bits = bits+1
   end
   1 << bits
end

def next_value(poly)
   #puts "#{poly}"
   # Find the smallest 2^n so that p(2^n)<0<p(2^(n+1))
   range1 = next_value_logarithmic(poly)
   range2 = range1*2
   # Do a binary search on the range 2^n.. 2^{n+1} to find value
   # v where p(v)<0<p(v+1)
   while range2-range1>1 do
       #STDERR.puts "#{range1} #{range2}"
       middle = (range1+range2)/2
       midval = evaluate(middle,poly)
       if midval < 0 then
           range1 = middle
       else
           range2 = middle
       end
   end
   range1
end

p = [1,0,0,-2]
denom1 = 1
denom2 = 0
numer1 = 0
numer2 = 1
(1..500000).each do |i|
   c = next_value(p)
   p = nextPoly(c,p)
   puts "#{i} #{c}"
end
puts '#{p}'

The more opaque script using polynomial shifting to try to avoid most multiplications.

require 'csv'

# Performance hack - evaluate a polynomial at 2**n
# Uses bit-shifting to improve performance
def evaluate_bits(n,coefficients)
    total = 0
    coefficients.each do |c|
       total = (total << n) + c
    end
    total
end

# p(x)=x^3-2
p = [1,0,0,-2]
max_index = 0
pn_1=0
qn_1=1
pn=1
qn=0
if ARGV.size>0
   # Read the CSV file for the first coefficients
   CSV.foreach(ARGV[0],col_sep: ' ') do |row|
     index = row[0].to_i
     coefficient = row[1].to_i
     max_index = index
     pn_1,pn = pn, pn*coefficient +pn_1
     qn_1,qn = qn, qn*coefficient +qn_1
     if index % 10000 == 0
         STDERR.puts "Finished coefficient #{index}"
     end
     puts "#{index} #{coefficient}"
   end
   # p(x)=(x*p_n + p_(n-1))^3 - 2*(x*q_n+q_(n-1))^3
   a = pn*pn*pn - 2*qn*qn*qn
   b = 3*(pn*pn*pn_1 - 2*qn*qn*qn_1)
   c = 3*(pn*pn_1*pn_1 - 2*qn*qn_1*qn_1)
   d = pn_1*pn_1*pn_1 - 2*qn_1*qn_1*qn_1
   if (a>0)
       p = [a,b,c,d]
   else
       p = [-a,-b,-c,-d]
   end
end

# Find n so that 2^n so that p(2^n)<0<p(2^{n+1})
# Returns pair n,p(2^n)
def next_value_logarithmic(poly)
   bits = 0
   while (value = evaluate_bits(bits+1,poly))<0 do
      bits = bits+1
      last_value = value
   end
   if bits == 0
     last_value = poly.inject(0){|sum,x| sum + x }
   end
   [bits,last_value]
end

# Given cubic p(x) and n and p(2^n), compute polynomial q(x)=p(2^n+x)
# p(2^n+x)=p(2^n)+xp'(2^n)+x^2p''(2^{n})/2+x^3p'''(2^{n})/6
# No multiplication, only bit shifts and addition;
# Total of eight operations
def shift_poly_bits(n,p_2ton,poly)
   a,b,c,d= poly
   d1 = p_2ton
   # Precompute 2^n*3*a
   a3 = (a+(a<<1))<<n
   # c1=3a*2^{2n}+2b*2^n + c
   c1 = (a3<<n) + (b<<(n+1)) + c
   # b1 = 3a2^n + b
   b1 = a3 + b
   [a,b1,c1,d1]
end

# Find integer m\geq 1 so that p(m)<0<p(m+1)
def next_value(poly)
   # Find the smallest 2^n so that p(2^n)<0<p(2^(n+1))
   bits,value = next_value_logarithmic(poly)
   # At all points, shifted(x)=poly(x+minimum)
   shifted_poly = shift_poly_bits(bits,value,poly)
   minimum = 1<<bits
   # Do a binary search
   while bits>0 do
       bits = bits-1
       midval = evaluate_bits(bits,shifted_poly)
       if midval < 0 then
           shifted_poly = shift_poly_bits(bits,midval,shifted_poly)
           minimum = minimum + (1<<bits)
       end
   end
   # returns value c wher poly(c)<0<poly(c+1)
   # and the polynomial poly(c+x)
   [minimum,shifted_poly]
end

((max_index+1)..10000000).each do |i|
   c,p2 = next_value(p)
   # p2(x) is p(c+x)
   p = p2.reverse.map {|v| -v }
   puts "#{i} #{c}"
   if i%10000== 0 then
       max = p.map { |c| c.abs() }.max
       STDERR.puts "Coefficient #{i} max bits=#{max.bit_length}"
   end
end

Closing in on n=10,000,000,  the program is dealing with cubic polynomials with coefficients of about 16,500,000 bits each. In base 10, these numbers would have more than 5,000,000 digits. My laptop takes about 6.7 minutes to compute each additional 10,000 coefficients, or about 25 additional coefficients per second.

The continued fraction of $\sqrt[3]{2}$ are conjectured to never repeat themeselves (to not be periodic).  That's quite a statement.  This is discussed in the paper by Lang and Trotter Continued Fractions of Some Algebraic Numbers. This is also sequence A002945 in the OEIS.

Here's the first 100 continued fraction digits:


Discrete Mathematics -Propositional Logic How to get

B↔C from B∧C . Using derivation method i.e Step by Step .

My attempt :
            i) B∧C Rule P
            ii)B Rule T (i) Simplification B∧C ⇒ B
            iii)C Rule T (i) Simplification B∧C ⇒ C

I'm stuck here ..

HINT (first 3 lines):

Suppose $B\land C$
Suppose $B$
$C$  (from 1)

I use Lukasiewicz/Polish notation and condensed detachment.  The axioms I will use are:

CpCqp.
CKpqp.
CKpqq.
CCpqCCqpEpq.

Here we go:

assumption 5 Kbc.
D2.5       6 b.
D1.6       7 Cqb.
D3.5       8 c.
D1.8       9 Cqc.
D4.9      10 CCcqEqc.
D10.7     11 Ebc.


Find the roots of a complex square polynomial I'm stuck with this exercise and would welcome some help. I've been asked to:

  Find the roots in $\mathbb{C}$ and factorize in $\mathbb{C[x]}$ the polynomial $x^2 + (3+2i)x + 5 + i$

I tried many things. The first attempt was to see if I could factorize it just by looking at it. Then, I tried the formula $x=\frac{-b \pm w}{2a}$, where $w^2= b^2-4ac$

I did that and got $w^2= 8i + 15$ and then wrote $w= c+di$, since it's a complex number, and wrote a system of equations:

$a^2-b^2= -15$

$2ab= 8$

$a^2 + b^2= 289$

With that system I get that $a=\pm\sqrt{137}$, with that information I get $b$ and then go back to my previous equation with $x$, but that is already far from the answer, that is: $x_1=-1+i$, $x_2=-2-3i$

There's something obviously wrong in my reasoning. Could someone help me spot it?
Thanks a lot.

You should have got
$$
b^2-4ac=(3+2i)^2-4(5+i)=9+12i-4-20-4i=8i-15
$$
Now, setting $8i-15=(r+is)^2$ you get
$$
\begin{cases}
r^2-s^2=-15 \\[4px]
2rs=8 \\[4px]
r^2+s^2=\sqrt{8^2+15^2}=\sqrt{289}=17
\end{cases}
$$
Thus $2r^2=2$ and $r=\pm1$, together with $s=4/r$, so the square roots are $w_1=1+4i$ and $w_2=-1-4i$.

You can compute the roots as
$$
\frac{-b+w_1}{2}
\qquad\qquad
\frac{-b+w_2}{2}
$$
Notice $-b$, not $-b^2$.

Just use the standard formula for solving a quadratic equation.

It works for complex coefficients the same as real.


Show that $3^{2008}$ + $4^{2009}$ can be written as product of two positive integers each of which is larger than $2009^{182}$. I came across this problem and i could not solve it despite many tries.
Show that $3^{2008}$ + $4^{2009}$ can be written as product of two positive integers each of which is larger than $2009^{182}$.
I think the question can be solved more easily..Can anybody suggest any other method.

We use the standard factorisation:

x4 + 4y4 =
(x2 + 2xy + 2y2)( x2 - 2xy + 2y2).

We observe that for any integers x, y,

x2 + 2xy + 2y2 = (x + y)2 + y2 ≥ y2,

x2 - 2xy + 2y2 = (x - y)2 + y2 ≥ y2,


Compute $\mathbb{E}[\text{max}(x,0) \text{max}(y,0)]$ where $(x,y)$ is jointly gaussian with given covariance and nonzero mean Is there a closed form expression for the expectation of $g(x,y) = \text{max}(x,0) \text{max}(y,0)$ where $(x,y)$ is jointly gaussian with the bivariate gaussian distribution, when the mean is not zero and the covariance is non singular?

I tried to apply Price's theorem that can be written in the following form:

$$\frac{\partial^2 \mathbb{E}[g(x,y)]}{\partial \rho^2} = \mathbb{E} \left[ \frac{\partial^4 g(x,y)}{\partial x^2 \partial y^2} \right] = \mathbb{E}[\delta(x) \delta(y)] = \frac{1}{2 \pi \sigma_x \sigma_y\sqrt{1 - \rho^2}} \text{exp}\left(-\frac{1}{2(1 - \rho^2)} \left(\frac{\mu_x^2}{\sigma_x^2} +  \frac{\mu_y^2}{\sigma_y^2} - \frac{2\rho \mu_x\mu_y}{\sigma_x \sigma_y} \right)\right)$$

Unfortunately, integrating the last equality with respect to the correlation coefficient doesn't seem to have a closed form in terms of known functions. I have posted the integral question earlier before here.

Moreover, it doesn't seem to be possible to find the expectation directly as:

$$\int_0^{\infty} \int_0^{\infty} xy f(x,y) dx dy$$

EDIT: Price's theorem can nicely handle the zero mean case with an elegant solution.

Since a claim has been made that the general case  is somehow beyond reach I decided to post it below. I am not going to present all the details of the computation because it took me half of a day to carry out it in Mathematica and I have not saved the intermediate steps. Yet let me assure you that I only used integration by parts and definitions of special functions. In some cases certain integrals were not reducible to known special functions so I just left those terms aside. Later it appeared that even those hard terms are all expressible through a generalized Owen's T function Generalized Owen's T function which is defined as a joint Gaussian probability in the following way:
\begin{equation}
T(h,a,b):= {\bf P}\left(X>h \quad \wedge \quad a X+b > Y > 0 \left. \right| X = N(0,1) , Y=N(0,1)   \right)
\end{equation}

Back to our question. Clearly the expectation in question equals:
\begin{eqnarray}
&&{\mathbf E}\left[\mbox{max}(X,0)\mbox{max}(Y,0)\right]=
\int\limits_{-\frac{\mu_1}{s_1}}^\infty
\int\limits_{-\frac{\mu_2}{s_2}}^\infty
\left(\mu_1+s_1 \xi\right)\left(\mu_2+s_2 \eta\right) \cdot \rho(\xi,\eta) d\xi d\eta
\end{eqnarray}
where
\begin{equation}
\rho(\xi,\eta) := \frac{1}{2\pi \sqrt{1-\rho^2}} \exp\left[
-\frac{1}{2} \frac{1}{1-\rho^2}
\left( \xi^2+\eta^2 - 2 \rho \xi \eta\right)
\right]
\end{equation}
is the joint pdf of a bivariate Gaussian distribution with means zero and variances one.
Now the onle thing we need to do is to compute four two dimensional integrals being expectations of $1$,$\xi$,$\eta$ and of $\xi \cdot \eta$. Here I only state the result and then as usual verify it numerically. We have:
\begin{eqnarray}
{\mathbf E}\left[\mbox{max}(X,0)\mbox{max}(Y,0)\right]=
\mu_1 \mu_2 I_{0,0} + \mu_1 s_2 I_{0,1} + \mu_2 s_1 I_{1,0} + s_1 s_2 I_{1,1}
\end{eqnarray}
where
\begin{eqnarray}
&&I_{0,0}:=\int\limits_{-\frac{\mu_1}{s_1}}^\infty
\int\limits_{-\frac{\mu_2}{s_2}}^\infty 1 \cdot \rho(\xi,\eta) d\xi d\eta=\\
&&T\left(-\frac{\mu_2}{s_2},\frac{\rho }{\sqrt{1-\rho ^2}},\frac{\mu_1}{\sqrt{1-\rho ^2} s_1}\right)+\frac{1}{4} \left(\text{erf}\left(\frac{\mu_2}{\sqrt{2} s_2}\right)+1\right)\\
&&I_{0,1}:=\int\limits_{-\frac{\mu_1}{s_1}}^\infty
\int\limits_{-\frac{\mu_2}{s_2}}^\infty \eta \cdot \rho(\xi,\eta) d\xi d\eta=\\
&&\frac{e^{-\frac{\mu_2^2}{2 s_2^2}} \left(\text{erf}\left(\frac{\mu_1 s_2-\mu_2 \rho  s_1}{\sqrt{2-2 \rho ^2} s_1 s_2}\right)+1\right)+\rho  e^{-\frac{\mu_1^2}{2
   s_1^2}} \text{erfc}\left(\frac{\mu_1 \rho  s_2-\mu_2 s_1}{\sqrt{2-2 \rho ^2} s_1 s_2}\right)}{2 \sqrt{2 \pi }}\\
&&I_{1,0}:=\int\limits_{-\frac{\mu_1}{s_1}}^\infty
\int\limits_{-\frac{\mu_2}{s_2}}^\infty \xi \cdot \rho(\xi,\eta) d\xi d\eta=\\
&&\frac{e^{-\frac{\mu_1^2}{2 s_1^2}} \left(\text{erf}\left(\frac{\mu_2 s_1-\mu_1 \rho  s_2}{\sqrt{2-2 \rho ^2} s_1 s_2}\right)+1\right)+\rho  e^{-\frac{\mu_2^2}{2
   s_2^2}} \text{erfc}\left(\frac{\mu_2 \rho  s_1-\mu_1 s_2}{\sqrt{2-2 \rho ^2} s_1 s_2}\right)}{2 \sqrt{2 \pi }}\\
&&I_{1,1}:=\int\limits_{-\frac{\mu_1}{s_1}}^\infty
\int\limits_{-\frac{\mu_2}{s_2}}^\infty \xi \eta \cdot \rho(\xi,\eta) d\xi d\eta=\\
&&-\rho  \cdot T\left(-\frac{\mu_2}{s_2},-\frac{\rho }{\sqrt{1-\rho ^2}},-\frac{\mu_1}{\sqrt{1-\rho ^2} s_1}\right)-\frac{\mu_1 \rho  e^{-\frac{\mu_1^2}{2 s_1^2}}
   \left(\text{erf}\left(\frac{\mu_2 s_1-\mu_1 \rho  s_2}{\sqrt{2-2 \rho ^2} s_1 s_2}\right)+1\right)}{2 \sqrt{2 \pi } s_1}+\frac{1}{4} \rho
   \left(\text{erf}\left(\frac{\mu_2}{\sqrt{2} s_2}\right)+1\right)-\frac{\mu_2 \rho  e^{-\frac{\mu_2^2}{2 s_2^2}} \text{erfc}\left(\frac{\mu_2 \rho  s_1-\mu_1 s_2}{\sqrt{2-2
   \rho ^2} s_1 s_2}\right)}{2 \sqrt{2 \pi } s_2}+\frac{\sqrt{1-\rho ^2} \exp \left(\frac{\mu_1^2 s_2^2-2 \mu_1 \mu_2 \rho  s_1 s_2+\mu_2^2 s_1^2}{2 \left(\rho
   ^2-1\right) s_1^2 s_2^2}\right)}{2 \pi }
\end{eqnarray}

rho =.; mu1 =.; mu2 =.; s1 =.; s2 =.;
myrho[x_, y_] :=
  1/(2 Pi Sqrt[1 - rho^2]) Exp[-1/
      2 1/(1 - rho^2) (x^2 + y^2 - 2 rho x y)];
T[h_, a_, b_] :=
  NIntegrate[(E^(-(b^2/2) - xi b h - 1/2 (1 + xi^2) h^2)) /(
     2 (1 + xi^2) \[Pi]) -
     b  /(2 Sqrt[2] Sqrt[ \[Pi]]) (
      xi  Erfc[(h + xi (b + xi h))/(Sqrt[2] Sqrt[1 + xi^2])])/ ((1 +
        xi^2)^(3/2)) E^(-(b^2/(2 + 2 xi^2))), {xi, 0, a},
    WorkingPrecision -> 20] + Erfc[h/Sqrt[2]] Erf[b/Sqrt[2]] 1/4;

{mu1, mu2, s1, s2} = RandomReal[{0, 1}, 4, WorkingPrecision -> 50];
rho = RandomReal[{0, Sqrt[s1 s2]}, WorkingPrecision -> 50];

l1 = NIntegrate[{1, y, x, x y} myrho[x, y], {x, -mu1/s1,
    Infinity}, {y, -mu2/s2, Infinity}];

l2 = {1/4 (1 + Erf[mu2/(Sqrt[2] s2)]) +
    T[-(mu2/s2), rho/Sqrt[1 - rho^2], mu1/(Sqrt[1 - rho^2] s1)],
   1/(2 Sqrt[
     2 \[Pi]]) (E^(-(mu2^2/(
        2 s2^2))) (1 +
         Erf[(-mu2 rho s1 + mu1 s2)/(Sqrt[2 - 2 rho^2] s1 s2)]) +
      E^(-(mu1^2/(2 s1^2)))
        rho Erfc[ (-mu2 s1 + mu1 rho s2)/(Sqrt[2 - 2 rho^2] s1 s2)]),
   1/(2 Sqrt[
     2 \[Pi]]) (E^(-(mu1^2/(
        2 s1^2))) (1 +
         Erf[(-mu1 rho s2 + mu2 s1)/(Sqrt[2 - 2 rho^2] s1 s2)]) +
      E^(-(mu2^2/(2 s2^2)))
        rho Erfc[ (-mu1 s2 + mu2 rho s1)/(Sqrt[2 - 2 rho^2] s1 s2)]),
    Sqrt[1 - rho^2]/(2 \[Pi]) E^((
     mu2^2 s1^2 - 2 mu1 mu2 rho s1 s2 + mu1^2 s2^2)/(
     2 (-1 + rho^2) s1^2 s2^2)) +
    1/4 rho (1 + Erf[mu2/(Sqrt[2] s2)]) - ( mu1 rho)/(
     2 Sqrt[2 \[Pi]]
       s1) (1 +
       Erf[(mu2 s1 - mu1 rho s2)/(Sqrt[2 - 2 rho^2] s1 s2)]) E^(-(
      mu1^2/(2 s1^2))) - ( mu2 rho )/(2 Sqrt[2 \[Pi]] s2)
      Erfc[(mu2 rho s1 - mu1 s2)/(Sqrt[2 - 2 rho^2] s1 s2)] E^(-(
      mu2^2/(2 s2^2))) -
    rho T[-(mu2/s2), -(rho/Sqrt[1 - rho^2]), -(mu1/(
       Sqrt[1 - rho^2] s1))]};
MatrixForm[{l1, l2}]

It would be nice to do some additional sanity checks like finding some limiting cases. I will attempt to do it asap.

Update: Let us check the case $\mu_1=\mu_2=0$. In here we have:
\begin{eqnarray}
{\mathbf E}\left[\mbox{max}(X,0)\mbox{max}(Y,0)\right] &=& s_1 s_2 I_{1,1}\\
&=&s_1 s_2 \left( -\rho \cdot T\left(0,-\frac{\rho}{\sqrt{1-\rho^2}},0\right) + \frac{\rho}{4} + \frac{\sqrt{1-\rho^2}}{2\pi}\right)\\
&=& s_1 s_2 \left(
\frac{\rho}{2\pi} \mbox{arctan}(\frac{\rho}{\sqrt{1-\rho^2}}) +
\frac{\rho}{4} + \frac{\sqrt{1-\rho^2}}{2\pi}\right)
\end{eqnarray}
as it should be.

The calculation in question is pretty straightforward and can be done using elementary methods. Assume for simplicity that the variables have variance one and mean zero. Then the joint pdf reads:
\begin{eqnarray}
\rho(x,y) = \frac{1}{2\pi \sqrt{1-\rho^2}} \exp\left[ -\frac{1}{2} \frac{1}{(1-\rho^2)} (x^2+y^2-2 \rho x y )\right]
\end{eqnarray}
We integrate over $x$ first.
\begin{eqnarray}
I(y):=\int\limits_0^\infty x \rho(x,y) dx &=& \frac{1}{2\pi \sqrt{1-\rho^2}} \int\limits_0^\infty \exp\left[-\frac{1}{2} \frac{1}{(1-\rho^2)} ((x-\rho y)^2 + y^2(1-\rho^2))\right]\\
&=&\frac{1}{2\pi \sqrt{1-\rho^2}}e^{-\frac{1}{2} y^2} \int\limits_{-\rho y}^\infty (x+ \rho y) e^{-\frac{1}{2} \frac{1}{1-\rho^2} x^2} dx \\
&=& \frac{1}{2\pi} \sqrt{1-\rho^2} e^{-\frac{1}{2} y^2 \frac{1}{1-\rho^2}} + \frac{\rho  e^{-\frac{y^2}{2}} y}{2 \sqrt{2 \pi }} + \frac{\rho  e^{-\frac{y^2}{2}} y \text{erf}\left(\frac{\rho  y}{\sqrt{2-2 \rho ^2}}\right)}{2 \sqrt{2 \pi }}
\end{eqnarray}
Now we multiply the result by $y$ and integrate the whole thing over $y\in(0,\infty)$. Even at the first glance it is clear that the integrals from the first two terms on the right hand side are doable it is only the last integral that might cause difficulties. However even that integral is doable and it reads:
\begin{equation}
\int\limits_0^\infty y^2 e^{-\frac{1}{2}y^2} Erf[a y] dy = \frac{1}{\sqrt{\pi}} \left[ \frac{2 a}{1+2 a^2} + \sqrt{2} \arctan(\sqrt{2} a)\right]
\end{equation}
The result was derived by differentiating with respect to the parameter $a$.
The final result is as follows:
\begin{eqnarray}
\left< max(x,0) max(y,0) \right> = \frac{\left(1-\rho ^2\right)^{3/2}}{2 \pi } + \frac{\rho }{4} + \frac{\rho  \left(\sqrt{1-\rho ^2} \rho +\tan ^{-1}\left(\sqrt{\frac{\rho ^2}{1-\rho ^2}}\right)\right)}{2 \pi }
\end{eqnarray}

rho =.;
myrho[x_, y_] :=
  1/(2 Pi Sqrt[1 - rho^2]) Exp[-1/
      2 1/(1 - rho^2) (x^2 + y^2 - 2 rho x y)];
rho = RandomReal[{0, 1}, WorkingPrecision -> 50];
NIntegrate[x y myrho[x, y], {x, 0, Infinity}, {y, 0, Infinity},
 WorkingPrecision -> 20]
(1 - rho^2)^(3/2)/(2 \[Pi]) + rho/4 + (
 rho (rho Sqrt[1 - rho^2] + ArcTan[Sqrt[rho^2/(1 - rho^2)]]))/(
 2 \[Pi])


Analytic continuation of the logarithm This is an example from Serge Lang Complex Analysis book, it says

  Let us start with the function $log(z)$ defined by the usual power
  series on the disc $D_0$ which is centered at $1$ and has radius $<1$
  but $>0$. Let the path be the circle of radius $1$ oriented
  counterclockwise as usual. If we continue $log(z)$ along this path,
  and let $(g,D)$ be its continuation, then near the point $1$ it is
  easy to show that \begin{equation} g(z)=log(z)+2\pi i. \end{equation}
  Thus $g$ differs from $f_0$ by a constant, and is not equal to $f_0$
  near $z_0=1$.

I can't get the expression of $g$ that Lang says "It's easy to show". I tried constructing explicitly a sequence $(f_0,D_0),...,(g,D_0)$ which is the analytic continuation of $(f_0,D_0)$ but I had no success.

Essentially what is happening is that you have ended up at a different point on the associated Riemann Surface, which can be visualized by a spiral.

In more concrete terms, consider
$$ \log(e^{i\alpha})\quad \text{ for } \alpha \in [0,2\pi)$$
Which is the values we get for the logarithm as we travel around the unit circle. If we decide to follow the natural convention and have $\log(e^{i\cdot0}) = \log(1) =0,$ then we can start considering the rotated values, and note that $\log(e^{i(2\pi-\epsilon)}) = i(2\pi-\epsilon) \to 2\pi i$ as we let $\epsilon \to 0.$