Prove that $n$ divides $\phi(a^n -1)$ where $a, n$ are positive integer without using concepts of abstract algebra I need to show that $n$ divides $\phi(a^n -1)$ where $a, n$ are positive integer without using concepts of abstract algebra

I know that $$a^n\equiv 1\pmod {a^n-1}$$

How do I proceed from there?

In general (for $\gcd (a,m)=1)$, if $a_k≡1 \mod m $, then $ord_m(a) ~∣~k.$ In particular, since $a^{ϕ(m)}≡1(\mod m)$, we have $ord_m(a)∣ϕ(m)$. We have $ord_{(a^n−1)}a=n.$

Proof: Let there exist $d< n$ such that $ a^d ≡ 1 \mod(a^n-1) $ .

=> $ (a^n-1)~ |~ (a^d -1)$ which is not possible since $d < n $.

=> $ord _{a^n−1} =n.$

=> $n~|~ \phi(a^n-1)$

The order of the multiplicative group mod $a^n-1$ is $\phi(a^n-1),$ which (together with your start) proves your statement, but I assume that since the word "group" is used, this is not the proof asked for.


Determine the Sylow $p$-subgroups of $SL(2, \mathbb{F}_5)$ I'm working on the following problem

  Let $H := SL_2(\mathbb{F}_5)$.
  Determine, for every primefactor $p$ of $|H|$, a Sylow $p$-subgroup.

What I have so far:

Let $G := GL_2(\mathbb{F}_5)$ and $G' := \mathbb{F}_5^\times$, then is $f: G \to G'$ where $A \mapsto \det(A)$ a group homomorphism, since for $A, B \in G$
$$f(A) \cdot f(B) = \det(A) \cdot \det(B) = \det(A \cdot B) = f(A \cdot B)$$
As $f$ is a homomorphism we know that $\ker(f) \triangleleft G$, where we see that
$$\ker(f) = \{A \in G \mid f(A) = 1\} = H \triangleleft G$$
hence $G_{/H} \to G'$ is an isomorphism according to the first isomorphism theorem.
This means that $G_{/H} \cong G'$ and since $|G'| = 4$ follows $|G_{/H}| = [G:H] = 4$.
So since
$$|G| = (5^2-1)(5^2-5) = 24 \cdot 20 = 480$$
we have
$$|G| = [G:H] \cdot |H| \implies |H| = \frac{|G|}{[G:H]} = \frac{480}{4} = 120$$
where we see that $|H| = 120 = 2^3 \cdot 3 \cdot 5$, so according to the first Sylow theorem, there are subgroups $K < H$ or order $2^3$, $3$ and $5$.

So far so good, but now I struggle to determine the Sylow $p$-subgroups for $p = 2, 3, 5$.
I know that I can use the third Sylow theorem to determine how many subgroups there must be, but I don't see how this would help me find a concrete subgroup.

Is there any systematic approach to find a concrete Sylow $p$-subgroup or is it a brute-force exercise which involves detailed knowledge of the properties of $H$?

As was already pointed out, the element of order $5$ is canonical. For the element of order $3$ I just tried some stuff out and noticed that
$ \left(
\begin{array}
.4 & 4 \\
1 & 0 \\
\end{array}
\right)
$ works.

The most interesting is the group of order $8$. You should keep in mind that there are few possible isomorphism classes: a group of order $8$ is either abelian, or isomorphic to one of $D_8$ (the dihedral group of order $8$) or $Q_8$ (the quaternion group). To start, you know that $\mathbb{F}_5^\times$ is cyclic of order $4$, generated by $2$ or $3 = -2$. This gives you the following elements of order $4$:
$ I = \left(
\begin{array}
.2 & 0 \\
0 & 3 \\
\end{array}
\right),
J = \left(
\begin{array}
 .0 & 2 \\
2 & 0 \\
\end{array}
\right)
$ and there is always the usual element of order $4$ (using that $4 = -1$):
$K = \left(
\begin{array}
.0 & 4 \\
1 & 0 \\
\end{array}
\right)
$. So now you have three elements of order $4$ all squaring to $-Id$. This should be reminiscent of the quaternion group. And indeed, you can check that the group generated by these three matrices is indeed $\langle I, J, K | I^2 = J^2 = K^2 = IJK = -1 \rangle = \{ \pm Id, \pm I, \pm J, \pm K \} \cong Q_8.$


What does it mean to not divide any term I am trying to solve the below problem(Please don't solve it)

  The sequence 1, 1, 1, 3, 5, 9, 17, 31, 57, 105, 193, 355, 653, 1201
  ... is defined by T1 = T2 = T3 = 1 and Tn = Tn-1 + Tn-2 + Tn-3.

  It can be shown that 27 does not divide any terms of this sequence. In
  fact, 27 is the first odd number with this property.

  Find the 124th odd number that does not divide any terms of the above
  sequence.

What does it mean does not divide any terms? Can you explain it to me, please?

"$27$ does not divide any terms" means it does not divide $31$ or $57$ or $105$ or ... and so on for all the numbers in that sequence. (I started that at $31$ because it clearly can't divide anything smaller than itself.)

That mean find the 124-th number (equal $a(124)$ for example) verify : $\forall n \in \mathbb{N}^{*} \quad a(124) \nmid T_n$


When a quadratic involving three primes is a perfect square How do we find all primes $p,q,r$ such that $p^2+q^2+rpq$ is a perfect square ?

with $r=7$ and $p=q$ we have the expression a perfect square

Generally before searching primes need to understand what formula solutions of this equation.
Generally speaking, this equation has a lot of formulas for the solution. Because it is symmetrical.

Write the formula can someone come in handy. the equation:

$Y^2+aXY+X^2=Z^2$

Has a solution:

$X=as^2-2ps$

$Y=p^2-s^2$

$Z=p^2-aps+s^2$

more:

$X=(4a+3a^2)s^2-2(2+a)ps-p^2$

$Y=(a^3-8a-8)s^2+2(a^2-2)ps+ap^2$

$Z=(2a^3+a^2-8a-8)s^2+2(a^2-2)ps-p^2$

more:

$X=(a+4)p^2-2ps$

$Y=3p^2-4ps+s^2$

$Z=(2a+5)p^2-(a+4)ps+s^2$

more:

$X=8s^2-4ps$

$Y=p^2-(4-2a)ps+a(a-4)s^2$

$Z=-p^2+4ps+(a^2-8)s^2$

For the particular case: $Y^2+XY+X^2=Z^2$  You can draw more formulas.

$X=3s^2+2ps$

$Y=p^2+2ps$

$Z=p^2+3ps+3s^2$

more:

$X=3s^2+2ps-p^2$

$Y=p^2+2ps-3s^2$

$Z=p^2+3s^2$

In the equation:  $X^2+aXY+bY^2=Z^2$ there is always a solution and one of them is quite simple.

$X=s^2-bp^2$

$Y=ap^2+2ps$

$Z=bp^2+aps+s^2$

$p,s$ - integers asked us.


$\langle x_n\rangle $ be a sequence such that $x_{(n+2)}=\frac{(x_{n+1}+ x_{n})}{2} $for all $n \in \mathbb{N}$, prove that $x_n$ is convergent If $\langle x_n\rangle $ is a sequence of positive real numbers such that $$x_{(n+2)}=\frac{(x_{n+1}+ x_{n})}{2}$$ for all $n \in \mathbb{N},\ $ let $x_1 <x_2$

then subsequence of odd terms is increasing and subsequence of even terms is  decreasing .But how to prove it mathematically? We have

$$x_{(n+2)}-x_n=\frac{(x_{n+1}- x_{n})}{2}= \frac{(x_{n}- x_{n-2})}{4}$$ How to proceed from here? Any hint please.

Suppose that $x_n <= x_2$ for all n less than or equal to k then $x_{k+1}=\frac{{x_k}+x_{k-1}}{2} <= \frac{2x_2}{2}=x_2$ so by induction sequence is bounded.Now suppose $x_{n-1} <= x_{n}$ for all n less than or equal to k,then $x_{k+1}=\frac{{x_k}+x_{k-1}}{2} => \frac{2x_k}{2}=x_k$ so by induction sequence is increasing.As it is bounded and increasing it is convergent.


Formal definition of the function $Sin$ Which is the formal definition of the function $Sin$, starting from axioms of real numbers ?

I never found it in any book.

Not its Taylor serie, that is based upon the intuitive definition of $Sin$; I want the formal definition of $Sin$ without confusing drawings of circles, angles and straight lines.

One formal definition is via complex analysis, as used, e.g., by metamath (df-sin):
$$
\sin z=\frac{e^{iz}-e^{-iz}}{2i}.
$$

This, in turn, rests on the definition of $\exp$ (df-ef):
$$
\exp z=\sum_{k=0}^\infty\frac{z^k}{k!}.
$$

For more details see the definition of sum (df-sum) which, in the case of infinite sums, is defined in terms of limits (df-clim).

Maybe I found an answer in this web page:

http://math2.org/math/algebra/functions/trig/sin-cos-definition.htm


Prove the divergence of the sequence $\left\{ \sin(n) \right\}_{n=1}^{\infty}$. I am looking for nice ways of proving the divergence of the sequence $\left\{x_n\right\}_{n=1}^{\infty}$ defined by
$$x_n=\sin{(n)}.$$
One (not so nice) way is to construct two subsequences: one where the indexes are picked such that they lie in the intervals
$$I_k=\left(\dfrac{\pi}{6}+2\pi(k-1),\dfrac{5\pi}{6}+2\pi(k-1)\right)$$
and one where they lie in
$$J_k=\left(\dfrac{7\pi}{6}+2\pi(k-1),\dfrac{11\pi}{6}+2\pi(k-1)\right).$$
If ${x_n}$ converges, then all its subsequences must converge to the same limit, but here the first subsequence has all its values in the interval $\left[ \frac{1}{2}, 1\right]$ while the second has its values in $\left[-1,-\frac{1}{2}\right]$. Contradiction.

Filling the details of this proof is rather tedious and not very elegant. Anyone has a better idea?

Equidistribution argument is very elegant, but it becomes a sledgehammer method when it comes to a question of mere convergence.

A simple argument can reveal the divergence of $(\sin n, n \geq 1)$. Let

$$ (x_n, y_n) = (\cos n, \sin n)$$

(I changed the notation for the sake of consistency of notation.) Then the application of the addition formula for trigonometric function or the rotation matrix gives

$$ \begin{align*}
x_{n+1} &= x_n \cos 1 - y_n \sin 1 \\
y_{n+1} &= x_n \sin 1 + y_n \cos 1.
\end{align*} $$

Now assume $(y_n)$ converges. Then since $\sin 1 \neq 0$, we have

$$ x_{n+1} = (y_{n+1} - y_n \cos 1) \cot 1 - y_n \sin 1$$

and hence $(x_n)$ also converges. Now let $(x_n, y_n) \to (\alpha, \beta)$. Then taking limit to the recursive formula we have

$$ \begin{align*}
\alpha &= \alpha \cos 1 - \beta \sin 1 \\
\beta &= \alpha \sin 1 + \beta \cos 1.
\end{align*} $$

Solving this system of linear equations give $(\alpha, \beta) = (0, 0)$. On the other hand, since

$$ x_n^2 + y_n^2 = 1, $$

we must have

$$ \alpha^2 + \beta^2 = 1,$$

a contradiction! Therefore $(y_n)$ cannot converge. ////

Of course, we can say much more on $(y_n)$. For example, we can show that the set of limit points of $(y_n)$ is exactly $[-1, 1]$, and the Cesaro mean of $(y_n)$ is 0 from Weyl's criterion.

(This proof is from the book Problems in Real Analysis Advanced Calculus on the Real Axis.)

A known resultis that if {x_n} and {y_n} are two sequences such that y_n is divergent and x_n\y_n converges to 0 then x_n also diverges.
Take x_n=sin (n) and y_n=n then the result follows.


Three vertices of a parallelogram have coordinates (-2,2),(1,6) and (4,3). Find all possible coordinates of the fourth vertex. Do I use the distance formula for the two points and use them to add to each other to get two parallel sides?

If three points are $P,Q,R$ then $R+(P-Q)$ gives a fourth vertex for a parallelogram. So pick the ordered pair $(P,Q)$ in all six ways, and that gives three parallelograms.

There are actually only three such paralellograms, some using my description being repeats (same vertices). So if $P,Q,R$ are the three given points (which are not collinear) the three parallelograms are those formed by using the given three vertices along with any one of the three choices $$P+Q-R,\ P+R-Q,\ Q+R-P$$ as the fourth vertex of the parallelogram.

Added note: In each case the subtracted point winds up being diagonally opposite the constructed point in that parallelogram. For example, if $X=P+Q-R,$ then also $X-P=Q-R$ as expected in a parallelogram labeled going around say counterclockwise in the order $X,P,R,Q.$ The equality of the vectors $X-P$ and $Q-R$ means they are parallel and point in the same direction, so that side $XP$ is parallel to side $QR.$ And also from $X=P+Q-R$ we get $X-Q=P-R$ showing the other pair $XQ,PR$ are parallel (the remaining pair of parallel sides of that parallelogram). Suggest draw a picture to see why for this case each of $XR$ and $PQ$ wind up as "diagonals" of the parallelogram formed.

Here, Three points are given,

$P(-2,\;2)$

$Q(1,\;6)$

$R(4,\;3)$

and take $S(x,\;y)$

Since in parallelogram

$\vec{PS}=\vec{QR}$,

Therefore,
$$(Sx−Px,Sy−Py)=(Rx−Qx,Ry−Qy)$$
$$⇒(x−(−2),y−2)=(4−1,3−6)$$
$$⇒x−(−2)=4-1\;\;\implies x=0$$
 $$⇒y−2=3−6\;\;\implies y=-1$$

Thus, coordinates of fourth vertex are $S(0,\;-1)$

I hope it' ll help.


$f(z) = |z|^2$ is complex differentiable only on $(0,0)$ Here's my proof attempt:

$$f(z) = |z|^2$$

When $z=x+yi$, we have:

$$f(z) = x^2+y^2 + 0i$$

Doing Cauchy Riemann

$$\frac{\partial u}{\partial x} = 2x = \frac{\partial v}{\partial y} = 0$$

only when $x=0$.
The second Cauchy Riemann condition is

$$\frac{\partial v}{\partial x} = 0 = -\frac{\partial u}{\partial y} = 2y$$

only when $y = 0$.

Therefore, the function can be differentiable only on $(0,0)$ but we haven't proven it yet, we must show that the partial derivatives are continuous on $(0,0)$, but they are. Therefore, $f(z)$ is only complex differentiable on $(0,0)$

Is everything all right?

It looks good except when you say "we must show that the partial derivatives are continuous". It would be enough to  prove that $f$ is differentiable as a real function of two variables at the origin.


Relation between diagonalizibilty of linear operator and its characteristic polynomial? In Hoffman-Kunze (Linear Algebra) there is this theorem :

The first two statements are equivalent, does this means if characteristic polynomial can be factored in the above way then T is diagonalizable?

But consider :

Characteristic Polynomial of A satisfies statement 2 but matrix A is not diagonalizable, So statement 1 and statement 2 are not equivalent. Am I missing something, Please Help.

First: The second statement has two clauses. It asserts that two things happen:

The characteristic polynomial splits; and
The dimension of the eigenspace of $\lambda$ equals the multiplicity of $\lambda$ as a root of the characteristic polynomial for every eigenvalue $\lambda$.

So the second statement is not just about being able to factor the characteristic polynomial.

Second: by my calculations, your matrix has characteristic polynomial (expanding along the second row)
$$\begin{align*}
\det(A-tI) &= \det\left(\begin{array}{ccc}
6-t & 3 & -8\\
0 & -2-t & 0\\
1 & 0 & -3-t
\end{array}\right)\\
&= (-2-t)\det\left(\begin{array}{cc}
6-t & -8\\
1 & -3-t
\end{array}\right)\\
&= (-2-t)\Bigl( (6-t)(-3-t) + 8\Bigr)\\
&= (-2-t)(t^2 -3t - 10) = -(t+2)^2(t-5).
\end{align*}$$

Let us write then $c_1=-2$, $d_1=2$, $c_2=5$, $d_2=1$, using the notation of the theorem.

It is true that it is not diagonalizable, because $W_1$, the nullspace of $A+2I$, has dimension $1$. But this means that it does not satisfy the second statement, because $\dim(W_1) = 1 \lt 2=d_1$. That is, $A$ does not satisfy the second statement in the theorem, because it has two satisfy two clauses and it only satisfies one of them.

$(\textrm{ii})$ In other words what it says is that $\textsf T$ is diagonalizable if and only if the dimension of each eigenspace associated with $c_i$ is equal to the multiplicity of the same eigenvalue $c_i$.

In your case, you forgot to check the dimension.


For what values of $p$ and $b$ is the vector $(b,8,b+7)$ a solution of this system? Tried to solve this - didn't work.

According to the exercise, the answer is $p = 1 \ , b = -5$. I am getting something way more different.

$$\begin{bmatrix}
1 & 3p & -1\\
-3 & p & -7
\end{bmatrix}\begin{bmatrix}
17 \\ 9
\end{bmatrix}$$

  For what values of $p$ and $b$ is the vector $(b,8,b+7)$ a solution of
  the system?

Well, I don't know - but it always helps to have the system in reduced row echelon form:

$$3r_1+r_2$$
$$\begin{bmatrix}
1 & 3p & -1\\
0 & 10p & -10
\end{bmatrix}\begin{bmatrix}
17 \\ 60
\end{bmatrix}$$

$$\frac{1}{10}r_2$$

$$3r_1+r_2$$
$$\begin{bmatrix}
1 & 3p & -1\\
0 & p & -1
\end{bmatrix}\begin{bmatrix}
17 \\ 6
\end{bmatrix}$$

Clearly, this system will have infinite solutions depending on one variable. It can be either the second or the third column:

If $p = 0$ then the second column is the one that is free. If $p \not = 0$ then the free one is the third.

For $p = 0$

$$\begin{bmatrix}
1 & 0 & -1\\
0 & 0 & -1
\end{bmatrix}\begin{bmatrix}
17 \\ 6
\end{bmatrix}$$

$$-r_2 \ \ , \ \ r_2 + r_1$$

$$\begin{bmatrix}
1 & 0 & 0\\
0 & 0 & 1
\end{bmatrix}\begin{bmatrix}
11 \\ -6
\end{bmatrix}$$

The solution would be

$$S = \{(x,y,z) = (11, t, -6)\} \ \forall t \in \mathbb{R}$$

$(b,8,b+7)$ cannot be a solution in this case, because if $p = 0$, we would need $b = 11$, but clearly $11 + 7 \not = -6$.

For $p \not = 0$

$$\begin{bmatrix}
1 & 3p & -1\\
0 & p & -1
\end{bmatrix}\begin{bmatrix}
17 \\ 6
\end{bmatrix}$$

$$\frac{1}{p}r_2$$

$$\begin{bmatrix}
1 & 3p & -1\\
0 & 1 & -1/p
\end{bmatrix}\begin{bmatrix}
17 \\ 6/p
\end{bmatrix}$$

$$(-3p)r_2+r_1$$

$$\begin{bmatrix}
1 & 0 & 2\\
0 & 1 & -1/p
\end{bmatrix}\begin{bmatrix}
-1 \\ 6/p
\end{bmatrix}$$

So

$$S = \{(x,y,z) = (-1-2t,(6+t)/p, t) \} \ \ \forall t \in \mathbb{R}$$

If you want $(b,8,b+7)$ to be a solution here we have to solve the system:

$$-1 - 2t = b\\
(6+t)/p = 8\\
t = b+7$$

The solutions:

$$b = 13/3\\
p = 13/6$$

So my answer is

$$b = 13/3\\
p = 13/6$$

Which is wrong.

What did I do wrong? What should I have done?

Is the system $Mv=u$ where $M$ is the matrix and $v$ is your solution vector (it isn't clear from what you have written, which presents as an undefined matrix product). i.e. the system is $b+24p-(b+7)=17$ and $-3b+8p-7(b+7)=9$

The first of these cancels nicely to give $p=1$ and the second then gives $b=-5$

(corrected for comment)

The solution of

$$\left\{ \begin{align} -1-2t & = b;\\
\frac{6+t}{p} & =8; \\
t & =b+7 \end{align} \right.$$

is

$$t=2,
b=-5,
p=1$$


Maximum timestep for RK4 I'm trying to find the maximum timestep that can be used when applying a RK4 numerical method to solve this system. I know how I would do this for single equation but have no idea how to solve it for a system of equations. Please can someone help?

$
\dfrac{d}{dt}
\begin{bmatrix}
y_1 \\
y_2
\end{bmatrix}
=
\begin{bmatrix}
y_2 \\
-1.16 \ \sin(y_1)
\end{bmatrix}
$

Edit:

In terms of a model equation in the form:

$
\dfrac{dy}{dt} = \lambda y
$

For the $4^{th}$ order Runge-Kutta method, the general value at a step $n+1$ is:
$
y_{n+1} = y_n \ \left[ 1 + (h \lambda) + \dfrac{1}{2} (h \lambda)^2 + \dfrac{1}{6} (h \lambda)^3 + \dfrac{1}{24} (h \lambda)^4 \right].
$

So for stability:

$
\left | 1 + (h \lambda) + \dfrac{1}{2} (h \lambda)^2 + \dfrac{1}{6} (h \lambda)^3 + \dfrac{1}{24} (h \lambda)^4 \right | \leq 1.
$

This is what I was saying, for a single equation, if I can rearrange my problem into the form of the model equation, I can just solve a 4th-order polynomial to determine my maximum stepsize because it must satisfy my above inequality.

But I'm confused as to how I would do it for the system of equations. The initial values of $y_1$ and $y_2$ are 2.7 and 0 respectively - at t = 0.

If we integrate your second equation only once, we get $\dfrac{dy_1}{dt} = 1.16cos(y_1) + C_1 t$. If I could work out this constant $C_1$ I would have something in the form of the model equation - as $C_1 t$ would dominate the oscillatory cosine term as $t$ gets large.

If we note that $\dfrac{dy_1}{dt} = y_2$:

$y_2 = 1.16cos(y_1) + C_1 t$

Still I'm no closer to finding $C_1$

As said in the comment (by user147263/Woodface), there is no real way to answer without further information. You could ask what the timestep is for an error of $10^{-6}$ or something. One also needs the integration interval.
Generally, as a fourth order method, the error is of the form
$$
Ch^4+D\mu/h
$$
where $\mu$ is the machine constant of the floating point type, $C,D$ are expressions in the ODE function and its derivatives. If the constants have a commensurable magnitude, the error expression has a minimum at $h=\sqrt[5]\mu\approx 10^{-3}$.

edit 2017: To incorporate some scale into that recommendation, chose $Lh=\sqrt[5]\mu\approx 10^{-3}$ for computations in double precision where $L$ is a Lipschitz constant, here $L=1.16$ (or its square root). Based on the shape of the stability region

one can postulate good behavior of the RK4 method for $Lh<2.5$, as the semi-circle of that radius is contained in the stability region. As the recommended step size from the error analysis is much smaller, you should get no stability problems from error magnification.

(added 2/2019) Testing the various recommendations: The described test problem is the second order physical pendulum equation
$$
\ddot y + 1.16\sin(y)=0, ~~ y(0)=2.7, ~~ y'(0)=0.
$$
translated into first order form.
The solution is about $p(t)=2.7\cos(0.6\,t)$ (closer is $2.9⋅\cos(0.5777⋅t)-0.2⋅\cos(3⋅5.777⋅t)$).
Insert this as exact solution into an MMS (method of manufactured solutions) formulation for the mapping $F[y]=\ddot y+1.16\sin(y)$, that is, $$F[y]=F[p] ~~ \text{  with } ~~ y(0)=p(0), ~~ \dot y(0)=\dot p(0) .$$
Trying out the solution and error against the exact solution for various step sizes gives the plots

The error relates to the second plot as $e(t)=c(t)e^{0.4t}h^4$ where $c(t)$ is the function depicted.
As expected from the stability analysis, for step sizes $h=1,..,2.5$ the numerical iteration stays somewhat in the region of the exact solution for some time, while the errors grow steadily. $h=2.7$ is visibly outside the stability region for the problem.
From the error analysis one would expect the optimal step size at $h=0.001$, however as one can see in the plot, only in the range $h=0.002,..,0.1$ do the errors follow the pattern predicted by the leading error term. For $h=0.5$ the higher order error terms have too much influence, for $h=0.001$ the error pattern is disturbed by the accumulation of floating point noise.

(added 1/15/2021) There is a simple modification of the integration loop that makes the addition of the update approximately doubly accurate. Initialize an update vector as zero and replace
        y[i+1,:]=y[i]+(k1+2*k2+2*k3+k4)/6;

with
        update += (k1+2*k2+2*k3+k4)/6;
        y[i+1,:]=y[i]+update
        update -= y[i+1]-y[i]

This means that update accumulates the floating point errors of the step updates and shifts them to the solution if they become large enough. This, surprisingly, not only improves the result for the very small step sizes, but also for the large step sizes. Above plots, the first with unchanged parameters, the second with smaller step sizes and the exponential factor reduced to $\exp(0.05·t)$ in $e(t)=c(t)e^{0.05·t}h^4$ look now like

While at $h=0.00025$ one can see the granularity of the floating point type, the values still follow the error profile of the step sizes from $h=0.01$ down to $h=0.0005$. This means that the error at the smallest step size is bounded by $0.07·(0.00025)^4·e^{0.05t}= 2.734375·10^{-16}·e^{0.05t}$. Smaller errors are almost not possible in the floating point precision.

For the 4th order Runge Kutta scheme, the good approximated timestep to use is usually equal to or less than 0.05.  Your problem is so simple that it can be easily solved in excel worksheet or it can also be easily solved with the 1st order Runge Kutta scheme.


Derivative of a linear transformation $f: \mathbb{R^n} \to \mathbb{R^m} $ In multivariable calculus we have that , $f: \mathbb{R^n} \to \mathbb{R^m}$ is differentiable at $a \in \mathbb{R^n}$, if there exists a linear transformation $\mu : \mathbb{R^n} \to \mathbb{R^m} $ such that $$\lim_{h \to 0} \frac{|f(a+h)-f(a)-\mu(h)|}{|h|} = 0$$

then $\mu$ is the derivative of $f$ at $a$ and we denote it by $Df(a)$. According to this definition it is easy to see that, if $f$ is a linear transformation, then $Df(a)=f$ for all $a \in \mathbb{R^n}$.

Having this definition in mind, I am confused by this simple example: Let $f: \mathbb{R} \to \mathbb{R}$ be defined by $f(x)=2x$, is a linear transformation,

I can't believe that  $Df(a)=f$, which sends any real $x$ number to $2x$

but clearly $Df(a)=2$, is the Jacobian $1 \times 1$ matrix, I kind of see that, it is just the same thing, this also sends any real $x$ number to $2x$.

What is happenning here ?!?!
Thanks for your help in advance

Think of derivative of $f$ at $x$ as a map that applies to vectors based at $x$. In one dimension we can say that $$Df(x)=(2)$$ a $1\times 1$ matrix which stretches these vectors  by the factor of $2$. The analog in two  dimensions is
$$Df(x) = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} $$ which does the same.

It is misleading to say that $Df(x)=f$. The original map $f$ operates on the points in $\mathbb R^n$, the derivative $Df(x)$ operators on the vectors based at $x$. We can write
$$Df(x)\,v = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}v $$
to indicate this. But
$$Df(x)\, x = \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix}x $$
is a notational confusion.

Same in one dimension: what we usually write as $f'(x)=2$ can be written as $Df(x)\,v = 2v $  for compatibility with multivariable notation.

Suppose $f:R^n\rightarrow R^m$ linear map

and $Df(x)=f$

Now $D^2f(x)=0$

Can you tell me where 0 should be?


Order of operations - why are they in the order they're in? I understand the order of operations, but why are they ordered the way they're ordered? Is there a particular reason why multiplication should have a higher precedence than subtraction, other than to prevent ambiguity?

Edit: I'm a curious software developer that's relatively lousy at math. A simple explanation that your grandma could understand would be very welcome. :-)

The precedence rules allow efficient representation of polynomial expressions in a concise normal expanded form. If you reversed the precedence of multiplication then the notation would be more efficient for factored polynomial expressions. But, generally, polynomial operations are much less efficient in factored form (compare, e.g. equality testing and addition).

The reason that polynomial rings are ubiquitous is their universality (they are free $\rm\:R$-algebras). So, for example, any polynomial identity that you prove true in $\rm\:\mathbb Z[x]\:$ will hold true over any ring, e.g. the binomial theorem.

To illustrate your example with substraction and multiplication, denote them as two variable functions: $S(a,b)=a-b,\ P(a,b)=a\cdot b$. Then to calculate $a-b\cdot c$(without any brackets), we would have $S(a,P(b,c))$. In order to calculate the result, we need to find $P(b,c)$ first.
There are many operations defined with the aid of the simpler ones: multiplication is defined using addition; powers are defined using multiplications. It is then natural to give priority to the elevated operations, since they depend on the simpler ones.
Think practical. In a warehouse there are 2000 bottles of Coca cola, in boxes of 20 pcs. Someone takes out 30 boxes. How many bottles are left? Answer: $2000-30\times 20$. If we calculate $2000-600=1400$ we get the practical result right(we go and count the remaining bottles to see that... ) . If we calculate substraction first we get $1970\times 30>2000$, and we see that something is wrong.

I think the third aspect is the most important, since mathematics is made to help modelling different things we encounter in real life. Not all maths is abstract. If you take the operations and change their priorities, you will get results which do not correspond with reality.


Express 99 2/3% as a fraction? No calculator My 9-year-old daughter is stuck on this question and normally I can help her, but I am also stuck on this! I have looked everywhere to find out how to do this but to no avail so any help/guidance is appreciated:

The possible answers are:

$1 \frac{29}{300}$

$\frac{269}{300}$

$9 \frac{29}{30}$

$\frac{299}{300}$

$1 \frac{299}{300}$

Whenever I try this I am doing $\frac{99.66} \times 100$ is $\frac{9966}{10000}$ and then simplify down, but I am not getting the answer. I get $\frac{4983}{5000}$? Then I can't simplify more — unless I am doing this totally wrong?

Hello All - just an update - I managed to teach this to her right now, and she cracked it pretty much the first time! Thanks sooooooo much for the answers and here is some of the working she did:

You can almost eyeball this, which is a useful technique for dealing with multiple choice quizzes against the clock.

If 100% = 1, then you're looking for a fraction slightly less than one. That lets you rule out all but two of the answers. How much less? A third of a percent. One percent is 1/100. Divide that by three to get 1/300. That points you at 299/300 being the answer.

Is it too hard to mentally calculate the numerator?

$99*3 + 2 =  300 - 3 + 2 = 299$


continuous but not absolutely continuous function How to show by definition (without the variation of function) that $f \colon [0,1] \to \mathbb{R}$
 $$ f(x) = \begin{cases} 0 \quad \mbox{if} \quad x=0  \\ x \cos \frac{\pi}{x}  \quad \mbox{if} \quad x \neq 0 \end{cases} $$
is not absolutely continuous?

Ok, another answer not using the variation... We have to show that for every $\delta > 0$ there exist mutually disjoint open intervals $\Delta_i = (a_i,b_i)$, $i=1,\ldots,n$, such that the sum of their lengths is $<\delta$ but $\sum_1^n|f(a_i) - f(b_i)|\ge 1$. So, let $\delta > 0$ be given. Choose $N\in\mathbb N$ odd (we need this later) such that $\frac{1}{N} < \delta$. Now, let
$$
\Delta_i = \left(\frac 1 {N(i+1)},\frac{1}{Ni}\right), \quad i=1,\ldots,n,
$$
where $n$ is such that $\frac 1 N\sum_{i=1}^n\frac 1 i\ge 1$. Now, we have that
$$
\sum_{i=1}^n\operatorname{Length}(\Delta_i) = \sum_{i=1}^n\left(\frac{1}{Ni} - \frac 1 {N(i+1)}\right) = \frac 1 N\left(1 - \frac 1 {n+1}\right) < \frac 1 N < \delta.
$$
But on the other hand,
\begin{align*}
\sum_{i=1}^n\left|f\left(\frac 1 {Ni}\right) - f\left(\frac 1 {N(i+1)}\right)\right|
&= \sum_{i=1}^n\left|\frac {\cos(Ni\pi)}{Ni} - \frac {\cos(N(i+1)\pi)}{N(i+1)}\right|\\
&= \sum_{i=1}^n\left|\frac {(-1)^{Ni}}{Ni} - \frac {(-1)^{N(i+1)}}{N(i+1)}\right|\\
&= \sum_{i=1}^n\left|\frac {1}{Ni} - \frac {(-1)^N}{N(i+1)}\right|\\
&= \sum_{i=1}^n\left(\frac {1}{Ni} + \frac {1}{N(i+1)}\right)\,\ge\,1.
\end{align*}
So, $f$ cannot be absolutely continuous.

An absolutely continuous function on a bounded interval is of bounded variation. The present function $f$, defined on $[0,1]$, is not of bounded variation (it oscillates too much) and is thus not absolutely continuous. To see this, for $n\in\mathbb N$ set $x_i := \frac 1 i$, $i=1,\ldots,n$. Then
$$
\sum_{i=1}^n|f(x_{i+1}) - f(x_i)| = \ldots\,\le\,\sum_{i=1}^n\frac 1 i,
$$
which tends to $\infty$ as $n\to\infty$.


Find Maximum Value of Expression I'm trying to learn about using the AM-GM inequality. I understand the basics of it so far, but I'm having a difficult time in terms of applying it to solve problems. I've encountered the problem below and just can't seem to find a way to solve it.

  Find the maximum value of the expression $$(1-x)(2-y)(3-z)(x+\frac{y}{2}+\frac{z}{3})$$
  given that $x<1$, $y<2$, $z<3$, and $x+\frac{y}{2}+\frac{z}{3}>0$.

So, how do you? It seems that expanding the expression above just makes it more complicated. Should I even try to use AM-GM for this?

Thanks in advance.

By AM-GM $$(1-x)(2-y)(3-z)\left(x+\frac{y}{2}+\frac{z}{3}\right)=6(1-x)\left(1-\frac{y}{2}\right)\left(1-\frac{z}{3}\right)\left(x+\frac{y}{2}+\frac{z}{3}\right)\leq$$
$$\leq6\left(\frac{1-x+1-\frac{y}{2}+1-\frac{z}{3}+x+\frac{y}{2}+\frac{z}{3}}{4}\right)^4=\frac{243}{128}.$$
The equality occurs for $$1-x=1-\frac{y}{2}=1-\frac{z}{3}=x+\frac{y}{2}+\frac{z}{3},$$ which says that we got a maximal value.


Find all holomorphic functions that satisfy a condition Find all holomorphic funtion $f:B(0,1)\mapsto B(1,4)$ s.t. $f(0)=3$ and $f(1/2)=1$

$B(a,r)$ is the open ball with centre a and radius r.

I think that maybe Schwarz lemma will help, but dont know how.
Thanks!

Try for example:
$$
f(z)=4\cdot\frac{2z-1}{z-2}+1
$$
which is also a bijection.


Three equations, three unknowns, and one constraint Suppose we have the following three equations:
$$
r_y = \frac{r_y}{2} + \frac{r_a}{2} \\
r_a = \frac{r_y}{2} + r_m \\
r_m = \frac{r_a}{2}
$$

We also have additional constraint for uniqueness:
$$
r_y + r_a + r_m = 1
$$

The solution is:
$$
r_y = r_a = \frac{2}{5}, \quad r_m = \frac{1}{5}
$$

I kindly as for any pointer how to solve such system of equations.

How to solve this can be a bit tricky,
I will suggest this approach. First, get rid of the fraction form. Instead have this form of equations.

your first equation is $r_{y} - r_{a} = 0$, then your second equation $-r_{y} + 2r_{a} + 2 r_{m} = 0$ and finally your third one $-2r_{a} + 2 r_{m} = 0$. Then put it in the form of a matrix. Then use gauss elimination process to solve. Hope that help.

Your first equation yields, after simplifying, $r_y=r_a$.  With your constraints, along with your identity for $r_m$ in the third equation yields
$$2r_a+\frac{r_a}{2}=1$$


How many pairs of dance partners can be selected from a group of $12$ women and $20$ men? How many pairs of dance partners can be selected from a group of $12$ women and $20$ men ?

Ans given : $P(20, 12)$

Shouldn't the answer be $20 × 12$ as the pair can be selected from any of the $12$ women and for each women there are $20$ men to choose for.

You first select 12 men from possible 20, that can be done in $\binom{20}{12}$ ways. Now these 12 men have to be paired with the 12 women. Each pairing is simply a bijective function from the set of 12 men to the set of 12 women. Number of such bijective mappings is $12!$. So in all
$$\binom{20}{12} \cdot 12!=P(20,12) \quad \text{ways}.$$

Let $W$ denote the set of women and $M$ denote the set of men. An arrangement in which each woman has exactly one male dance partner is the same thing as an injection $W \rightarrow M$. So compute:

$$|M^{\downarrow W}| = |M|^{\downarrow |W|} = \prod_{w = 0}^{|W| - 1}(|M|-w).$$

Notation. whenever $A$ and $B$ are sets, the notation $A^{\downarrow B}$ means the set of all injections $B \rightarrow A,$ and whenever $a$ and $b$ are natural numbers, $a^{\downarrow b}$ denotes the falling factorial. Now just plug your numbers in.


The Laplace transform of the first hitting time of Brownian motion Let $B_t$ be the standard Brownian motion process, $a > 0$, and let $H_a = \inf \{ t : B_t > a \}$ be a stopping time. I want to show that the Laplace transform of $H_a$ is
$$\mathbb{E}[\exp(-\lambda H_a)] = \exp (-\sqrt{2\lambda} H_a)$$
by considering the martingale $$M_t = \exp \left(\theta B_t -\frac{1}{2}\theta^2 t\right)$$

There's an obvious argument to follow here: assuming the optional stopping theorem applies, we have
$$1 = \mathbb{E}[M_{H_a}] = \mathbb{E} \left[ \exp \left(\theta a - \frac{1}{2}\theta^2 H_a\right) \right] = \exp(\sqrt{2\lambda} a) \mathbb{E} \left[ \exp(-\lambda H_a) \right]$$
where $\theta = \sqrt{2\lambda}$. This is exactly what we wished to show. However, as far as I can tell, the hypotheses of the optional stopping theorem are not satisfied here. Here is the statement I have:

  If $(X_n)$ is a martingale and $T$ is an a.s. bounded stopping time, then $\mathbb{E}[X_T] = \mathbb{E}[X_0]$.

I think not all is lost yet. $M_t > 0$ for all $t$, so the martingale convergence theorem applies, and $M_t \to M_\infty$ a.s. for some integrable random variable $M_\infty$. For each $t$, $H_a \wedge t = \min \{ H_a, t \}$ is a bounded stopping time, so certainly $\mathbb{E}[M_{H_a \wedge t}] = \mathbb{E}[M_0]$. But,
$$\mathbb{E}[M_{H_a \wedge t}] = \mathbb{E}[M_{H_a} \mathbf{1}_{\{H_a \le t\}}] + \mathbb{E}[M_t \mathbf{1}_{\{H_a > t\}}]$$
and clearly what one wants to do is to take $t \to \infty$ on both sides. But here's where I get stuck: I'm sure I need a convergence theorem here in order to conclude that the equation remains valid in the limit.

Now, $0 < M_{H_a} = \exp \left(\theta a - \frac{1}{2} \theta^2 H_a \right) \le \exp(\theta a)$, so the dominated convergence theorem applies, and so
$$\lim_{t \to \infty} \mathbb{E}[M_{H_a} \mathbf{1}_{\{H_a \le t\}}] = \mathbb{E}[M_{H_a} \mathbf{1}_{\{H_a < \infty\}}]$$
and I believe Fatou's lemma gives me that
$$\liminf_{t \to \infty} \mathbb{E}[M_t \mathbf{1}_{\{H_a > t\}}] \ge \mathbb{E}[M_{\infty} \mathbf{1}_{\{H_a = \infty\}}]$$
but I think what I need is the equality
$$\lim_{t \to \infty} \mathbb{E}[M_t \mathbf{1}_{\{H_a > t\}}] = \mathbb{E}[M_\infty \mathbf{1}_{\{H_a = \infty\}}]$$
and as far as I can tell neither the monotone convergence theorem nor the dominated convergence theorem applies here. Is there anything I can do to rescue this line of thought?

Use the fact that $0\leqslant M_t\mathbf{1}_{\{H_a>t\}}\leqslant\exp\left(\theta a−\frac12\theta^2t\right)$.

I don't think you need to worry about the case $H_a = \infty$ - for the simple reason that this is an event of probability zero.


In any direction of the sphere, What is the distance from a point to a circle? In any direction of the sphere, What is the distance from a point to a circle?Is there a vertical relationship between the planes of its circle?

There is a point $A$ and a circle $C$ on the sphere. Draw a big circle $D$ from point $A$. There are countless small circles $c$ that pass through point $A$ and tangent to $D$. Some of them will intersect with $C$. The intersection point is $B$. What is the shortest small arc from point $A$ to $B$? Its plane is perpendicular to the $C$ plane, regardless of the direction of $D$?

I guess they are always perpendicular to each other.

On the sphere, my work requires me to find the distance from the outer point of the circle to the circle, and the distance must be in a certain direction.

Conclusion:

Mr. Aretino has proved that when AB is the shortest, AB and C are not vertical. (2) What is the mathematical significance of this conclusion? Are there different shortest directions on the sphere? What is the relationship between this and the shortest distance from a point on the plane to a line in different directions?

@Aretino gave the following interesting answer, but he did not give the shortest AB plane and C plane is vertical, hope he or others can give this conclusion.

Pictorial shortest

Aretino: I thought it was quite evident that the plane of arc $AB$ is not perpendicular to the plane of circle $BE$. In any case I changed again my construction, to show the dihedral angle between those two planes.

Aretino: If circles $C$ and $D$ are perpendicular then of course arc $AB$ lies along circle $D$. You can experiment by yourself with the interactive construction to see if there are other cases.

Aretino：You didn't answer my question: If $D$ is parallel to $C$, is $AB$ the shortest when $AB$ is perpendicular to $C$?

Mr. @Aretino has proved that when AB is the shortest, AB and C are not vertical.


Integer solutions to $x^2+y^2=N$? This is not homework; but I am completely lost on how to go about handling this especially for large N

if you want integer solutions, then the primes of the form $4k+3$ dividing $N$ must occur to an even power.  To see this you must show that

i) the product of a sum of two squares is a sum of two squares

ii) an odd prime $p$ is the sum of two squares iff $p\equiv1\mod 4$

i) $(a^2+b^2)(c^2+d^2)=(ac-bd)^2+(ad+bc)^2$ (think of the norms of complex numbers)

ii)wikipedia entry with some proofs

Ok, here's my attempt at this. The wikipedia page got better results though :(

define f(x,y) = x^2+y^2
define g(n) = N
rewrite the equation to look like f(x,y) = g(n)
Now use a pullback from category theory. You go backwards in the functions.
It starts like a pair (k,k), where k=k.
then left side of the pair tries to find x and y, while right side tries to find n.
the left side is more complicated. First consider a+b expression. You can split number N to two parts so that a+b = N (this N comes from left side of pair). Any splitting is ok. Think of range [0..N] and split it to [0..a] and [a..N]. Choose freely both a and b, but so that a+b=N.
Now you have chosen the a and b. These are numbers representing the area of a square. Now to find x and y, you just move from the area of square to the length of the side of the square. $x=\sqrt a$, $y=\sqrt b$ This will result in x : R, y : R. These are not in $N$, unless you choose a and b in certain way. So the process of choosing a and b fails if length of square's side is not integers.
Problem with algorithms solving this is that there are many different ways to choose the splitting, especially when the numbers are large. And then later in the process, many of those possible solutions (a,b) will fail to materialize.
There is one additional trick available which lets us narrow the solutions. The right side g(n)=N, can be rewritten as if it were a constant. Split $g : N \rightarrow N$ to $! : N \rightarrow 1$ and $g_1 : 1 \rightarrow N, g_1=N$. Now this might seem like technicality, but it gives us one big advantage - the g can only give us single element of $N$. So all values of n are in the same area inside N in right side. If we consider tuples (x,y,n) which are the solutions to the equation, large number of them are the same, and we only need to consider (x,y) as a shorthand notation for (x,y,N). This is only necessary because we chose to use a function g originally and did not regognize it to be a constant.
Now there is also intersection operation on powersets involved. To find out which (x,y,n) tuples are solutions of the equation, the projections (x,y,n) |-> (x,y) and (x,y,n) |-> (n) need to be "inverted" using the pullback. Left and right side of the equation gave us "areas" of values of x, y and n which are involved, now combining those areas gives us the solution. But then we end up with two (x,y,n) (powerset) tuples, and their intersection is needed to find the solutions of the equation. The pair (k,k) needs to be considered at this stage when both (x,y,n) tuples are combined - the pairs create equivalence class on the tuples which allows us to find several/all solutions to the equation. Jumping from left side to right side and back to left side of the equation might end up in different elements, and if we started from a solution to the equation, the new elements are again a solution to the equation.
Now going further than this likely requires breaking $\sqrt a$ and $\sqrt b$ to something smaller and discovering more of it's properties. (I will still need to read the wikipedia page properly to see how they did it)


What is the formal definition of the set $E^n$? I'm currently in a non-linear optimization course, and I can't seem to find, even in the book, the definition of $E^n$, which is clearly some set. Please see the picture below context.

$\textbf{E}^n$ is used to denote Euclidean $n$-space, that is $\mathbb{R}^n$ equipped with the Euclidean norm $\Vert\cdot\Vert:\mathbb{R}^n\to\mathbb{R}^+$, $ \Vert\textbf{x}\Vert=\sqrt{\textbf{x}\cdot\textbf{x}}$, and distance function $d(\textbf{x},\textbf{y})=\Vert\textbf{x}-\textbf{y}\Vert$

The notation $\textbf{E}^n$ is used to distinguish a Euclidean space from a non-Euclidean real spaces. For example, $\textbf{H}^n$ is used to denote a real hyperbolic space, which may be constructed from $\mathbb{R}^{n+1}$ with the distance function $d(\textbf{x},\textbf{y})=\text{arcosh}{B(\textbf{x},\textbf{y})}$.

A set is a well-defined collection of distinct objects. ... The most basic properties are that a set can have elements and that two sets are equal (one and the same) if and only if every element of each set is an element of the other; this property is called the extensionality of sets.


Double limit question Let $f: X \times Y \rightarrow \mathbb{R}$ be a continuous function where $X \subseteq \mathbb{R}$ and $Y \subseteq \mathbb{R}$. Suppose that

$(x_{n},y_{n}) \rightarrow (x,y)$ in $\mathbb{R}^{2}$,
$(x_{n},y_{n}) \in  X \times Y$, and
$\lim _{y_{n} \rightarrow y} f(x',y_{n}) = \infty$ for all $x' \in X$.

Is it true that there exists an $N>0$ such that $f(x_{n},y_{n}) > 0 $ for all $n \geq N$?
Seems intuitive but I am a bit stuck.

No, this is not necessarily true. The function $f(x,y)=\frac{1}{y}-\frac{1}{x^2}$ is continuous in $(0,1)^2$, and for $x_n=y_n=\frac{1}{n}$, we have $f(x_n,y_n)<0$ for $n\geq 2$, while for $x'\in(0,1)$, we have $f(x',y_n)=n+1-\frac{1}{x'^2}\to\infty$.
Edit: To have a counterexample with $f$ bounded below, for $(x,y)\in (0,1)^2$, consider the function $$f(x,y)=\left\{\begin{array}{c l}\frac{1}{y}-\frac{1}{x}-1, & 0<y<x<1 \\ -1, &\text{otherwise}, \end{array}\right.$$ and choose  $x_n=y_n=\frac{1}{n}$.

I think it is not true. Suppose f(x,y)=-x+ 1/|y-1|. let y=1, yn=1+1/n, xn=2n.


Property of multiplication of ideals in $\mathcal{O}_K$ Let $\mathfrak{a}, \mathfrak{b}$ be two coprime ideals of $\mathcal{O}_K = \mathbb{Z}[\sqrt{-d}]$ such that $\mathfrak{a}\mathfrak{b} = (n)$ for some $n \in \mathbb{Z}$. Does $\mathfrak{a}^m = (u)$ imply $\mathfrak{b}^m = (u)$, $u\in \mathbb Z$? I get that the order of $\mathfrak{a}$ and $ \mathfrak{b}$ must be the same since they are inverses in the ideal class group but I'm not sure why $\mathfrak{a}^m = \mathfrak{b}^m$.

I don't think this is true.  Hopefully I haven't misunderstood your question.  Consider $d = -1$.  Let $\mathfrak p = (1+i)$, so $\mathfrak p^2 = (2)$.  Let $\mathfrak a = \mathfrak p$, and $\mathfrak b = \mathfrak p^3$.  Then $\mathfrak a \mathfrak b = \mathfrak p^4 = (4)$ is the ideal generated by an integral, and also $\mathfrak a^2 = (2)$ is the ideal generated by an integral.  But $\mathfrak a^2 \neq \mathfrak b^2$.


Solving and interpreting $f(x+y)=f(x)+f(y)+x^2y$ for all $x,y \in \mathbb{R} $. The question reads,
 Find all differentiable functions $f$ such that $$f(x+y)=f(x)+f(y)+x^2y$$ for all $x,y \in \mathbb{R} $. The function $f$ also satisfies $$\lim_{x \rightarrow 0}\frac {f(x)}x = 0$$

To solve the problem I wrote the expression for $\frac{df}{dx}$ using first principle and found that $\frac{df}{dx} = x^2.$(Due to the given conditions.)

Using this and calculating $f(0)$ as zero I got the implication that $f(x)$ must be $\frac{x^3}{3}.$

But,
Clearly the calculated $f(x)$ does not satisfy the required condition for the problem. Thinking about where I had committed the mistake, I realized that $f(x+y)=f(x)+f(y)+x^2y$ cannot be true for all real $x,y$.

So how is it that these operations that implied $f(x)=\frac{x^3}{3}$ went wrong?

And why is it that while doing these operations I could not 'see' that the conditions cannot be satisfied for all real $x$?

EDIT:Reading the comments, I want to make the clarification that I realise the fact that there can not be any $ f(x) $ that satisfies these conditions using a different method. However I fail to understand why the method elaborated in the question does not reflect this fact to me??

There cannot be any function satisfying this equation for the simple reason that $f(x+y)-f(x)-f(y)$ does not change if you interchange $x$ and $y$. So if such a function exists we must have $x^{2}y=y^{2}x$ for all $x,y \in \mathbb R$ which is absurd.
Now as to what went wrong in your approach: From $x^{2} =-1$ you can deduce that $x^{4}= 1$ but $x^{4} =1$ is not the final answer. In fact there is no real number  $x$ such that $x^{2} =-1$. The simple answer to your question is converse of result is not always true. If you got $f(x)=\frac {x^{3}} 3$ from the given equation it doesn't mean that there is a solution. You should always go back to the original equation and check if the function you obtained is indeed a solution. If it is not then you have not arrived at any contradiction because converse of result is not always true.

First notice by letting $x=y=0$ we get $f(0)=0$. Further we have $$f(x+y)-f(x) = f(y) +x^2y$$ now if $y\ne 0$ we have $$\lim _{y\to 0} {f(x+y)-f(x)\over y} = \lim _{y\to 0} \big({f(y)\over y} +x^2\big)$$
so we have $f'(x) =x^2$ and thus, for some real $c$ we have $$f(x) ={x^3\over 3}+c$$
Since $f(0)=0$ we get $c=0$. As always we have to chek it buy pluging it in starting equation:
$${(x+y)^3\over 3} = {x^3\over 3}+{y^3\over 3} +x^2y\implies y^2x=0$$
for all $x$ and $y$ which is clearly nosene.


In 3D find a point on plane with given plane angle, distance and three other points I have three points $x, y, z$ in 3D which are not on the same line. With this points we can build a plain $P$. Is there any way to get point $z'$ on $P$ which is at a distance of $d$ from $y$ and angle between $xy$ and $yz'$ is $\alpha$? I'm looking for some algorithm for that.

You could start thinking the particular case when $Z=0$; i.e. when you are in $\mathbb{R}^2$. Put $y$ at the origin and draw all the points with distance $d$ from the origin (a circle with radius $d$ and center at $(0,0)$). Does this gives you a clue?

Say $\vec x, \vec y,\vec z, \vec {z’} $ are the position vectors of the respective points, with $$\vec{z’}=(a,b,c) $$ I’ll show you how you can get $3$ equations in the $3$ unknowns: $a,b,c$.
Firstly, we know $x,y,z’$ are coplanar, and so $$ (a,b,c) \cdot (\vec {xy} \times \vec {yz})=0 \hspace{1 cm} \mathbf{(1)} $$ Also, $|\vec{yz’} | = d$, so $$|(a,b,c)-\vec y | = d \hspace{1 cm} \mathbf{(2)}$$
Lastly, taking into account the angle, $$\cos\alpha = \frac{\vec{xy}\cdot\vec{yz’}}{|\vec{xy}| |\vec{yz’}|} \\ = \frac{(\vec y-\vec x)\cdot (\vec{z’} -\vec y)}{|\vec y-\vec x| d }
\\ \implies (a,b,c)\cdot (\vec y-\vec x) = d\cos\alpha |\vec y-\vec x| + \vec y \cdot(\vec y-\vec x) \hspace{1 cm} \mathbf{(3)} $$
Just solve these equations to get $a,b,c$ and consequently $\vec{z’}$.


Monty hall problem extended. I just learned about the Monty Hall problem and found it quite amazing. So I thought about extending the problem a bit to understand more about it.
In this modification of the Monty Hall Problem, instead of three doors, we have four (or maybe $n$) doors, one with a car and the other three (or $n-1$) with a goat each (I want the car).

We need to choose any one of the doors. After we have chosen the door, Monty deliberately reveals one of the doors that has a goat and asks us if we wish to change our choice.

So should we switch the door we have chosen, or does it not matter if we switch or stay with our choice?

It would be even better if we knew the probability of winning upon switching given that Monty opens $k$ doors.

I decided to make an answer out of my comment, just for the heck of it.

$n$ doors, $k$ revealed

Suppose we have $n$ doors, with a car behind $1$ of them. The probability of choosing the door with the car behind it on your first pick, is $\frac{1}{n}$.

Monty then opens $k$ doors, where $0\leq k\leq n-2$ (he has to leave your original door and at least one other door closed).

The probability of picking the car if you choose a different door, is the chance of not having picked the car in the first place, which is $\frac{n-1}{n}$, times the probability of picking it now, which is $\frac{1}{n-k-1}$. This gives us a total probability of $$ \frac{n-1}{n}\cdot \frac{1}{n-k-1} = \frac{1}{n} \cdot \frac{n-1}{n-k-1} \geq \frac{1}{n} $$

No doors revealed
If Monty opens no doors, $k = 0$ and that reduces to $\frac{1}{n}$, which means your odds remain the same.

At least one door revealed
For all $k > 0$, $\frac{n-1}{n-k-1} > 1$ and so the probabilty of picking the car on your second guess is greater than $\frac{1}{n}$.

Maximum number of doors revealed
If $k$ is at its maximum value of $n-2$, the probability of picking a car after switching becomes $$\frac{1}{n}\cdot \frac{n-1}{n-(n-2)-1} = \frac{1}{n}\cdot \frac{n-1}{1} = \frac{n-1}{n}$$
For $n=3$, this is the solution to the original Monty Hall problem.

Switch.

@DwB

Your explanation would work if you chose your door AFTER you were shown a door that had a goat behind it.

Since you make your choice while there are 3 choices, the probability you choose the right door is 1/3 no matter if he shows you a goat or not.

If you choose one of the goats at first, you have a probability of 2/3 of choosing one of those right?  So the beauty of this problem is that if you choose a goat first, and switch you will always get the car.

Think about it, if you choose one of the goats first, then the other goat will always get shown.  If you switch then you will always switch to the car.

You want to choose a goat first to win this game by SWITCHING.(2/3)
You want to choose a car fiRst to win this game by NOT SWITCHING.(1/3)

4 doors...

You choose a car at first(1/4)
You choose a goat at first(3/4)

He opens 2 doors with a goat behind them, so your first choice and the last door is left.

If you chose the car first and switched now, you would lose.

If you chose any of the 3 goats initially, and now switched, you would win the car.

It just comes down to the fact you have a better chance of picking one of the goats initially, so assuming he shows all the other "bad choices" to you, it is better to switch your answer you chose originally.

The probabilities you win by switching your choice, assuming that Monty opens all the other doors accept for the original one you picked and one other one.

3 doors (2/3)
4 doors (3/4)
5 doors (4/5)
6 doors (5/6)
7 doors (6/7)
8 doors (7/8)

n doors [(n-1)/n] = 1 - 1/n

as n approaches infinity 1/n becomes 0, and the general probability = 1 - 0 = 1.

So if we were to play this game with an infinite amount of doors, and switch our choice, we would win 100% of the time!


Weakly convergence in C(X) for metric compact spaces Let $X$ be a compact metric space.
Let $f_n$ be a sequence of functions in $C(X)$. Show that if $f_n$ converges to $0$ weakly then $|f_n|$ converges to $0$ weakly.
Actually, I need to show then $f_n$ converges to $0$ weakly iff $\sup ||f_n||$ is bounded (uniform norm) and $f_n$ converges to $0$ pointwise.
But I can show the direction: weakly convergence implies $\sup$ norm is bounded by the first argument.
I'm familiar with Reisz Representation Theorem.

Thank you in advance.

Assume that $f_n \to 0$ weakly. It can be shown that every weakly boundes sequence is norm bounded in any Banach space. So $\sup ||f_n ||<\infty,$ Moreover for any $x\in X$ define regular Borel measure $$\mu_x (A) =\begin{cases} 1 \mbox{ iff } x\in A \\ 0 \mbox{ iff } x\notin A\end{cases}. $$
Then by the weak convergence we have $$f_n (x) =\int_X f(u) \mu_x (du) \to 0$$
hence $f_n (x) \to 0 $ pointwise.
On the other hand if $\sup ||f_n ||$ is finite and $f_n \to 0 $ pointwise then by the Lebesgu'e dominated convergence theorem we get $$\int_X f_n (u) \mu (du )\to 0 $$ for any regular borel measure on $X.$ Hence $f_n \to 0$ weakly.


Is the disjunction of these two false statements true? Let $\mathcal{L}$ be the formal system of statement calculus. Let $A$ be a formula of $\mathcal{L}$. Then each of the three formulas
\begin{gather*}
A \vee (\sim A) \\
A \implies [(\sim A) \implies A] \\
(\sim A) \implies [A \implies (\sim A)] \\
\end{gather*}
is a theorem of $\mathcal{L}$.
By case analysis,
$$[(\sim A) \implies A] \vee [A \implies (\sim A)]$$
is a theorem of $\mathcal{L}$. But this seems intuitively incorrect.

For example, each of the statements
\begin{gather*}
(\forall x \in \mathbb{N})[(x < 3) \implies (x \ge 3)]\\
(\forall x \in \mathbb{N})[(x \ge 3) \implies (x < 3)]
\end{gather*}
is false. However, the disjunction of these two statements is true, by the argument above.

Edit. I guess the example was not very good.

Suppose you flip a coin. It is a fair coin and it cannot stand up on its side. It lands at your feet on a solid, flat surface.

Let $A$ be the statement that the coin shows heads. Then $(\sim A)$ is the statement that the coin shows tails. The statement
\begin{gather*}
\text{"if the coin shows heads, then it shows tails,}\\
\text{or}\\
\text{if the coin shows tails, then it shows heads"}
\end{gather*}
is true, even though both disjuncts are false.

Maybe this question no longer belongs on math.stackexchange.com.

$$ (\forall x \in \mathbb{N})[x < 3 \implies x \ge 3] \lor
(\forall x \in \mathbb{N})[x \ge 3 \implies x < 3] $$
is not true, and your analysis doesn't imply that it ought to be.

What is true by your argument is
$$ (\forall x \in \mathbb{N})\bigl[(x < 3 \implies x \ge 3) \lor
(x \ge 3 \implies x < 3)\bigr] $$
but you can't move the quantifiers into the middle of its structure and expect it to remain true.

In general, $(A\to B)\lor(B\to A)$ is always true in classical logic, no matter whether $A$ and $B$ are negations of each other or not. Perhaps even more intuition-challenging, $(A\to B)\lor(B\to C)$ is always true too.

Response to edit: I think you're being confused by two different uses of "if ... then" in mathematical English.

On one hand in casual non-formal mathematical English, "if it's heads, then it's tails" would mean something like

  In every relevant situation where it's heads, it is also the case that it's tails.

(and one then hopes that the context makes it clear which situations are "relevant").
On the other hand, in (classical) formal logic, the formula "${\rm heads}\Rightarrow{\rm tails}$" means

  In the particular situatation we're looking at, it is not the case that it is heads yet it isn't tails.

Which is the same as saying

  The particular situation we're looking at is not a counterexample to sense (1).

Confusion creeps in because "${\rm heads}\Rightarrow{\rm tails}$" is often pronounced "if heads then tails", as if its meaning were the same as in (1), but it is not when formulas are involved.

The sense in which you can prove that $(A\Rightarrow \neg A)\lor(\neg A\Rightarrow A)$ is true is (2) and only (2). It stops being true if you try to reinterpret the $\Rightarrow$s as if they meant (1).

Furthermore it it looks like you're also confusing yourself by the difference between "is true" and "has a proof" when you say that "this or that" is true but "this" and "that" are both false.

What you mean here is that you believe you have a proof of "this or that". But clearly it is not true because the definition of "or" is that "this or that" is true exactly when at least one of "this" and "that" is true -- and you have just argued that they are both false.

So when you ask,

  How can "this" be false and "that" be false and "this or that" be true?

the answer is that they can't and you're mistaken about at least one of those parts. Same as if you have an apparent proof that concludes $2+2=3$: you shouldn't be asking "why is $2+2$ not $4$?" but "what is wrong with this thing that looks like a proof but obviously can't be?"

$ [A \to \neg A] \leftrightarrow [\neg A \vee \neg A] \leftrightarrow \neg A $

$ [\neg A \to A] \leftrightarrow [A \vee A] \leftrightarrow A $

So, you can see that
$ [A \to \neg A] \vee [\neg A \to A]$ is the same as $ A \vee \neg A $

So this is true:

$ \forall x : P(x) \vee \neg P(x) $

But, and this is where your intuition went down the wrong path, this is not generally true:

$ [\forall x : P(x)] \vee [\forall x : \neg P(x)] $ as your number example showed.


$a+b+c+ab+bc+ca+abc=1000$ find the minimum value of $a+b+c$ Given $a+b+c+ab+bc+ca+abc=1000$.

Find the minimum value of $a+b+c$.

Now we are considering $a$, $b$, $c$ to be integers and here in lies the pertinent problem as I could not get an answer in integers but in fractions.

Without loss of generality, suppose that $a\leq b\leq c$.  Also, write $N:=1001$.  Since $$(a+1)(b+1)(c+1)=N\,,$$ we have that $$|a+1|\,|b+1|\leq N\,.$$  This clearly means
$$|a+1|+|b+1|\leq N+1\,.$$
(If $x$ and $y$ are positive integers with $xy\leq N$, then $(x-1)(y-1)\geq 0$, which implies that $x+y\leq xy+1\leq N+1$.)
Thus,
$$-(a+1)-(b+1)\leq |a+1|+|b+1|\leq N+1\text{ or }a+b\geq -N-3\,.$$
Since $c+1$ is clearly positive, we have $c+1\geq 1$, which means $c\geq 0$.  That is,
$$a+b+c\geq -N-3\,.$$
The equality holds iff $(a,b,c)=(-N-1,-2,0)$.

If $a$, $b$, and $c$ are required to be nonnegative, then by the AM-GM Inequality, we have
$$\frac{(a+1)+(b+1)+(c+1)}{3}\geq \sqrt[3]{(a+1)(b+1)(c+1)}=\sqrt[3]{1001}>10\,.$$
That is, $a+b+c>27$, or $a+b+c\geq 28$.  As $(a,b,c)=(6,10,12)$ works, the minimum value of $a+b+c$ is indeed $28$.  (Unlike the first part of this answer, $1001$ can't be replaced by an arbitrary positive integer $N$.  For each $N$, a different analysis is required.)

I am assuming here $a=b=c$ since $1000$ is the cube of $10$.

$$(a+1)(b+1)(c+1) = 1000$$

now, $(a+1)^3 = 1000$;
$$a+1 = 10\iff
a=9$$

so, $a+b+c = 3*9 = 27$.

Just that we can get the integer value but I cannot say $27$ is the minimum value for $a+b+c$.


Evaluation of the double integral $\int_{[0,1]×[0,1]} \max\{x, y\} dxdy$ Evaluate:  $$\int_{[0,1]×[0,1]} \max\{x, y\} dxdy$$

I am totally stuck on it. How can I solve this?

In general, the integral
$$I_n=\int_{[0,1]^n}\max\{x_1,\cdots,x_n\}\,\mathrm{d} x_1\cdots\mathrm{d}x_n$$
can be viewed as an integral of the largest order statistic, say $X^{(1)}$.

Let $x^{(1)}=\max \{x_1,\cdots,x_n\}$, the distribution of $X^{(1)}$ will be $$F_{X^{(1)}}(x)=F_{X_1}(x)\times\cdots\times  F_{X_n}(x)=x^n.$$
Hence
$$I_n=\int_{[0,1]^n}\max\{x_1,\cdots,x_n\}\,\mathrm{d} x_1\cdots\mathrm{d}x_n=\int_{[0,1]}x\,\mathrm{d}F_{X^{(1)}}(x)=\int_0^1 nx^n\,\mathrm{d}x=\frac{n}{n+1}.$$


How do I solve $\lim_{x\to +\infty}\frac {\ln(x-1)^2}{x} $? I just can't get past the $x-1$. The $2$ I know how to get rid of.

$$\lim_{x\to +∞}\frac {\ln\left((x-1)^2\right)}{x}  $$

$$\lim_{x\to \infty} \frac{\ln(x-1)^2}{x}$$

It is not hard to show $\ln x< x$

$\ln(x-1)^2=4\ln\sqrt{x-1}\leq 4\sqrt{x-1}\leq 4\sqrt x$

So the limit is smaller than

$$\lim_{x\to \infty} \frac{4\sqrt{x}}{x}=\lim_{x\to \infty} \frac 4{\sqrt x}=0$$

Now the following is true because square root must be non negative

$-4\sqrt x\leq ln(x-1)^2$

So the limit is bounded by $0$

By applying L'Hôpital's rule you have:
$$\lim_{x\rightarrow \infty} (\ln(x-1)^2)'=\lim_{x\rightarrow \infty} \frac{2}{x-1}=0$$
and $$\lim_{x\rightarrow \infty}(x)'=1$$
Therefore the limit $$\lim_{x\rightarrow \infty}\frac{(\ln(x-1)^2)'}{x'}=\frac{0}{1}=0$$
exists and it is equal to
$$\lim_{x\to +∞}\frac {ln(x-1)^2}{x}$$


How to derive the gradient and Hessian of the Volumetric barrier $v$ I'm reading the monograph Convex Optimization: Algorithms and Complexity . In Section 2.3, the author discusses an algorithm of Vaidya. He makes the following definition:

Definition 1:  Let $A \in \mathbb{R}^{m \times n}$ where the $i^{t h}$ row is $a_{i} \in \mathbb{R}^{n},$ and let $b \in \mathbb{R}^{m} .$ We
consider the logarithmic barrier $F$ for the polytope $\left\{x \in \mathbb{R}^{n}: A x>b\right\}$
defined by
$$
F(x)=-\sum_{i=1}^{m} \log \left(a_{i}^{\top} x-b_{i}\right)
$$

Definition 2:  The volumetric barrier $v$ is defined by
$$
v(x)=\frac{1}{2} \log \operatorname{det}\left(\nabla^{2} F(x)\right)
$$

Note that we have $$
\nabla^{2} F(x)=\sum_{i=1}^{m} \frac{a_{i} a_{i}^{\top}}{\left(a_{i}^{\top} x-b_{i}\right)^{2}}
$$
Now, the author claims that we can easily prove the following results:
$$
\nabla v(x)=-\sum_{i=1}^{m} \sigma_{i}(x) \frac{a_{i}}{a_{i}^{\top} x-b_{i}}
$$
$$
\nabla^{2} v(x) \succeq \sum_{i=1}^{m} \sigma_{i}(x) \frac{a_{i} a_{i}^{\top}}{\left(a_{i}^{\top} x-b_{i}\right)^{2}}
$$
where we introduce the leverage score
$$
\sigma_{i}(x)=\frac{\left(\nabla^{2} F(x)\right)^{-1}\left[a_{i}, a_{i}\right]}{\left(a_{i}^{\top} x-b_{i}\right)^{2}}
$$
(Here, it seems that the author does not explain the notation $[a,a]$, but I guess that it refers to the dot product)

Unfortunately, I have no idea how these results are shown. Can anyone help?

Denote the elementwise/Hadamard product for two matrices of equal dimensions as $$A\odot B$$
and the trace/Frobenius product as $$A:B = \operatorname{Tr}(A^TB)$$
For typing convenience, define some auxiliary variables
$$\eqalign{
y &= Ax-b,\quad&r = y^{\odot-1} \quad({\rm hadamard\, inverse}) \\
Y &= \operatorname{Diag}(y), &R = \operatorname{Diag}(r) \\
}$$
and their differentials
$$\eqalign{
R\,Y &= I &\implies R\,dY = -Y\,dR \\
r\odot y &= {\tt1} &\implies r\odot dy = -y\odot dr \\
Yr = Ry &= {\tt1} &\implies R\,dy = -Y\,dr \\
dy &= A\,dx \\
dr &= (RY)dr \\
   &= R(-R\,dy) \\
   &= -R^2A\,dx  \\
}$$
Using these variables, the gradient ($g$) and hessian ($H$) of the log barrier can be calculated.
$$\eqalign{
F &= -{\tt1}:\log(y) \\
dF &= -{\tt1}:r\odot dy = -r:dy = -A^Tr:dx \\
\frac{\partial F}{\partial x} &= -A^Tr \;\doteq\; g \\
\\
dg &= -A^Tdr = A^TR^2A\,dx \\
\frac{\partial g}{\partial x} &= A^TR^2A \;\doteq\; H \\
}$$
The following quantities
$$\eqalign{
dH &= 2A^TR\,dR\,A \\
K &= AH^{-1}A^T \\
w &= \operatorname{diag}(K) \\
W &= \operatorname{Diag}(w) = I\odot K \\
}$$
will be useful in calculating the gradient ($p$) and hessian ($Q$) of the volumetric barrier.
$$\eqalign{
 v &= \tfrac{1}{2}\log\det H \\
dv &= \tfrac{1}{2}\,d\operatorname{tr}\log H \\
   &= \tfrac{1}{2}\,H^{-T}:dH \\
   &= H^{-1}:A^TR\,dR\,A \\
   &= \left(AH^{-1}A^T\right):\operatorname{Diag}(r\odot dr) \\
   &= w:(r\odot dr) \\
   &= -(r\odot w):(R^2A\,dx) \\
   &= -A^TR^3w:dx \\
\frac{\partial v}{\partial x} &= -A^TR^3w \;\doteq\; p \\
\\
dp &= -A^T(3R^2dR)\,w - A^TR^3dw \\
   &= -3A^TR^2W\,dr
    + A^TR^3\operatorname{diag}\left(AH^{-1}dH\,H^{-1}A^T\right) \\
   &= 3A^TR^2WR^2A\,dx + 2A^TR^3\operatorname{diag}(KR\,dR\,K) \\
   &= 3A^TR^2WR^2A\,dx + 2A^TR^3(K\odot K)(r\odot dr) \\
   &= 3A^TR^2WR^2A\,dx + 2A^TR^3(K\odot K)R\,dr \\
   &= 3A^TR^3(YWY)R^3A\,dx - 2A^TR^3(K\odot K)R^3A\,dx \\
   &= A^TR^3\big(3YWY-2K\odot K\big)R^3A\,dx \\
\frac{\partial p}{\partial x}
   &= A^TR^3\big(3YWY-2K\odot K)R^3A \;\doteq\; Q \\
}$$
So there are the gradients and hessians in matrix notation.

The author's expression for the leverage is unclear. My best guess is
$$\eqalign{
\sigma &= \operatorname{Diag}\left(RAH^{-1}A^TR\right) = RWR \\
}$$
then, for example you could write the above gradient and hessian as
$$\eqalign{
p &= -A^T\sigma r \\
Q &= 3(A^TR\sigma RA) - 2A^TR^3(K\odot K)R^3A \\
}$$
The use of the $\succeq$ symbol suggests that he purposely omitted the coefficient of $3$ and the negative $(K\odot K)$ term; perhaps they weren't important for his subsequent calculations.

Vaidya proves these as Lemmas 1 and 2 in (P.M. Vaidya, 1996). We reproduce the derivation of the gradient below.

For ease of notation, define $\rho_i(x) := \frac{a_i a_i^{\top}}{(a_i^\top x-b_i)^2}$ and $H(x):=\nabla^2F(x)$. We use the following approximation: $\det(I+tA) = 1+t\cdot{\rm Tr}(A)+o(t^2)$. Let $h$ be an arbitrary direction vector.

\begin{align}
   2\cdot v(x+th) = \log\det[H(x+th)]&= \log\det\bigg[\sum_i \frac{a_i a_i^{\top}}{(a_i^\top (x+th)-b_i)^2}\bigg]\\
           &=\log\det\bigg[\sum_i \frac{a_i a_i^{\top}}{(a_i^\top x-b_i)^2}\cdot \bigg(1-\frac{2ta_i^\top h}{(a_i^\top x-b_i)}+o(t^2)\bigg)\bigg]\\
           &= \log\det\bigg[H(x)-\sum_i \rho_i(x) \cdot\bigg(\frac{2ta_i^\top h}{(a_i^\top x-b_i)} + o(t^2)\bigg) \bigg]\\
           &= \log\det[\ H(x)\ ]\\
&+\log\det\bigg[  I-\sum_i H(x)^{-1/2}\rho_i(x)H(x)^{-1/2} \cdot\bigg(\frac{2ta_i^\top h}{(a_i^\top x-b_i)}+o(t^2)\bigg) \bigg]\\
           &= \log\det[\ H(x)\ ]+\log\bigg[ 1-\sum_i\frac{2ta_i^\top h}{(a_i^\top x-b_i)}\cdot
 {\rm Tr}(H(x)^{-1/2}\rho_i(x)H(x)^{-1/2})+o(t^2)\bigg]\\
           &= \log\det[\ H(x)\ ] - \sum_i\frac{2ta_i^\top h}{(a_i^\top x-b_i)}\cdot
 {\rm Tr}(H(x)^{-1/2}\rho_i(x)H(x)^{-1/2})+o(t^2).\tag{1}\label{eq_res_1}
\end{align}

Next we show that ${\rm Tr}(H(x)^{-1/2}\rho_i(x)H(x)^{-1/2})=\sigma_i(x)$. To this end observe
\begin{align}
{\rm Tr}(H(x)^{-1/2}\rho_i(x)H(x)^{-1/2})={\rm Tr}(H(x)^{-1}\cdot\rho_i(x))&=\frac{{\rm Tr}(H(x)^{-1}a_i a_i^{\top})}{(a_i^\top x-b_i)^2}\\
&=\frac{a_i^{\top}H(x)^{-1}a_i}{(a_i^\top x-b_i)^2}\\
&=\sigma_i(x).
\end{align}

Now we can compute the following directional derivative
\begin{align}
\lim_{t\to 0}\ \frac{   v(x+th)-v(x)}{t}=-\sum_i \sigma_i(x)\cdot
 \frac{a_i^\top h}{a_i^\top x-b_i}
\end{align}

Since $\nabla v(x)$ is the unique vector which, for all $w\in \mathbb{R}^n$, satisfies $\langle \nabla v(x), w\rangle = \frac{d v(x+tw)}{dt}\big|_{t=0}$, and as the choice of $h$ was arbitrary, we can conclude that
$$\nabla v(x) := -\sum_{i}\sigma_i(x) \cdot\frac{a_i}{a_i^\top x-b_i}.$$


If $x'(t)$ is bounded for $x\geq0$ show that $\lim_{t\to\infty}{x(t)}=0$ Assume that $x(t)$ is nonnegative for $t\geq0$ and $\int_0^\infty{x(t)\;dt}<\infty$. If $x'(t)$ is bounded for $x\geq0$, then show that $\lim_{t\to\infty}{x(t)}=0$.

I started with contradiction taking the limit is finite and strictly positive. Then it is easy to derive the contradiction. But the limit may be oscillatory too or diverging to $+\infty$.

Any help would be apreciated. Thanks in advance.

If $x(t)\not\to 0$, then there exists a $\delta>0$ such that there exists a sequence $t_n\to\infty$ with $x(t_n)\ge \delta$. Taking a subsequence we can assume $t_{n+1}-t_n\ge \frac\delta M$. We have $|x'(t)|\le M$ so $x(t)\ge \frac\delta2$ for $|t-t_n|<\frac\delta{2M}$ and so $$\int_{t_n}^{t_{n+1}} x(t)dt \ge \frac{\delta^2}{4 M}.$$ Hence the integral must be infinite.

Incomplete proof (missing the case when $x(t)$ has no limit)

By contradiction, let $L \gt 0$ such that $\lim_{t \to \infty} x(t) = L$.

Then, by definition $\forall$ $\varepsilon \gt 0$ exists $k \in \mathbb{R^+}$ such that
$$\mid x(t) - L \mid \lt \varepsilon \;\;\;\;\;\;\;\; \forall \;\;t \geq k$$
Let $\varepsilon \in (0, L)$, then
$$0 \lt L - \varepsilon \lt x(t) \;\;\;\;\;\;\;\; \forall \;\; t \geq k$$
$$\Rightarrow 0 < \int_k^\infty (L - \varepsilon)dt < \int_k^\infty x(t)dt < \infty$$
But this is a contradiction since $L - \varepsilon$ is a positive constant so $\int_k^\infty (L - \varepsilon)dt$ is not bounded.


Joint probability density function and independence The joint probability density function of the random variables $X_1$ and $X_2$ is given by$f_{x_1 x_2}(x_1,x_2)= \left\{\begin{matrix}
(1/6) \cdot x_1x_2,1\leq x_1 \leq 3 \quad and \quad 1\leq x_2\leq 2\\
0, otherwise\end{matrix}\right.$

And let the variable Y be defined as $ Y = X_1 X_2^2$

How do I find out if $X$ and $Y$ are independent and if they are uncorrelated? And how do i find he mean of $E[Y]$ ?

Hint 1: If $X_1$ and $X_2$ are independent then $\mathbb E(X_1X_2^2)=\mathbb E(X_1)\cdot \mathbb E(X_2^2)$

Hint 2: $f_{X_2}(x_2)=\int_1^3 \frac16\cdot x_1\cdot x_2 \, \ dx_1$

Hint 3: $\mathbb E(X_2^2)= \int_1^2 x_2^2 \cdot f_{X_2}(x_2)\, \ dx_2$


How to understand compactness? How to understand the compactness in topology space in intuitive way?

Maybe you should think about compactness, as something that takes local properties to global properties. For example, if $f:K\rightarrow \mathbb{R}$ is continuous, $K$ is compact, and $f(x)>t_x>0$ for all x, then you can find $t>0$ such that $f(x)>t>0$ for all x - so from $f(x)>t_x>0$ point wise, you know that $f>t>0$ as a function. (This is a simple consequence of Weierstrass theorem in $[a,b]\rightarrow \mathbb{R}$)

Usually we find some property that is true for every "small" enough open sets, then use compactness to reduce the case to finitely many open sets and use induction to show that the property is true for all of the space. This is at least how I understand compactness.

As the commenters below this message wrote (and I didn't emphasize enough), we usually use compactness to reduce infinite problems\conditions\restraints to a finite subset that cover the entire space, and then use some argument that works only for finite cases (like induction, taking max\min, take finite sums etc). In my example above, we wanted to find a minimum over all the lower bounds, but in the infinite case this is usually just an infimum (and can be zero), but when we reduce to a finite case there is a minimum.

In this way we can think of compactness as something that let us use some finite argument on infinite covers (and many times to transfer some property from the cover to the entire space).

For me, the compactness of a topological space means that it has enough points to provide exact solutions to continuous equations. More precisely,

  compactness = Any equation that can be approximated by a consistent system of $\leq$ inequalities of continuous functions has a solution.

For instance, being a solution to the equation $x^2 = 2$ is equivalent to being a simultaneous solution to the infinite system of inequalities $\lbrace |x^2-2| \leq \frac 1 n \rbrace_{n\in\mathbb N}$. This system is consistent in the sense that every finite subsystem has solutions; after all, we can approximate $\sqrt 2$ by rational numbers arbitrarily well. Then, it is the compactness of bounded intervals of real numbers that tells us that there exists a simultaneous solution of all of these inequalities, and hence of the equation itself.

In contrast, trying to encircle $\sqrt 2$ as a rational number fails, you can enclose it by smaller and smaller intervals, but when you try to catch it with an infinite number of intervals, it "falls through the cracks".

The principle discussed above is also known as the finite intersection property. The inequalities, or "conditions", define a collection of closed sets and the exact solution we seek is contained in the intersection of infinitely many closed sets. If every finite intersection of these closed sets is non-empty, i.e. if every finite collection of conditions can be fulfilled, then one could expect that all of the conditions can be satisfied simultaneously. This is precisely what compactness ensures.

Other examples:

Peano's theorem about the existence of solutions to ordinary differential equations relies on compactness to construct a good candidate.
A commonly used technique for showing the existence of solution for partial differential equations is the calculus variations. In this context, the weak compactness of the unit ball in a Hilbert space is often used to get a weak solution to the PDE which is then upgraded to a strong solution.


Prove that between two roots of $f(x)$ there is a root of $g(x)$ Let $f(x),g(x)$ be differential functions, and $f'(x)g(x)\neq f(x)g'(x)$ for all $x\in\mathbb R$. Prove that between two roots of $f(x)$ there is a root of $g(x)$.

I guess this has to do with Rolle's theorem. I saw that when $f'(x)=0$, $g(x)\neq0$ and when $f(x)=0$, $g'(x)\neq0$, but I didn't manage to prove the conjecture. Thanks for any help!

Suppose that $a,b$ are roots of $f$ with $a<b$ and $g(x)\neq0$ for $x\in (a,b)$. Consider the function $h(x)=\dfrac{f(x)}{g(x)}$ in $[a,b]$ to derive a contradiction(is well defined, differentiable on [a,b] with $h(a)=h(b)$...).

If $g(x) \neq 0$ for all $x \in (a,b)$ then $h(x)=\frac{f(x)}{g(x)}$ is well definied and differentiable. Also $h’(x)=\frac{fg’-f’g}{g^2}=0$ by hypothesis and $h(x)=c$ with $c \in \mathbb{R}\setminus\{0\}$ and $f(a)=cg(a)=0=cg(b)$ then $g(a)=0= g(b)$ contradiction
Sorry, my english is too bad.
Must be $fg’-f’g=0$ and $a,b$ consecutive roots of $f$ and $g(a)$ and $g(b)$ not both zero


After switching a lamp on and off infinitely many times in one minute, is it on or off? So we have a lamp. It's switched on. let's represent its state of being switched on with associating it with $1$ and being off with $-1$. after half a minute passes, you turn it off, after another quarter of a minute passes you turn it on and so on.

Now this process will take : $1/2+1/4+1/16+...=1$ minute

and this is the states of the lamp put into a sequence:

$(1,-1,1,-1,1,...)$

After a minute of switching it on an off passes, will it be on or off?

I'm well aware that the limit of this sequence does not exist, however, my intuition dictates that after a minute passes, it is either on or off.

So does this lamp have a final state(on or off) which is impossible for us to know? or it does not have a finals state(it seems to me, it must have one!)?

The apparent paradox in this question arises from the conflict between the following two ideas:

Our intuition about physical lamps.  We expect them to have certain properties, such as being on or off, but not both or neither.
A complicated function that oscillates between two values at an increasing pace, whose oscillations become arbitrarily rapid.

It is not possible to have a physical lamp whose on/off state follows the complicated function.  This is the resolution to the paradox.  Thus we must either give up (1) or (2).  If we give up (1), then there is no reason to believe that the state must be either on or off.  The mathematical function has no limit.

Assuming the wiring is 100% efficient and the speed of the electrical current is instantaneous (which is just as illogical as the idea of unreal numbers such as infinite), there would be no final flip, as the time it takes to switch it is always divided by half.

However, in a real-world application, where wiring has latency and electrical current has a set speed, when the time between switches gets low enough that the switch is being turned on then off before any current could pass the wires (assuming any force could act that fast), the overall state will always be off.


Show that $|a-b|<\frac{|b|}{2}$ implies $\frac{|b|}{2}<|a|$ Pretty simple, for $a,b \in \mathbb R$, show that $|a-b|<\frac{|b|}{2}$ implies $\frac{|b|}{2}<|a|$.  I can see this graphically on the number line, but I can't seem to show it algebraically.
I'm think it involves the triangle inequalities:

$$|x+y|\le|x|+|y|$$
$$|x-y|\ge||x|-|y||.$$

First, rewrite $|a-b|<\cfrac{|b|}{2}$ as $|b-a|<\cfrac{|b|}{2}$.

Now, I'll use your second version of the triangle inequality (without the extra bars) and we obtain
$$|b|-|a| \leq |b-a|<\frac{|b|}{2}$$
which means
$$|b|-|a| < \frac{|b|}{2}$$
Finally, add $|a|$ and substract $|b|/2$ to both sides of the latter to get the desired result.


Prove that if $p$ is a prime, $a$ is an integer, and $p$ divides $a^2$, then $p$ divides $a$. I think I need to use the Fundamental Theorem of Arithmetic but that only applies to $a > 1$ so I think I need to do it by cases.

Case 1) $a = 0$.

Case 2) $a = 1$.

Case 3) $a > 1$.

Case 4) $a < 0$.

Cases 1 and 2 are easy.  I think Case 4 will follow from Case 3 but I struggle with Case 3.

Case 3) $a > 1$. Then by Fundamental Theorem of Arithmetic, $a$ is equal to a unique product of primes. Then, $a^2$ is a product of those primes squared. Then, because $p$ divides $a^2$, $a^2 = pk$ for some $k$ (an integer). So that product of primes squared equals $pk$. Now how do I argue that one of those primes must be $p$?  Once we see that one of those primes is $p$ (i.e. $p$ is a prime factor of $a$), we can say that $p$ divides $a$.

So my questions are:
How do I argue that one of those primes must be $p$?
Can the proof be done without resorting to four cases?

Assume that $p \nmid a$. Then, $a=p \cdot q +r$ for some $r >0$,  s.t. $r \neq 0 (modp) $.

Since,  ${p \mid a^2 } \Rightarrow {a^2 \equiv 0 (mod p)}$

So, $a^2=(p \cdot q +r)^2=p^2 q^2+2pqr+r^2 \equiv r^2 (modp)\neq 0 (modp)  \Rightarrow p \nmid a^2$ Contradiction.

Let the prime factorisation of a be as follows :
 $a = p_1p_2 \dots p_n$, where $p_1,p_2, \dots p_n$ are primes, not necessarily distinct.
Therefore, $a_2 = (p_1p_2 \dots p_n)(p_1p_2 . . . p_n) = p_{1^2}p_{2^2} \dots p_{n^2}$.

Now, we are given that $p$ divides $a{^2}$. Therefore, from the Fundamental Theorem of Arithmetic, it follows that $p$ is one of the prime factors of $a{^2}$. However, using the uniqueness part of the Fundamental Theorem of Arithmetic, we realise that the only prime factors of $a{^2}$ are $p_1p_2 \dots p_n$. So $p$ is one of $p_1, p_2,\dots, p_n$.

Now, since $a = p_1p_2 \dots p_n$, $p$ divides $a$.


Subextension of a finitely generated extension of fields If $E/K$ is a finitely generated field extension and $F$ is an intermediate field how can I prove that $F/K$ is finitely generated?

The result is easy if you suppose that the big extension $E/K$ is algebraic ( equivalently: finite) and Benjamin has given you a proof.

But it is also true in complete generality: if $E/K$ is a field extension and if $E=K(x_1,...,x_n)$ is a finitely generated extension, then any intermediate field $K\subset F\subset E$ is also finitely generated .

The difficulty is that some or all of the $x_i$'s might be transcendental over $K$.
Even in the case of a purely transcendental extension $K\subset K(X_1,...,X_n)$ the situation is quite complicated and it is not true  for $n\gt 1$ that $F$ must be  purely transcendental too  : this is the failure of Lüroth's theorem in higher dimensions.

A proof of finite generation of $F$  in the general non-algebraic case is surprisingly difficult to locate in the literature.  The only reference I could find is Theorem 24.9  in Isaacs's book.

First of all if $E/K$ is finitely generated this means that $E = K(a_1,\ldots a_n)$ where the $a_i$ are algebraic over $K$. Since $F$ is an intermediate field, you have the containment

$$K \subseteq F \subseteq E.$$

We now need the result the following result:

    If $E = K(a_1,\ldots a_n)$, then $[E:K]$ finite.

$\textbf{Proof:}$ Since $a_1$ is algebraic over $F$, $[K(a_1) : K]$ is finite. Since $a_2$ is algebraic over $K$, it is algebraic over $K(a_1)$ because $K \subset K(a_1)$. Hence $[K(a_1,a_2):K]$ is finite by the dimension counting formula. Continuing in this fashion we see that $[E:K] = [K(a_1, \ldots a_n):K]$ is finite, proving our claim.

Since $[E:K]$ is finite, the dimension counting formula implies that $[E:F]$ and $[F:K]$ are finite. In particular this means that $[F:K]$ is finite so that $F/K$ is finitely generated.

Alternatively if you know linear algebra, $F$ being an intermediate field is a $K$ - vector subspace of a finite dimensional $K$ - vector space $E$, hence is finite dimensional as well.


Does this probability problem make sense? I just got a probability problem as follows.

  Every person in a region has a probability of getting infected by HIV
  of 0.1. If the total population of that region is 1000 persons, find
  how many person approximately get infected?

What I don't understand is how the probability for each person correlate to the population. Is this problem correct? I don't know  which concept to be applied here. Independent event, mutually exclusive event, etc,... I am really confused!

There is some ambiguity, but I assume the question intends for you to see that with probability 0.1, a person is infected by HIV where the probability of each person getting infected is independent of each other. So the number of people infected is the expected value of a binomial distribution where $n = 1000$ and $p = 0.1$. Then the answer is simply $np = 100$.

Another way to see this is to define a random variable $X_{i}\sim Ber(0.1)$ where $X_{i} = 1$ if the $i$th person is infected or $0$ otherwise. Then by linearity of the expected value,

$$\mathbb{E}\left(\sum_{i=1}^{1000}X_{i}\right) = \sum_{i=1}^{1000}\mathbb{E}(X_{i}) = \sum_{i = 1}^{1000}0.1 = 1000(0.1)$$

You can use the Central limit theorem to do the approximation.
Central Limit Theorem


Prime ideal and nilpotent elements If $\mathfrak p \subset R$ is a prime ideal, prove that for every nilpotent $r \in R$ it follows that $r \in \mathfrak p$.

The only hint that my tutor gave me was to use induction. Can someone explain what he means by this?

Thanks for the help!

Hint $\ $ Induction on $\rm\,n\,$ shows $\rm\ b = a_1\cdots\, a_n\in P\,$ prime $\rm\: \Rightarrow\:$ some $\rm\ a_i\in P.\:$

Yours is the special case $\rm\: b=0,\ \  a_i = a,\ $ i.e. $\rm\: a^n\! = 0\in P.$

You may prove the the nilradical$=\{x\in R:x^{n}=0,n\in\mathbb{N}\}=\cap P$ where $P$ is a prime ideal of $R$. That should complete the proff, hint use Zorns lemma.


A bounded sequence cannot be divergent. True or false "A bounded sequence cannot be divergent." Is this statement true?
As far as I know a bounded sequence can either be convergent or finitely oscillating, it cannot be divergent since it cannot diverge to infinity being a bounded sequence. Contrapositively, can I say an unbounded sequence can either be divergent or infinitely oscillating?

Yes you can.

State that your sequence is unbounded. Suppose it has no upper bound (the same logic is to be applied if it has no lower bound).

So:

$$\forall x \in \mathbb R, \exists n \in \mathbb N, u_n > x$$

Work a bit with this, and the definition of limit:

$$ \exists a \in \mathbb R,  \forall \epsilon \in \mathbb R, \exists n_0 \in \mathbb N, n > n_0 \implies |u_n - a | < \epsilon $$

The definition of limits tells you that for every $\epsilon$, there is a rank $n_0$ such that $(u_n)_n$ is bounded for $n>n_0$. Since there is a finite number of terms $u_n$ for $0 \le n\le n_0$, the sequece is also bounded for $n \le n_0$.

This is a contradiction with the undounded hypothesis, and you can then conclude that the "convergent" hypothesys is wrong.


My dilemma about $0^0$ We know that $0^0$ is indeterminate.

But if do this:

$$(1+x)^n=(0+(1+x))^n=C(n,0)\cdot ((0)^0)((1+x)^n) + \cdots$$

we get
$$(1+x)^n=(0^0)\cdot(1+x)^n$$

So, $0^0$ must be equal to $1$.

What is wrong in this?
Or am I mistaking $0^0$ as indeterminate?

Other threads are somewhat similar but not exactly the one I am asking.

There's a common yet subtle misconception among mathematics students that algebraic laws are prior to the numerical values that they describe. For example, "why" is it that:

$$5(3+2)=5\cdot 3 + 5\cdot 2\tag1$$

The tendency is to say, oh, this is true because of the distributive law: $a(b+c)=ab+ac$. But actually, equation (1) is true because it's true, not because of an abstract law. That is, you can actually calculate the left hand side and the right hand side and check that they're equal. And it's the fact that that always works that justifies the abstract law.

In your case, the binomial theorem is considered true because it always works when you plug in actual numerical values. You know how to calculate $(2+3)^8$ without the binomial theorem, but the result you get is consistent with the binomial theorem. To compute an expression using a general theorem, and then conclude that a specific numerical expression has a specific value is backwards reasoning.

If we're starting from the point of view that $0^0$ is undefined, then when we apply the binomial theorem to $(0+a)^n$ and realize that it's asking us to compute $0^0$, the conclusion is not that the binomial theorem is true all of the time, and that therefore $0^0$ must have a certain value, the conclusion is that therefore the binomial theorem does not apply to $(a+b)^n$ when $a$ or $b$ is zero. If we want it to, we have to come up with a definition for $0^0$, and then re-prove the binomial theorem, making sure that our proof is consistent with our new definition.

And, in fact, if we define $0^0=1$, then it's possible to do that. Your reasoning almost amounts to a proof that given that definition, the binomial theorem works, but it's important to recognize that you're verifying that a given definition leads to a given theorem, you are not concluding that a definition is "true" as a consequence of a theorem.

When you establish the Binomial theorem, you multiply homogeneous factors $(a+b)$ together and never get any $0$ exponent.

$$(a+b)^1=a+b\\
(a+b)^2=a^2+2ab+b^2\\
(a+b)^3=a^3+3ab^2+3a^2b+b^3\\
\cdots$$

These are introduced artificially to form a compact expression, and this formula cannot serve to "prove" the value of $0^0$.


number of ways to put 4 black,4 white,4 red balls in 6 different boxes The question says:in how many ways we could put 4 black,4 white,4 red balls in 6 different boxes?

boxes are distinguishable,black balls are identical,red balls are identical,and white balls are identical.

one or more box can remain empty.

i know number of ways to put k indistinguishable balls to n distinguishable boxes.but i don't know how to attack this problem because 3 different kind of balls.

Hint: Because colors are distinguishable, the answer is just how you put 4 red balls in 6 boxes times 4 white balls in 6 boxes, times 4 black balls in 6 boxes. Each color has the same number of balls, so they have the same number of combinations, so the answer is the number of ways to put 4 balls in 6 boxes, cubed (one time for each color, so three times multiplied)

I will write out the approach that @Asimov posted.

Let's consider the 4 red balls. I can put red ball #1 into any of the six boxes. I can also put red ball #2 into any of the six boxes. And so on for red ball #3 and red ball #4. Therefore, I the 4 red balls into 6 boxes in 6^4 different ways.

Since, I also have 4 white and 4 black balls, I also have 6^4 possibilities for each of those.

This brings the total number of ways for all three to (6^4) * (6^4) * (6^4) or 6^12.


How to prove that if $D$ is countable, then $f(D)$ is either finite or countable? Is there anybody who can help me to prove that if $D$ is countable, and $f$ is a function whose domain is $D$, then $f(D)$ is either finite or countable?

Recall that $f$ is a collection of ordered pairs such that if $\langle a,b\rangle$ and $\langle a,c\rangle$ are both in $f$ then $b=c$.

Furthermore, $f(D)=\{b\mid\exists a\in D:\langle a,b\rangle\in f\}$. Since for every $a\in D$ there is at most one ordered pair in $f$ in which $a$ appears in the left coordinate, we can define an injective function from $f(D)$ into $D$.

Suppose $D=\{d_n\mid n\in\mathbb N\}$, then for $b\in f(D)$ define $\pi(b)=d_n$ where $n=\min\{k\in\mathbb N\mid f(d_k)=b\}$. You should verify that this is indeed an injective function.

Recall that if we have an injective function from $A$ into $B$ then $|A|\leq |B|$ and if $|B|$ is countable then $A$ must be countable (or finite). From here the proof is about done.

In fact this depends on your definition of countable or finite: If your definition of countable or finite is "has an injection from $A$ into $\mathbb N$" you can prove that this is equivalent to "there is a surjection from $\mathbb N$ onto $A$" via a similar argument to what I wrote above.

Using the second definition you can simply argue:

Since $D$ is countable (or finite) there is some $g\colon\mathbb N\to D$ which is a surjection, therefore $f\circ g\colon\mathbb N\to f(D)$ is a surjecitve function and therefore $f(D)$ is countable (or finite).

A complete account of the proof of the equivalent definitions of countability can be found here.

You can't have more elements in the image of a set under a function then in the domain-that's basically the point of Gaston's post above. Here's another way to express it so that the proof is clearer: Consider the Cartesian product definition of a mapping from a countable set D into a set E where the image of f is a subset of E. (I know,duh-but when you're trying to understand in mathematics why something is true, it's worthwhile to state the obvious so you make sure you understand the definitions.) A function by definition is a set of ordered pairs where no 2 different ordered pairs have the same first member. So ask yourself something and the proof will be clear: Can you construct such a set of ordered pairs representing f:D ----> E if f(D) is uncountable?


Can $|m\alpha+n\beta|$ be made arbitrarily small? I wondered is that always true that if $\alpha$ and $\beta$ are non-zero real numbers, then can we make $|m\alpha+n\beta|$ arbitrarily close to zero, for some non-zero integers $m$ and $n$. My guess would be "no", but I couldn't come up with a simple counter example. Sorry in advance if there is a very simple one. (That's easily seen to be true when $\alpha$ and $\beta$ are rational, since it can be made equal to zero! But, I somehow feel like it shouldn't be true for all reals. I first tried to prove it's true considering density of rationals, but I think I failed, that's why I believe it's not true.) Thanks a lot!

[Standard Argument]:  Suppose, to the contrary, that you can not get near zero (non-trivially).  We want to prove that the ratio $\frac{\alpha}{\beta}$ is rational.  Look at the set $$S = \{m\alpha + n\beta, m,n \in \mathbb Z\}$$ Clearly that set is closed under addition and closed under multiplication by integers.  We note that if there are any real accumulation points for S then 0 is an accumulation point. (Pf: differences between elements of S are also elements of S and if some sequence drawn from S converged to L, say, the elements in that sequence would draw arbitrarily close to each other.).  Thus if S does not have 0 as an accumulation point there is a least positive element of S, call it A.  But then all elements of S are multiples of A (else the division algorithm would hand us a smaller positive element of S).  in particular $\alpha$ = MA and $\beta$ = NA for integers M and N.  But then $\frac{\alpha}{\beta}$ = M/N .

This is not true in general.

Take $\alpha=\beta=1$ for a counterexample. You can't find a nonzero sequence of elements of the set converging to zero.

However it is true if you take for example $\alpha=1$ and $\beta=\sqrt{2}$.

$$G=\{m \alpha + n \beta | (n,m) \in \mathbb Z^2\}$$ is a subgroup of the reals. It is a classical result that the subgroups of the real are either discrete or dense.


if a bookmaker quotes pay off odds of 99 to 1 against a particular horse... if a bookmaker quotes pay off odds of 99 to 1 against a particular horse winning a race, does that suggest that the chance that the horse will win is 1/100, less than 1/100 or more than 1/100?

I tried doing this on my own but saying "Oh it's subjective so it depends on whether or not we trust the bookmaker's intuition" and then I realized I was messing up the definitions between chance and odds. Can someone please help me this?

Thank you so much in advance.

odds against = unfavorable occurrences : favorable occurrences,

whereas probability = favorable occurrences / total occurrences,

thus odds against of $\;99:1 = \dfrac1{99+1} = \dfrac1{100}$

99 to 1 odds just means money is staked more in favor of other horses.  This does not necessarily reflect the actual probability of a horse winning a race.

If you assumed the odds were a literal interpretation of the actual statistical outcome then 99 to 1 would mean 1/100 probability.
1/100 = 1/100

You can find more info here about it in the link below.

Odds


Independence of axioms What does it really mean for a set of axioms to be independent? In particular, say there are two persons $A$ and $B$, who assume and do not assume the axiom of choice respectively. So, $A$ can prove that every vector space has a basis. However, if $B$ takes an arbitrary vector space and gives it to $A$, then $A$ can produce a basis. So can't $B$ just use the basis that $A$ just produced? But then, every vector space would have a basis even if the Axiom of Choice were not assumed.

For a set of axioms $T$ to be independent it means that if $\varphi\in T$, then $\varphi$ is not a logical consequence of $T\setminus\{\varphi\}$.

In the case of the axiom of choice, and the rest of set theory, $\sf ZF$, you're mistaken. If we are in a universe without the axiom of choice, we cannot even ensure that $\Bbb R$ has a basis as a vector space over $\Bbb Q$. You are thinking about "passing the reals to another person", but you're invariably changing the universe in which you are working when you assume choice.

Even worse, set theory has taught us that given any set, it can be a countable set in a slightly larger universe of mathematics. And indeed every countable vector space has a basis. So given any vector space, we can always go to a larger universe where it has a basis. This is really not the point. The point is that in a given universe of mathematics, a vector space may or may not have a basis. Assuming different background theory means that you might be working in a different universe with more—or less—sets.

Here is a simpler example.

What does it mean that $\sqrt2$ "exists"? You are thinking about "all the mathematical universe", but if you are just given an ordered field $F$, you don't know a priori that $2$ has a square in $F$. It might be the rational numbers, where this is false and it might be the real numbers where this is true.

To make matters worse, we can still talk about $\sqrt2$ even if it doesn't exist, as it is the unique positive solution for $x^2-2=0$.

So does $\sqrt2$ exist? Again, we come to the problem of "what is mathematical existence?", and the answer is that existence if relative to some fixed universe of interest. In $\Bbb Q$, $\sqrt2$ does not exist; but in $\Bbb R$ it does exist.

Going back to set theory, even if you have the same vector space in two different universes of set theory, its subsets might be different. In particular, it might be that in one universe it has a basis and in another it does not. (And to add insult to injury, there is also the matter of confusing the object with its definition, which is a whole other story and makes matters far more difficult to grasp intuitively.)

I have the impression the other answers don't go to the root of the misunderstanding so I will try to contribute this viewpoint.  The real problem here is realist views concerning the existence of mathematical objects.  If sets and vector spaces are really out there, then once person A produces a basis for a vector space, what's to stop person B from using it?  To understand what is going on you have to step back and realize that such presumed existence is a fairy tale fed by beliefs in the so-called intended interpretation. Once viewed from such a distance the OP may be able to follow the remarks concerning the dependence of properties of sets on the universe one is in.


Find the recursive definition for the number of strings on 0, 1, 2 avoiding the substring 012? This is the question $a(n)$ the number of strings on $0, 1, 2$ avoiding the substring $012$ and the answer is $$a(n)=3a(n−1)−a(n−3)$$  with $$a(0)=1,a(1)=3,a(2)=9$$ My question is how to you get this recursive function? I have an exam on Thursday and I am struggling with the concept of finding the recursive definition for strings. Can someone explain how this works and give few examples to make it clear? Thank you.

Given a string of length $n-1$, you can add any of $0$, $1$, or $2$ to the end of it to get a string of length $n$, so you get $3 a(n-1)$ strings. However, if the last two elements of the original string were $01$, then adding $2$ is not allowed. How many such strings are there? Well, they are found by noting that after removing the $01$ you have an arbitrary string of length $n-3$. So you must subtract $a(n-3)$ to avoid counting these.

More visually,
\begin{align*}
  \underbrace{...0}_{n-1} &\quad\Rightarrow\quad ...00,\ ...01,\ ...02 \\
  \underbrace{...1}_{n-1} &\quad\Rightarrow\quad ...10,\ ...11,\ ...12 \\
  \underbrace{...2}_{n-1} &\quad\Rightarrow\quad ...20,\ ...21,\ ...22
\end{align*}
so that every length $n-1$ string gives nine length $n$ strings. If the original string didn't contain $012$, neither will the new one, except for the last entry on the second row. If that length $n-1$ string ended in $01$, then we will have created a $012$ in the new string. But if that string ended in $01$, then the other digits in the string (there are $n-3$ of them) are arbitrary. So there are $a(n-3)$ ways in which we could have erroneously added the $012$. We must subtract that to get the right answer.

So for example $a(3) = 3a(2) - a(0) = 26$, $a(4) = 3a(3) - a(1) = 75$.

$a_n=2a_{n-1}+b_{n-1}$

$b_n=b_{n-1}+c_{n-1}+a_{n-1}$

$c_n=a_{n-1}+b_{n-1}$

Explanation:

-First relation:

$2a_{n-1}$ indicates sequences starting by 1... or 2... while the relation $a$ is for unconstrained sequences

$b_{n-1}$ indicates 0... while $b$ is a sequence right-bounded by a $0$

-Second relation:

$c_{n-1}$ indicates 01... while $c$ is any sequence right-ended by $0...1$

$b_{n-1}$ indicates 00... while $a_{n-1}$ indicates 02...

-Third relation:

$a_{n-1}$ indicates 011... and $b_{n-1}$ points to 010...

Example:n=3 (xxx)

Naive enumeration says $S=3*3*3-1=26$ where the subtracted case is 210

Recursive enumeration:

$a_3=2a_2+b_2=2(2a_1+b_1)+b_2=2(2(2a_0+b_0)+b_1)+b_2=8a_0+4b_0+2b_1+b_2$

$b_1=b_0+c_0+a_0=3$

$b_2=b_1+c_1+a_1$

$c_1=a_0+b_0=2$

$a_1=3$

$b_2=3+2+3=8$

$a_3=8+4+8+6=26$

Example2:n=4 (xxxx)

-From informations above, $a_3=26,a_2=9,b_2=8,a_1=3,b_1=3$

$a_4=2a_3+b_3=2*26+b_3$

$b_3=b_2+c_2+a_2=8+c_2+9$

$c_2=b_1+a_1=3+3=6$

$b_3=23$

$a_4=52+23=75$ which matches previous results.