PYTHON RK2 (Midpoint Method) Good evening, I am writing code for a Numerical Analysis Project, but I am having difficulty iterating the RK2 (Midpoint Method) Correctly. I am using Python to do it, could anyone take a look at my code and see what I am doing wrong?

from my understanding when running the code something, is wrong with the array y that I create to put the iteration points into that will be later used for plotting.

Meaning in the array the initial point y[0] and y[1] are populated but after that, all values just go to zero. I'm not sure what is exactly wrong.

Here is my Code:

    import packages needed
    from __future__ import division
    `import numpy as np`
#===============================

def Midpoint_meth(def_fn, a, b, N, ya):

    """
        Test the Midpoint Method to solve
        initial value problem y'=f(t,y) with t in [a,b] and y(a) = ya.
        Step size h is computed according to input number of mesh points N
    """

    f = def_fn #input definining function

    h = (b-a)/N # developing step size h, from input values divided by N

    t = np.arange(a, b+h, h) #array intialized to hold mesh points t
    y = np.zeros((N+1,))      #array to hold Midpoint Method approximated y values

    y[0] = ya   #intial condition

    #iterative method
    for  i in range(0, N):

        tau = t[i]      #current mesh point t
        w = y[i]        #current value y(t)

        k1 = f(tau,w)
        h1 = h/2.0

        # next iteration using midpoint method
        y[i + 1] = y[i] + h * f(tau + h1, w + (h1*k1) )

        return (t, y)

trying to show:

$$\ w_i=\alpha$$
$$\ w_{i+_1}= \ w_i + hf\biggl((\ t_i + {\frac h2}, \ w_i + {\frac h2} f(\ t_i, \ w_i)\biggr) $$

You need to be careful with indentations in python. It does not matter how many empty lines you put between them, lines with the same indentation level belong to the same code block.

In your case, the return statement is on the same indentation level as the body of the loop. Thus you insert y[0] on initialization, y[1] during the first execution of the loop body and directly return after that computation.

In general it helps to use a debugger or insert print statements (that you later remove) at important way points to follow the way of the execution. For instance in the loop

print 'loop',i, t[i], y[i]

def RK2(f, y0, tspan):
    N = len(tspan)
    h = (tspan[N-1] - tspan[0])/(N - 1)
    d = len(y0)
    y = np.zeros((N, d), dtype=type(y0[0]))
    y[0] = y0;
    for j in range(0, N-1):
        tj = tspan[j]
        tnp1 = tspan[j+1]
        yj = y[j]
        fj = f(yj, tj)
        ybar = yj + fj*h
        fjbar = f(ybar, tnp1)

        y[j+1] = yj + 0.5*(fj + fjbar)*h
    return y

```


sum of infinite roots? What is the sum of this infinite series of roots:
$$\sqrt{1+\sqrt{2+\sqrt{3+\sqrt{4 + \cdots+\sqrt{\infty}}}}}$$

This is an interesting expression because the increase created by the addition of the next term is offset by the square root introduced with the next term. the key is determining what the relationship between these two is exactly

I know it converges on an irrational number around 1.75 as I have tried computing it for a few small iterations. But I was wondering how to algebraically simplify this expression.

I have tried many things and ways to manipulate the expression but to no avail.

This is the Nested Radical Constant
, which does converge, but whose closed-form expression is still unknown. See also Somos's Quadratic Recurrence Constant
 for a similar constant, involving their product.

$\newcommand{\+}{^{\dagger}}%
 \newcommand{\angles}[1]{\left\langle #1 \right\rangle}%
 \newcommand{\braces}[1]{\left\lbrace #1 \right\rbrace}%
 \newcommand{\bracks}[1]{\left\lbrack #1 \right\rbrack}%
 \newcommand{\ceil}[1]{\,\left\lceil #1 \right\rceil\,}%
 \newcommand{\dd}{{\rm d}}%
 \newcommand{\ds}[1]{\displaystyle{#1}}%
 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}%
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}%
 \newcommand{\fermi}{\,{\rm f}}%
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}%
 \newcommand{\half}{{1 \over 2}}%
 \newcommand{\ic}{{\rm i}}%
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}%
 \newcommand{\isdiv}{\,\left.\right\vert\,}%
 \newcommand{\ket}[1]{\left\vert #1\right\rangle}%
 \newcommand{\ol}[1]{\overline{#1}}%
 \newcommand{\pars}[1]{\left( #1 \right)}%
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}%
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,#2\,}\,}%
 \newcommand{\sech}{\,{\rm sech}}%
 \newcommand{\sgn}{\,{\rm sgn}}%
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}%
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
Hint:
$$
a_{1} = \root{1 + \root{a_{2}}}\,,\quad
a_{2} = \root{2 + \root{a_{3}}}\,, \ldots\,,
a_{n} = \root{n + \root{a_{n + 1}}}
$$
Then, $\ds{a_{n + 1} = \pars{a_{n}^{2} - n}}^{2}$. $\tt\mbox{Calculate}$ $a_{1}$.


Is the equation of a circle a function? We know, for a circle of radius 1 centred at the origin: $$ y^2 + x^2 = 1$$
Now when I saw the graph of the given equation- any x-value in within the domain has two images, but how can that be?
Since I've read that in a function any pre-image cannot have two images.

And now the above can be equivalently written as, $$y = \sqrt{1-x^2} $$ But now, surprisingly(for me), any x-value doesn't have two images!
But why is that?

Because there is also $$y=-\sqrt{1-x^2}.$$

The circle with equation $x^2+y^2=1$ it's not a graph  of a function.

We can understand it by another algebraic way.
$$\left(\frac{3}{5},\frac{4}{5}\right)$$ and
$$\left(\frac{3}{5},-\frac{4}{5}\right)$$
placed on the graph and this is a contradiction with a definition of the function.

Something is a function if, and only if, there is only one value of y for each value of x. That's the definition of function.

Why is so? Because the idea of function is that you can relate two values in a determinist way, so that you know the value of the dependent variable if you are given the value of the independent one. If however there are two
(or more) different possible values for y for a given value of x, then y doesn't actually depend exclusively on x, ie, it isn't a "function of x".


In a group of order $pq^2$, if p divides $|Z(G)|$, then G is abelian I want to prove that in a group of order $pq^2$, if $p$ divides $|Z(G)|$, then $G$ is abelian.

I know that in this case, the order of $G/Z(G)$ must be $q^2$.

If $G/Z(G)\simeq \mathbb{Z}/{q^2}\mathbb{Z}$, then we are done, because $\mathbb{Z}/{q^2}\mathbb{Z}$ is cyclic.

But what happens if $G/Z(G)\simeq \mathbb{Z}/{q}\mathbb{Z}\times \mathbb{Z}/{q}\mathbb{Z}$? I think I'm missing something trivial..

You do not know that $G/Z(G)$ must have order $q^2$. In fact you are try to prove $G$ is abelian, which implies $Z(G) = G$ so you are actually trying to show $G/Z(G)$ has order $1$.

Note that since $p$ divides $|Z(G)|$, a sylow $p$, $S_p$ is central. Thus $1 \to S_p \to G \to G/S_p \to1$ is a short exact sequence with $S_p$ central and $G/S_p$ nilpotent (order $q^2$ is abelian). It follows that $G$ is nilpotent, and hence is the product of its Sylow subgroups $G = S_p \times S_q$ and $S_q$ has order $q^2$ and is hence abelian.

Clearly $|Z(G)|=p$ and index$[G:Z(G)]=q^2$, so $Z(G)$ is normal Hall-subgroup. By the theorem of Schur-Zassenhaus it has a complement $H$, so $G=HZ(G)$, $H \cap Z(G) =1$. Hence $|H|=q^2$ and $H$ is abelian. But then $G=HZ(G)$ must be abelian.

Note without Schur-Zassenhaus: in general if $G/Z(G)$ is abelian, then $G$ is nilpotent. And hence $G$ is the direct product of its Sylow subgroups. In this case it follows that $G$ is abelian. By the way note that you must assume that $p \neq q$, otherwise the statement of the OP is not true.


Area of a trapezoid with perpendicular diagonals, embedded in a triangle Let $ABED$ be a trapezoid as in

If $AB \parallel DE$, $AE \perp BD$, $AB = 10$, $DE = 4$ and $\angle ACB = 45°$, what's the area of $ABED$?

I must also mention that this is for an elementary geometry course, so I have restricted myself to that approach --I have not attempted anything related to calculus or analytic geometry for the same reason.

I have tried the following: (1) Using triangle similarity to get the height of the trapezoid (2) Using sine or cosine law to the same end (3) Trying to get the length of its diagonals with similar approaches as above (4) Using some weird formula for the area of a triangle by multiplying its three sides and dividing it by the circunradius. None of the above worked. Every time I get equations with more than one variable, and can't figure out a way to get a system that would help me solve them.

Thank you all for any help!

The hint.

Let $FE=x$, $DF=y$, $L\in AB$ such that $DL||EB.$

Thus, $$x^2+y^2=16,$$
$$AF=2.5x,$$ $$BF=2.5y$$ and since $DL=EB$, $AL=10-4=6$ and $\measuredangle ADL=45^{\circ},$ by low of cosines we obtain:
$$AD^2+LD^2-2AD\cdot LD\cos45^{\circ}=AL^2$$ or
$$6.25x^2+y^2+6.25y^2+x^2-2\sqrt{(6.25x^2+y^2)(6.25y^2+x^2)}\cdot\frac{1}{\sqrt2}=36.$$
After solving of this system use $$S_{ABED}=\frac{1}{2}AE\cdot DB.$$

I got $xy=\frac{160}{21}$ and $S_{ABED}=\frac{140}{3}.$

If $\measuredangle FAB=\theta^{\circ}$ then $AE = 14 \cos \theta$, $DB = 14 \sin \theta$, and the area we require is $98 \sin \theta \cos \theta$. I can furthermore construct such a diagram for any $\theta$ (construct $AB$ then $F$ then $E$, $D$ and $C$).

The problem is underspecified.

eta: Apologies, I was not seeing the other part of the specification. The angle at $C$ is given. That makes the problem solvable.


How to prove that $\frac{d}{dx} \left(\sin^2(x)+\cos^2(x)\right)=0$ I have seen in a mathematics.stackexchange.com thread that to prove that $\sin^2x+\cos^2x=1$ one have to prove that the derivative of $\left(\sin^2(x)+\cos^2(x)\right)$ is $0$.

But how can we prove it?

$$
\begin{align}
\frac{d}{dx} (\sin x)^2 & = 2\sin x\cdot\frac{d}{dx} \sin x = 2\sin x\cos x \\[15pt]
\frac{d}{dx} (\cos x)^2 & = 2\cos x\cdot\frac{d}{dx}\cos x = 2\cos x\cdot(-\sin x)
\end{align}
$$

Now add those together.

I am assuming only the trigonometric definitions of $\sin$ and $\cos$ functions.
 $\sin^2x+\cos^2x=1$ can be proved$^1$ by showing that $\dfrac{d}{dx}(\sin^2x+\cos^2x)=0$. This identity is equivalent to pythagorean theorem. Pythagorean theorem in turn can be proved with the use of calculus, e.g. see here($\angle EDC$ tends to $\angle BCA$ as $\Delta x$ tends to $0$).

$^1$ By applying chain rule we can show that $\dfrac{d}{dx}f(x)=0$ where $f(x)=(\sin^2x+\cos^2x)$. We know if the derivative of a function is $0$ then the function is a constant.

So $\dfrac{d}{dx}(\sin^2x+\cos^2x)=0 \implies (\sin^2x+\cos^2x)= constant\ \  \forall\ x$.
At $x=0$, $=(\sin^20+\cos^20)=(0+1)=1$.

Now  we have computed the value of $(\sin^2x+\cos^2x)$ for a certain value of $x$(namely $x=0$), this value will remain same for every other value of $x$, since $(\sin^2x+\cos^2x)$ is a constant function.
  So $(\sin^2x+\cos^2x)=1\ \forall\ \ x$.


Can $S^n$ be defined without reference to an $n+1$ dimensional embedding space? It seems like the most common definition is the set of all vectors in $\mathbb{R}^{n+1}$ which are unit length. Is this a necessary definition?

If you just want something homeomorphic to $S^n$, you can use the one-point compactification of an open ball in $\Bbb R^n$.

Here's an equivalent definition: $S^n$ is the unique compact, simply connected $n$-dimensional manifold of constant, positive sectional curvature $+1$. That gives you the sphere with the metric you'd get from the embedding in $\mathbb{R}^{n+1}$.


Why is the gradient of this expectation intractable? I am learning policy gradient from slides of Stanford CS231 reinforcement learning

\begin{align}
\tau &= (s_0, a_0, r_0, s_1, a_1, r_1, ...) \\
J(\theta)&=\mathbb{E}_\tau [r(\tau)] \\
&=\int_\tau r(\tau) p(\tau;\theta)d\tau \\
\nabla_\theta J(\theta) &= \int_\tau r(\tau)\nabla_\theta p(\tau;\theta)d\tau
\end{align}

Can anyone tell me why the last integral is intractable?

You said about the slide below from CS231n?
(http://cs231n.stanford.edu/slides/2017/cs231n_2017_lecture14.pdf)

enter image description here

  Why is the gradient of this expectation intractable?

This is because (in most case) the dimension is too high.

If you calculate the integral by MC, you have to uniformly sample from the high dimensional space. In most case, it's intractable.

It is not the integral itself not being mathematically solvable or something, you have to understand it w.r.t. context (i.e. Monte Carlo Simulation).

If i'm recalling it right, this is the math basis of Reinforce Algorithm, in which case you want to  maximize the the object function through gradient update with Monte Carlo Simulation -- however, here comes the problem, in the last integral, there is no p term readily available s.t. you can write it into expectation form, so the missing piece here is only the p term, but you can do some math and extract such p term in the integral and transform it into some expectation  w.r.t. policy distribution.

After you have the nice expectation form of gradient for update in Monte Carlo setting, you are good to go.


How to show steps for 11x - 7 $\equiv$ 0 (mod 13)? I know the answer is $x = 3$, but I'm having trouble showing the steps for it. All I have is $11x \equiv 7 (\mod 13)$.

There are general methods here based on Euclid's division algorithm, and these always work. You can look that up.

Sometimes there are short cuts you can spot, as here. Don't forget that there are negative numbers in congruence classes. So here, modulo $13$ $$11x\equiv -2x$$ and you then only need to divide by $2$. So $7\equiv -6$ or $7\equiv 20$ get you to an even residue.

Whence the equation can be rewritten as $-2x+6\equiv 0$ with $x\equiv 3$ by simple division ($2$ and $13$ are coprime), or you could write $-2x-20\equiv 0$, from which you obtain $x\equiv -10\equiv 3$.

The problem 11x − 7 ≡ 0 (mod 13) is equivalent to 11 x0 − 7 = 13 x1 where x0, x1 ∈ . One way to solve the linear equation would be to express the original variables (x0, x1) in terms of parameter variables (t0, t1) where each of the parameter variables take on any value from . To
find such as expression you can judiciously introduce the parameter variables and corresponding equation one at a time.

After generating the additional equations (the ones with parameter variables) you can take them plus the original equation as a system on linear equations. The system of equations can be solved using back substitution. After each back substitution step if the resulting equation has a greatest common factor then the greatest common factor should be removed before continuing.

The PDF document "Showing the Steps to Solve 11x − 7 ≡ 0 (mod 13)" applies these ideas in a step-by-step fashion. The following diagram is a snap shop of the PDF page:


$M= \{ A \in Mat_{2 \times 2}{\mathbb{R}}| \det(A)=1 \}$ is homeomorphic to $S^{1} \times \mathbb{R}^{2}$ Let's consider a group $M$ (under multiplication) of all matrices $A$ of size $2 \times 2$ over $\mathbb{R}$ so that $\det(A)=1$. How to show that the group is homeomorphic to the $S^{1} \times \mathbb{R^{2}}$?

Topology on $M$ is induced by the norm $||A||=\sqrt{x_{1}^{2}+x_{2}^{2}+x_{3}^{2}+x_{4}^{2}}$, where $A = \begin{pmatrix}
x_{1} && x_{2}\\
x_{3} && x_{4}\\
\end{pmatrix}$
so it's the same as considering matrix as a point in $\mathbb{R^{4}}$. According to $Q = S^{1} \times \mathbb{R}^{2}$,, the topology on $Q$ is induced by the standart one from $\mathbb{R}^{4}$.

The common idea is to start with considering the $x_{1} x_{4} - x_{2} x_{3}=1$, but i can not find some rigorous ways how to conclude that it's precisely homeomorphic to $Q$.

Any help would be much appreciated.

The group $M$, which I am going to rename $G$, acts transitively on $\mathbb{R}^2 \setminus \{(0, 0)\}$ which is clearly homeomorphic to $\mathbb{R} \times S^1$. The stabilizer of the vector
$$
\begin{pmatrix} 1 \\ 0 \end{pmatrix}
$$
is the set of matrices of the form
$$
\begin{pmatrix} 1 & * \\ 0 & 1 \end{pmatrix}
$$
which is clearly homeomorphic to $\mathbb{R}$.

If you've seen group actions, that's a complete answer.

EDIT: I translated this into a language not using group actions in my previous answer, but made quite a mistake (the equations I wrote always admit a trivial solution).

This is not true !
Let $x_1=a+b$, $x_2=u+v$, $x_3=u-v$ and $x_4=a-b$. Then, $a^2-b^2-u^2+v^2=1$. Let $b=r\cos t$, $u=r\sin t$, then $a^2+v^2=1+r^2$. Let $a=\sqrt{1+r^2}\cos s$ and $v=\sqrt{1+r^2}\sin s$. Therefore, $\mathrm{SL}(2,\mathbb{R})$ can be parametrized by $t,s\in\mathbb{S}^1$ and $r\in\mathbb{R}^+$, and is diffeomorphic to $\mathbb{S}^1\times\mathbb{S}^1\times\mathbb{R}^+$.


What went wrong? [One-dimensional-inverse-square-law] Intrigued by this question,
one-dimensional inverse square laws,
I started to try to find an answer and came up with what follows. However, I calculated the derivatives to double check myself, and this does not work. However, it seems straightforward enough that I do not see my mistake.

Question: Is there a closed form solution to all

$$ \ddot x = \frac{k}{x^2}  \quad (x(0)=x_0), \  \dot{x}(0) = v_0, \ k \in \mathbb{R}.$$

Here is my very naive two cents. Multiplying both sides by $\dot{x}$ and obtain,

$$ \dot{x} \ddot{x} =  \frac{k \dot{x}}{x^2}.$$

Integrating both sides,  we obtain
\begin{equation} \tag{1} \frac{ (\dot{x})^2}{2} = -\frac{k}{x} + C. \end{equation}

Here it seems natural to re-write this as,
\begin{equation} \tag{2} \dot{x} = \sqrt{2} \sqrt{C - \frac{k}{x}}. \end{equation}
I do not know if that is fully acceptable. However, if it is we then have via integration one more time, or rather letting WolframAlpha integrate one side,

$$x(t) = \sqrt{2} x \sqrt{C - \frac{k}{x}} - \frac{ k \log \left( 2 \sqrt{C} x \sqrt{C - \frac{k}{x}} + 2Cx - k \right)}{\sqrt{2C}} + \tilde{C}. $$

Note, that both $C$ and $\tilde{C}$ are determined by initial conditions. Using the given initial conditions and (1), we can solve for $C$, to find that $C = \frac{v_0^2}{2} + \frac{k}{x_0}.$ Similarly we could solve for $\tilde{C}$, but before I did that I discovered my solution was wrong.

If you substitute my "solution" back into the differential equation, you find (much to my dismay)

$$ \ddot{x} = \frac{k}{x^2 \sqrt{2c - \frac{k}{x}}}. $$

Insight on my  mistake and/or a proper way to solve this is greatly appreciated.

Your manipulations are quite nice, and I enjoyed reading them, but I thought I'd point out they can be put in the language of conservation laws. You started with a differential equation

\begin{align*}
\ddot x(t) = F(x(t)) && x(0) =x_0 && \dot x(0) = v_0
\end{align*}

where $F(x)$ can be thought of as a "force field" on some 1D domain in $\mathbb{R}$. Now let $V = - \int F$ be a potential function (the choice of antiderivative turns out not to affect anything).  Basically, you took $\ddot x = F(x)$, wrote it as $ \dot x \ddot x = \dot x F(x) $ and used the chain rule to identify that with $\frac{d}{dt}\left( \frac{ \dot x^2}{2}  \right)= - \frac{d}{dt} V(x)$ or, in other words,
$$ \frac{d}{dt} \left( \frac{\dot x^2}{2} + V(x) \right) = 0$$
which says that the kinetic energy $K = \frac{\dot x^2}{2}$ and the potential energy $V(x)$ add to a constant $E$ which is fruitfully interpreted as the total energy of the system
$$ \frac{\dot x^2}{2} + V(x) = E.$$
We can calculate $E$ straight away from the initial conditions:  $E = \frac{1}{2} v_0^2 + V(x_0)$. Solving for the velocity, we get
$$ \dot x = \pm \sqrt{2(E-V(x(t))) }$$
which is a 1st order, automomous (in particular, separable) ODE. According as the intial velocity $v_0$ is positive or negative, we can safely assume that $\dot x(t)$ will retain its sign for a short time afterwards and remove the $\pm$ sign. For instance, if we suppose $v_0 > 0$, then, for sufficiently small $t$, we apply the usual technique of separating variables:
$$ t = \int_0^t \ ds =  \int_0^t \frac{ \dot x(s) }{ \sqrt{ 2(E-V(x(s)))} } \ ds = \int_0^{x(t)} \frac{1}{ \sqrt{2(E-V(x))}} \ dx$$
which yields (assuming the integral can be solved) an equation relating $x(t)$ and $t$. In the case of the inverse square attractive force, we have $F(x) = \frac{-k}{x^2}$ whence $V = -\frac{k}{x}$ is a potential function so the integral becomes
$$ t = \int_0^{x(t)} \frac{1}{ \sqrt{2\left( E+ \frac{k}{x} \right)}} \ dx $$
which is a bit beyond me, but not beyond most integration packages.

With the non bound states (K>0, K<0 and E>0) the solutions are not periodic. The complicated logarithmic relation between t and x, derived in previous posts, applies to these cases. If K<0 (attractive inverse square case)and E<0 then we can have bound states oscillating about the origin; i.e. we get periodic behavior. (don't worry about negative energy this just means it takes work to remove the particle from its bound state to infinity which is the reference zero of energy). The solution for this case follows the same method as reported by mikeF; when E= -|k|/x_a , where $x_a$ is the amplitude of the oscillation in the potential well. For a particle of mass m the integral for t can be done explicitly using the techniques shown earlier. The answer for the attractive inverse square law is

$$\gamma t =   [arctan( \sqrt{\frac{x}{x_a-x}}) - \frac{\sqrt{x(x_a-x)}}{x_a}].$$

where $\gamma = \sqrt {\frac{2|k|}{m.x_a^3}}$ and $x_a=\frac{-2|k||x_o|}{mv_o^2|x_o|-2|k|}= \frac{|k|}{|E|}$

We can use this to find the period, T, of oscillation in such a 1-d potential: If t is the time to go from x=0 to $x = x_a$ then $T= 4t= 2\pi.\frac{1}{\gamma}= \sqrt{\frac{m.x_a^3}{2|k|}}.$

This is the one dimensional analogue of Kepler's third law which says that the period of oscillation squared is proportional to the amplitude cubed and is a consequence of the inverse square law.


How to prove a matrix not invertible? How can I prove that A*B is not invertible when A (mXn) and B (nXm)

while (m>n)?

Hint: A Matrix is invertible iff it is square and has full Rank

HINT: Complete $A$ to $\tilde A$ by adding  on the right $m-n$ columns consisting of $0$'s, and $B$ to $\tilde B$ by adding below $m-n$ rows consisting of $0$'s. Observe that $A\cdot B = \tilde A \cdot \tilde B$.
Now clearly $\det \tilde A = \det \tilde B = 0$, so $\det AB = 0$.


How would you interpret the following statement involving "a.e."? Here is an edited fragment from an exercise:

  Let $(X, \mathcal A, \mu)$ be a measure space, $(f_n)$ be a sequence of such and such functions. If $f(x)= \lim f_n(x)$ exists for almost every $x\in X$ then $f$ has such and such properties (in particular $f$ is measurable).

I'm confused about the role of $f$. Is it already given in the statement? E.g. if in the proof, one would set $f$ to zero on some convenient set of measure zero, it wouldn't be OK? (A proof which I encountered does exactly that.)

The question probably has little to do with measure theory but that a general "if something shows up in the assumption, must it be fixed from there on?" but I've added the context just in case.

If the measure space is not complete, then following statement is, in general, not correct.

  Let $(f_n)_{n \in \mathbb{N}}$ be a sequence of measurable functions. If $f(x) = \lim_{n \to \infty} f_n(x)$ exists for almost every $x \in X$, then $f$ is measurable.

Just pick a non-measurable set $N$ such that there exists $M \supseteq N$ measurable with $\mu(M)=0$ (such a set $N$ exists if the measure space is not complete). Then $f_n(x) := 0$ converges for almost every $x \in X$ to $f(x) := 1_N(x)$, but $f$ is not measurable.

However, the following statement holds true:

  Let $(f_n)_{n \in \mathbb{N}}$ be a sequence of measurable functions. If $f(x) = \lim_{n \to \infty} f_n(x)$ exists for almost every $x \in X$, then $f$ has a measurable modification, i.e. there exists a measurable function $\tilde{f}$ such that the set $\{x; \tilde{f}(x) \neq f(x)\}$ is contained in a $\mu$-null set.

Note that the function $\tilde{f}$ satisfies in particular $$\tilde{f}(x) = \lim_{n \to \infty} f_n(x)$$ for $\mu$-almost all $x \in X$.

Changing $f$ on a set of measure zero does not affect almost convergence to it. The converging sequence $f_n$ may or may not converge for those newly assigned values but it does not really care, as long as it maintains convergence on the complement of a measure-zero set.


Derivation of Lagrange-Charpit Equations I am working through the derivation of the Lagrange-Charpit equations presented in this Wikipedia article: http://en.wikipedia.org/wiki/Method_of_characteristics#Fully_nonlinear_case

I am interested in the "fully" nonlinear case. Where we have:

$$F(x_1,...,x_n,u,p_1,...,p_n)=0  $$
And $$ p_i=\frac{\partial u}{\partial x_i} $$

I am fine with the derivation up until the point where they say that (Also, $\dot{x_i}=dx_i/ds$):

$$\sum_i(\dot{x}_idp_i-\dot{p_i}dx_i)=0 $$

Follows from taking the exterior derivative of:

$$ du-\sum_ip_idx_i=0  $$

From the little I know about exterior derivatives, it seems like doing this would give (for the two dimensional case):

$$0=\left(\frac{\partial p_2}{\partial x_1}-\frac{\partial p_1}{\partial x_2}\right)dx_1\wedge dx_2 $$

Because $ddu=0$ and the anti-symmetry of the wedge product.  I don't see how this result leads to the one given.  Could someone help me out? I feel like there is something very simple that I am missing.

$ \newcommand{\pd}[2]{ \frac{\partial #1}{\partial #2} } $
Attending to the positive answer to my request from the OP, consider the non-linear 1st order PDE:

$$F(x_i,u,p_i) = 0, \quad i = 1,\ldots, n, \quad p_i = \pd{u}{x_i}, \quad u = u(x_1,\ldots,x_n). \tag{1}$$

Assume $F \in \mathcal{C}^1_{x_i}$ and compute the partial derivative of eq. $(1)$ with respect to $x_i$:

$$\pd{F}{x_i} + \pd{F}{u}\pd{u}{x_i} + \pd{F}{p_i}\pd{p_i}{x_i} = 0,$$

or equivalently:

$$\pd{F}{x_i} + p_i \pd{F}{u} + \pd{F}{p_i}\pd{p_i}{x_i} = 0,$$

which is a quasilinear PDE for $p_i$ which can be readily solved leading the set of equations known as Lagrange-Charpit equations:

  $$ \color{blue}{\large{\frac{\mathrm{d}x_i}{\pd{F}{p_i}} = -\frac{\mathrm{d}p_i}{\pd{F}{x_i}+p_i \pd{F}{u}} = \frac{\mathrm{d}u}{\sum_ip_i \pd{F}{p_i}}}}   $$

where the last equality comes from the fact $\sum_i p_i \mathrm{d}x_i = \mathrm{d}u. $

Hope this helps.

Cheers!

let the general pde be f(x,y,z,p,q)......1
since z depends on (y)and(x) then,
dz=dz/dx.dx+dzdy.dy
dz=pdx+qdy...............................2
let another relation be Q(x,y,z,p,q)......3
on solving 1 and 3 we get the value of p and q
to determine Q we differentiate w.r.t x and y.
then,eliminate dp/dx


How i show this beautiful inequality :$\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}\geq \frac{3} {2}(\frac{1}{\sqrt{3}})^{n-m}$? let $m,n$ be integers, show that if $ n>m\geq 0 $ :

$$\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}\geq \frac{3}
{2}\left(\frac{1}{\sqrt{3}}\right)^{n-m}$$

where real $x,y,z > 0 $ and $xy + yz + zx = 1$

Thank you for your help .

Despite the attempts in the other answers and comments, and the two ``proofs'' given there, the inequality does not hold in general. We give two proofs for its failure:

First Proof (involving huge numbers): Set $x=3/7$, $y=4/7$, $z=37/49$. Then $xy+yz+zx=1$. For $m=7$, $n=8$ set $A=\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m}$ and $B=\frac{3}{2}(\frac{1}{\sqrt{3}})^{n-m}$. Then
\begin{equation}
A^2-B^2=-\frac{5464419604082977128654242570410694589510448147711713}{929486260504473222256638487813651283882185173994428050}<0.
\end{equation}

Second proof (more conceptual): Suppose that $0<x<y<z$. Then, with $n=m+1$,
\begin{equation}
\lim_{m\to\infty}
(\frac{x^n}{x^m+y^m}+\frac{y^n}{y^m+z^m}+\frac{z^n}{z^m+x^m})
=
\lim_{m\to\infty}
(\frac{x}{1+(y/x)^m}+\frac{y}{1+(z/y)^m}+\frac{z}{1+(x/z)^m})=z
\end{equation}
Now set $x=\frac{1}{3}$, $y=\frac{2}{3}$, $z=\frac{7}{9}$. Then $xy+yz+zx=1$, but $z<\frac{3}{2}\frac{1}{\sqrt{3}}$. So the stated inequality does not hold for all sufficiently large $m$.

Remark: The same question was asked here at MathOverflow, and I gave a similar counterexample there.

By the conditions, we have:
$$y =\frac{1-xz}{x+z}\ge0\quad\longrightarrow \quad xz\le1, z \le 1/x$$
Now, without loss of generality, assume $z\ge y\ge x$. We therefore have that the expression $f(x,y,z)$ in question satisfies:
$$f(x,y,z) \ge \frac{3}{2} \frac{x^n}{z^m}\ge \frac{3}{2}x^{n-m}$$

Now it remains to be shown that $x\ge 1/\sqrt{3}$. This is true since the minimal distance from the origin to the hyperboloid is exactly $1$, so that:
$$x^2+y^2+z^2 \ge 1 \quad\longrightarrow \quad 3x^2 \ge 1$$
Thus completing the proof.


How many words can be formed from 'alpha'? How many words can be formed from the word 'alpha'? The letter 'a' may be used twice but the other letters may only be used once.

There are no restrictions on whether or not they're real words, just combinations of letters. So, I need to add together:

$5$ letter words $+$ $4$ letter words $+$ $3$ letter words $+$ $2$ letter words $+$ $1$ letter words

There are, $(5!/2!)=60$, 5 letter words. 120 total but then we divide out by duplicate words because we have two a's.

$1$ letter words $=4$

The part I am having trouble with is the duplication of 'a'. I can't seem to figure out how to find $2,3,$ and $4$ letter words. For 2 letter words, there were so few examples I just wrote them out and saw that there were 13. So,

$(5!/3!)-[(2!)3+1]=13$

20 possibilities and then we need to subtract out the ones that contain 2 of p,l, and h, as well as the second aa.

I'm having a hard time believing my work for 2 letter words is accurate and attempting to do 3 and 4 letter words suggests that it isn't. Let me know if there is anything I can to do clear up what is being asked. Thanks in advanced.

Let us look for example at $4$-letter words. They are of two types: (i) words with at most one $a$ and (ii) words with two $a$'s.

For Type (i), there are $4!$. For Type (ii), we choose the location of the $a$'s in $\binom{4}{2}$ ways. For each such way, the remaining two places can be filled in $(3)(2)$ ways.

The same strategy works for $3$-letter words. There are $(4)(3)(2)$ ways to have at most one $a$. For the words with two $a$'s, the locations can be picked in $\binom{3}{2}$ ways, and for each such way the remaining slot can be filled in $3$ ways.

Note that if we had more non-duplicated letters, the analysis would be much the same.

Break each case in two parts: (i.e. words with all distinct letters + words with two repeated a's)

No. of 4 letter words formed from 'alpha' = No. of 4 letter words formed with all distinct letters (4C4 * 4!) + No. of 4 letter words formed with two a's already present (3C2 * 4!/2!).

Similarly,
No. of 3 letter words = (4C3 * 3!) + (3C1 * 3!/2!)

No. of 2 letter words = (4C2 * 2!) + (3C0 * 2!/2!)

No. of 1 letter words = (4C1 * 1!)

NOTE:
a) To form words with distinct letters we choose from ['a', 'l', 'p' and 'h'], hence 4Cx; then we arrange them in x! ways.

b) To form words with two a's already present, we need to choose from ['l', 'p' and 'h'], hence 3Cx; and arrange them in x!/2! ways.


Count the number of permutations of certain cycles type Suppose we have $n$ elements, assume there is a permutations over $k$ elements among the $n$ elements so $n-k$ are fixed. Let that the permutation over the k elements is represented by permutation cycles so the length of all permutation cycles $=k$.

As an example: Suppose we have the following permutation

$$ x = \left( {\begin{array}{c}
   x_1 =  \left( {\begin{array}{c}
   1 \\
   2 \\
   \end{array} } \right) \\
   x_2 = \left( {\begin{array}{c}
   3 \\
   4 \\
   5 \\
   \end{array} } \right) \\
   x_3 = \left( {\begin{array}{c}
   6 \\
   7 \\
   \end{array} } \right) \\
   8 \\
   9 \\
   \vdots \\
   15 \\
  \end{array} } \right)$$

My question: What is the number of permutations we can construct from the $n$ elements where each permutation should consists of the same cycles type?

Addition: I know that the number of $k-$cycles in the symmetric group $S_n$ is $\binom{n}{k}(k-1)!$ but I don't know what to do for the constraint asking that each permutation cycle has the same length in all permutations!

Hint:  The number of distinct $k$-cycles is $P^n_k\cdot \dfrac 1k=\dfrac{n!}{(n-k)!}\cdot \dfrac 1k$.

To do your example,  we would get, for permutations of type $(2,3,2)$ in $S_{15}$:

$P^{15}_2\cdot \dfrac 12\cdot P^{13}_3\cdot \dfrac 13\cdot P^{10}_2\cdot \dfrac 12=105\cdot572\cdot45=2702700$.

Now I need to divide by $2$, since I have double counted the two $2$-cycles.

So $\dfrac12\cdot2702700=1351350$.

See here, or here, for a good explanation.

Let's take your example of a cycle structure corresponding to the partition $3^1 + 2^2 + 1^8 \vdash 15$. I think the easiest way to handle the $a_i$ cycles of a given length $i$ is to select the lot (e.g. for the two two-cycles select four elements) and then repeatedly force the lower unselected element to be in the next $i$-cycle.

So we initially have 15 elements. We select three for the $3$-cycle in $\binom{15}{3}$ ways, leaving 12. Now we select four for the two $2$-cycles in $\binom{12}{4}$ ways, assign the lowest to one cycle along with one of the $\binom{3}{1}$ remaining ones; then assign the lowest to the next cycle along with the $\binom{1}{1}$ remaining one. Overall we get $$\left[\binom{15}{3} \binom{2}{2} \right] \left[ \binom{12}{4} \binom{3}{1} \binom{1}{1} \right] \left[ \binom{8}{8} \binom{7}{0} \binom{6}{0} \binom{5}{0} \binom{4}{0} \binom{3}{0} \binom{2}{0} \binom{1}{0} \binom{0}{0}\right]$$

In general, if we have $k$ items remaining to assign and $a_i$ cycles of length $i$ the term is $$\binom{k}{a_i i} \prod_{j=0}^{a_i - 1} \binom{(a_i - j) i - 1}{i - 1}$$


How can i convert the matrix 1 to matrix 2? A= 0 1 1 1 1 1 1 1 1 1          0 1 0 0 0 0 0 0 0 0
   1 0 1 1 1 1 1 1 1 1          0 0 1 0 0 0 0 0 0 0
   1 1 0 1 1 1 0 1 1 1          0 0 0 1 0 0 0 0 0 0
   1 1 1 0 1 1 1 0 0 0 -------> 0 0 0 0 1 0 0 0 0 0
   1 1 1 1 0 1 1 0 0 0          0 0 0 0 0 1 0 0 0 0
   1 1 1 1 1 0 0 1 1 0          0 0 0 0 0 0 0 1 0 0
   1 1 0 1 1 0 0 1 1 1          0 0 0 0 0 0 0 0 1 0
   1 1 1 0 0 0 1 0 1 1          0 0 0 0 0 0 1 0 0 0
   1 1 1 0 0 1 1 1 0 1          0 0 0 0 0 0 0 0 0 1
   1 1 1 0 0 1 1 1 1 0          0 0 0 0 0 0 0 0 0 0

Here a Boolean matrix let call it A matrix and A(NxN) and here N=10. 0 mean there is no connection between nodes(as it can be seen there is no self connection also) and 1 means there is connection between nodes.

Here the first local node is node1 and when a node is local then must be blocked so can not be selected (connected) again

    Node 1 connect  node 2, then node 2 is local (must be blocked)
    2 connect 3, now 3 is local (must be blocked)
    3 connect 4, 4 is local (must be blocked)
    4 connect 5, 5 is local (must be blocked)
    5 connect 6 ,6 is local (must be blocked)
    6 connect 8, 8 is local (must be blocked)
    8 connect 7, 7 is local (must be blocked)
    7 connect 9, 9 is local (must be blocked)
    9 connect 10,10 is local and stop here as all number are local.here 10 can not connect 1 about one was local

After connections matrix should be as the matrix at the end of figure.
How can i write it in matlab

Here is one way to make the matrix you describe, without further details i cannot see if it really matches what you are looking for but you should be able to edit it yourself.

%Set things for the start:
i = 1;
blocked = 1;
used = [];
B = zeros(size(A));
for i = 1:size(A,1)
    for j = 1:size(A,2)
        % Test if node i can be connected with node j
        if A(i,j) == 1 && ~any(blocked == j)&& ~any(used == i)
            % Connect node i with node j
            B(i,j) = 1;
            % Block node j
            blocked = [blocked j];
            % To make sure we only have one entry per row
            used = [used i];
        end
    end
end

You must apply some of the method to find the rank of the matrix either row exchange or column exchange method, after few steps your upper part will be just converging towards identity metrix of six order. Then manipulate for the last three rows.


Combinations for Screen lock For iPhone / Android I was wondering if there was a general for formula to calculate the combination of the password lock for the current smart phones

The following is the condition

We must use four nodes or more to make a pattern at least.
Once anode is visited, then the node can't be visited anymore.
You can start at any node.
A pattern has to be connected.
Cycle is not  allowed.

If using 4 as the minimum string for the password with 9 nodes , the result is 389112.

Is there anyway to estimate the number of combinations for 16 nodes, 25 nodes and so on?

Since each dot can be the starting point for the pattern, the number of possibilities are as follows:

4 dots combination (since this is the minimum required): 9 dots (starting points) ^ (raise to the power) 4 = 6,561 patterns
5 dots combination: 9 dots ^ 5 = 59,049 patterns
6 dots combination: 9 dots ^ 6 = 531,441 patterns
7 dots combination: 9 dots ^ 7 = 4,782,969 patterns
8 dots combination: 9 dots ^ 8 = 43,046,721 patterns
9 dots combination: 9 dots ^ 9 = 387,420,489 patterns

Total possible number of patterns = 435,847,230 patterns


Partial reciprocal sum How can I show that $$\sum_{k=1}^{n}\frac{1}{n+k}\leq\frac{3}{4}$$ for every integer $n \geq 1$?
I tried induction, estimates with logarithms and trying to bound the sum focusing on the larger terms or things like $\frac{1}{n+1}+\frac{1}{n+2}\leq\frac{2}{n+1}$ but nothing seems to work. Do you have any suggestion? Thanks

We have
$$e^x = \lim_{n\to\infty}\left(1+\frac xn\right)^n\ge1+x$$
by Bernoulli's inequality. This implies that $e^{-x} \ge 1-x$, so
$$x \le -\ln(1-x).$$
Letting $x=\frac{1}{n+k}$ forms
$$\frac{1}{n+k}\le -\ln\left(\frac{n+k-1}{n+k}\right)$$
$$\frac{1}{n+k}\le\ln(n+k)-\ln(n+k-1)$$
The sum on the right telescopes giving
$$\sum_{k=1}^{n}\frac{1}{n+k}\le \sum_{k=1}^{n}\ln(n+k)-\ln(n+k-1)=\ln(2n)-\ln(n)=\ln(2),$$
by which $\ln(2)<{3}/{4}$ and we are done.

First, note that $f(n)=\sum_{k=1}^n\frac{1}{k+n}$.  When $n$ is a positive integer greater than 1, this is a special case of $g(n)=\int_{n+1}^{2n+1}\frac{1}{\lfloor x\rfloor}dx$
$$\frac{dg}{dn}=\frac{d}{dn}\int_{n+1}^{2n+1}\frac{1}{\lfloor x\rfloor}dx=2\frac{1}{\lfloor 2n+1\rfloor}-\frac{1}{\lfloor n+1\rfloor}=\frac{2\lfloor n+1\rfloor-\lfloor2n+1\rfloor}{\lfloor1+2n\rfloor\lfloor1+n\rfloor}$$
For positive $n$, the numerator of this is $1$ if the fractional part of $n$ less than $\frac{1}{2}$, and $0$ is the fractional part is greater than $\frac{1}{2}$, and the denominator is positive.  As such, $g(n)$ is an increasing function for all positive $n$.
If a function $g(n)$ is increasing over the domain $(1,\infty)$, then it has an upper bound of $\lim_{n\rightarrow\infty}g(n)$
$$\lim_{n\rightarrow\infty}g(n)=\lim_{n\rightarrow\infty}\int_{n+1}^{2n+1}\frac{1}{\lfloor x\rfloor}dx=\lim_{n\rightarrow\infty}\int_{1}^{2n+1}\frac{1}{\lfloor x\rfloor}dx-\int_{1}^{n+1}\frac{1}{\lfloor x\rfloor}dx+\ln(2n+1)-\int_1^{2n+1}\frac{1}{x}dx-\ln(n+1)+\int_1^{n+1}\frac{1}{x}dx=\lim_{n\rightarrow\infty}\int_{1}^{2n+1}\frac{1}{\lfloor x\rfloor}-\frac{1}{x}dx-\int_{1}^{n+1}\frac{1}{\lfloor x\rfloor}-\frac{1}{2}dx+\ln(\frac{2n+1}{n+1})=\lim_{n\rightarrow\infty}\int_{1}^{2n+1}\frac{1}{\lfloor x\rfloor}-\frac{1}{x}dx-\lim_{n\rightarrow\infty}\int_{1}^{n+1}\frac{1}{\lfloor x\rfloor}-\frac{1}{2}dx+\lim_{n\rightarrow\infty}\ln(\frac{2n+1}{n+1})=\gamma-\gamma+\ln(2)=\ln(2)$$
Here, $\gamma$ is the Euler Mascheroni constant, defined as $\lim_{m\rightarrow\infty}\sum_{i=1}^m\frac{1}{i}-\ln(m)$.
We found that $g(n)$, and therefore our sum, has an upper bound of $\ln(2)$.  As $\ln(2)\approx0.69<\frac{3}{4}$, we have proven our sum is less than $\frac{3}{4}$ for all $n>1$


The unit group of a polynomial ring and the unit group of the coefficient ring Show that if $R$ is an integral domain then $R[X]^\ast=R^\ast$. Show that this is false if $R$ is not an integral domain.

Note1: this is not homework.Perhaps I must have mentioned that my biggest problem with attempting this question was the interpretation of $R[X]^\ast$ and $R^\ast$. It's for this reason that I noted this below yesterday.

Note2: this question actually comes from the notes of Iain Gordon(http://www.maths.ed.ac.uk/~igordon/).

$R^*$ here means the group of units of the ring $R$.

Suppose $f(x)\in R[x]$ is invertible; write it as $f(x)=a+xg(x)$ and suppose $b+xh(x)$ is the inverse. Then
$$
ab+x(ah(x)+bg(x)+xg(x)h(x))=1
$$
and, comparing alike terms, we get $ab=1$. Therefore $a$ is invertible.

Until now we have not used the fact that $R$ is a domain. Suppose it is and that $f(x)$ has positive degree. Then $f(x)k(x)$ has positive degree: indeed the leading coefficient of the product is the product of the leading coefficients, which cannot be zero. Thus an invertible polynomial must have degree $0$ and, by what we saw before, it is an invertible constant.

Note that we are identifying constant polynomials in $R[x]$ with $R$.

In order to find a counterexample, we look at the simplest nondomain, that is, $R=\mathbb{Z}/4\mathbb{Z}$. The first case we can try is $1+2x$: the constant term must be invertible, and the leading coefficient must be a zero divisor, by the arguments above.

Now
$$
(1+2x)^2=…
$$

The general result that, if $R$ is not a domain, then the set of invertible polynomials contains non constant polynomials is false. See https://math.stackexchange.com/a/30390/62967, where it is proved that a polynomial $a_0+a_1x+\dots+a_nx^n$ is invertible if and only if $a_0$ is invertible and $a_i$ is nilpotent for $i>0$.

So if $R$ is not a domain but it has no (nontrivial) nilpotent element, then the invertible polynomials are just the invertible constants. A ring with such a property is the product of two fields, for instance.

egreg's argument above shows that any unit in $R[x]$ is also a unit in $R$.  If $R$ has no multiplicative identity, then neither does $R[x]$.

Since the degree-polynomials form a subring of $R[x]$ that is isomorphic to $R$, then $R[x]$ is commutative if and only if $R$ is commutative. Similarly, if $a, b \in R$ are non-trivial divisors of 0, then $a, b \in R[x]$ are likewise non-trivial divisors of 0.

Summarizing, if $R$ lacks any of the properties that distinguish a ring from an integral divisor, then $R[x]$ also lacks that property.


Physical (Quantum Mechanical) Significance of completeness of Hilbert Spaces. I'm not sure if the question is very 'mathematical',but I'm asking any way. I have a basic knowledge of quantum mechanics and I'm studying Hilbert spaces.

I was wondering what is the physical significance of the completeness of a Hilbert Space wrt QM. i.e. why is the completeness required?

How do mathematical properties of Hilbert spaces or Banach spaces play a role in Quantum Mechanics?

There's an MO post with some good thoughts on the role of completeness from a mathematical point of view. (I especially like Andrew Stacey's idea of "going to a party hosted by the Square Integrables in their posh mansion, but spending the whole time hanging out with the Schwartz family" :-)

With a view to quantum mechanics, note that the Riesz representation theorem (which establishes the bra-ket correspondence) depends on completeness. You can trace the role of completeness in this proof, which uses the orthogonal decomposition theorem, whose proof in turn uses completeness.

The history section of the Wikipedia article on Hilbert spaces also has some relevant information.

completeness in hilbert space means that the metric associated with inner product converges,let draw a 1/x as a function on a graph,you will see its area converges to zero.

as x approaches to infinity 1/x goes to zero


Is there any way to solve recurrence equations with variable coefficients? So far I have done some problems that are best solved using generating functions. These mostly contain variable coefficients.
A simple one is $H(n) = (n+2)H(n-2)$. I have found solutions to these equations using mathematical induction, which requires a bit of conjecturing (by checking the result for initial values) and then proving it.
But what about bigger equations,* is there a definite way of solving them and obtain a simple formula (without resorting to generating functions)?

edit:
*Functions like $H(n) =f_1(n)H(n-1) + f_2(n)H(n-2) + \cdots + f_k(n)H(n-k)$. Where $f_1, f_2,\dots, f_k$ are functions of $n$.

$n\to n+2$ :

$H(n+2)=(n+4)H(n)$

$n\to 2n$ :

$H(2n+2)=(2n+4)H(2n)$

$H(2(n+1))=2(n+2)H(2n)$

$H(2n)=\Theta_1(n)\prod\limits_n(2(n+2))$ , where $\Theta_1(n)$ is an arbitrary periodic function with unit period

According to http://en.wikipedia.org/wiki/Indefinite_product#Rules ,

$H(2n)=\Theta_1(n)2^n\Gamma(n+2)$ , where $\Theta_1(n)$ is an arbitrary periodic function with unit period

$H(n)=\Theta(n)2^{\frac{n}{2}}\Gamma\left(\dfrac{n}{2}+2\right)$ , where $\Theta(n)$ is an arbitrary periodic function with period $2$


How to derive the formula for the $n$th moment of $X\sim N(0,1)$ My approach is to derive the moment generating function for $X\sim N(0,1)$ so that its $n$th derivative at $t=0$ will be the moment of $X\sim N(0,1).$

However, after finding
$$M_X(t)=e^{\frac{t^2}{2}}$$

I am struggled to express the function in MacLauren Series ... which means I am unable to derive the $n$th moment from this approach...

Could someone please shows me how to expand this moment generating function into a MacLauren Series?

Many thanks!

Also, I would also appreciate if someone could show me other methds in finding the solution.

Here we assume $n$ is even since for odd $n$ the $n$th moment is obviously $0.$ (Note in response to comments. I'm not saying the integral of EVERY odd function is zero.)
\begin{align}
& \frac 1 {\sqrt{2\pi}} \int_{-\infty}^\infty x^n e^{-x^2/2} \, dx = \sqrt{\frac 2 \pi} \int_0^\infty x^n e^{-x^2/2} \,dx \\[10pt]
= {} & \sqrt{\frac 2 \pi} \int_0^\infty x^{n-1} e^{-x^2/2} (x\,dx) = \sqrt{\frac 2 \pi} \int_0^\infty \sqrt{2u\,}^{\,n-1} e^{-u}\, du \\[10pt]
= {} & \frac{2^{n/2}}{\sqrt\pi} \int_0^\infty u^{(n-1)/2} e^{-u}\, du = \frac{2^{n/2}}{\sqrt\pi} \Gamma\left( \frac{n+1} 2 \right) \\[10pt]
= {} & \frac{2^{n/2}}{\sqrt\pi} \cdot \frac{n-1} 2 \cdot \frac{n-3} 2 \cdot \frac{n-5} 2 \cdots \frac 1 2 \Gamma\left( \frac 1 2 \right) \\[10pt]
= {} & (n-1)(n-3)(n-5) \cdots 1.
\end{align}

Here I have used the functional equation $\Gamma(\alpha+1) = \alpha\Gamma(\alpha)$ and the fact that $\Gamma\left( \frac 1 2 \right) = \sqrt\pi.$


Proving whether or not a certain recursively defined sequence is a Cauchy sequence or not Question:
Let $\{x_{n}\}$ be a recursively defined sequence defined as:
$$x_{1} = \frac{1}{2}$$
$$x_{n+1} = \frac{1-x_{n}}{4}$$
I want to show first that $|x_{n+1}-x_{n}| \le \frac{1}{4}|x_{n}-x_{n-1}|$. Using this, I have to prove that the sequence is Cauchy and then find its limit. Here is what I have done:
Part 1:
$$|x_{n+1}-x_{n}| = \left|\frac{1-x_{n}}{4} - \frac{1-x_{n-1}}{4}\right| \\
=\left|\frac{-x_{n}}{4} + \frac{x_{n-1}}{4}\right|
= \frac{1}{4}|-x_{n}+x_{n+1}|$$
But, I do not know how to go from this equality to the inequality that the question has. How may I manipulate the equation so that the $=$ changes to a $\le$?
Part 2:
Now as for the second subpart where I have to show that the sequence is Cauchy, I am not sure where to start, so I started with the definition of Cauchy sequence, which is as follows:
A sequence $\{a_{n}\}$ is called a Cauchy sequence if $\forall \epsilon>0 \exists      N \in \mathbb{N}$ s.t. if $m,n \ge N$, then
$$|a_{m}-a_{n}| < \epsilon$$
But the problem I face is that if I try to use this definition and try to apply it here, I am getting that:
$$x_{m}-x_{n} = \left|\frac{-x_{m-1}}{4}+\frac{x_{n-1}}{4}\right|$$
But after this, how can I use my previous result to prove that this is always bounded by $\epsilon$?
Could someone help me?

Let $f : [0,1] \rightarrow [0,1]$ be the function defined for all $x \in [0,1]$ by $$f(x)=\frac{1-x}{4}$$
First, $f$ is well-defined because it is decreasing over $[0,1]$, and $f(0), f(1) \in [0,1]$. Because $x_1 = 1/2 \in [0,1]$, you deduce by an easy induction that the sequence $x_{n+1} = f(x_{n-1})$ is well-defined and takes all its values in $[0,1]$.
For any integer $n \geq 1$, the MVT applied to the function $f$ between $x_n$ and $x_{n-1}$ gives $$|x_{n+1}-x_n|=|f(x_n)-f(x_{n-1})| = \frac{1}{4}|x_n -x_{n-1}| \quad \quad (*)$$
(because $f' \equiv -1/4$ over $[0,1]$). This is the first part.
Now, for every integers $q>p$, you can see by an easy induction (using the equality $(*)$), that
$$|x_q-x_p|= \frac{1}{4^{p-1}}|x_{q+1-p}-x_1| \leq \frac{1}{4^{p-1}}$$
This proves that the sequence is Cauchy, and you are done.

You have found $|x_{n+1}-x_n| =\frac14|-x_n+x_{n-1}| = \frac14|x_n-x_{n-1}|$, which satisfies the inequality. If you insist, write $|x_{n+1}-x_n| \le |x_{n+1}-x_n| = \frac14|x_n-x_{n-1}|$ and now you have the inequality.
Using this we see that for any $n$:
$$|x_{n+1}-x_n|\le\frac14|x_n-x_{n-1}| \le \frac1{4^2}|x_{n-1}-x_{n-2}| \le\cdots\le\frac1{4^{n-1}}|x_2-x_1|=\frac1{2^{2n-2}}\cdot\frac38 = \frac3{2^{2n+1}}$$
The "$\cdots$" can be made rigorous via induction on $n$.
Now suppose $m > n$. Then
\begin{align}|x_m-x_n| &\le |x_m-x_{m-1}| + |x_{m-1}-x_{m-2}| + \cdots + |x_{n+1}-x_n|\\
&=\sum_{k=n}^{m-1}|x_{k+1}-x_k|\\
&\le\sum_{k=n}^{m-1}\frac{3}{2^{2k+1}}\\
&<\sum_{k=n}^{\infty}\frac{3}{2^{2k+1}}\\
&=\frac {3}{2^{2n+1}}\div\left(1-\frac14\right)\\
&=\frac1{2^{2n-1}}\end{align}
Using this upper bound, it should be easy to find the $N \in \mathbb N$ for any arbitrary $\epsilon >0$.


The vectors $p, q, r$ and $s$ satisfy the identities $p+q+r+s = 0$, and $\|p\| = \|q\| = \|r\| = \|s\|$. Prove that $p\cdot q = r\cdot s$ I got asked this question in class today and I am unsure on where to start!
The vectors $p, q, r$ and $s$ satisfy the identities $p+q+r+s = 0$, and $\|p\| = \|q\| = \|r\| = \|s\|$.
Prove that $p\cdot q = r\cdot s $.

Rewrite the formula to get $$p+q = -r-s$$
Then compute the norm of this sum $$\|p\|^2+2p\cdot q+\|q\|^2=\|r\|^2+2r\cdot s+\|s\|^2$$
cancelling the norms (which are all equal) from both sides gives $$2p\cdot q=2r\cdot s$$


general formula for $\log_x(y)$ when $y$ is negative I'm looking for a general formula for solving a problem of the form $\log_x(y)$ when $y<0$. It seems like the formula is $\frac{\ln(|y|)+\pi i}{\ln(x)}$, but I would like to know how this is derived, or a proof for it. I got this this from looking at the answer of wolfram alpha, so this formula may not be correct.

The complex logarithm function is defined as

$$\text{Log}\,z:=\log|z|+i\arg z\;\;,\;\;0\neq z\in\Bbb C$$

with $\;\log\;$ the usual, real logarithm.

But, as we know, the argument of a complex number is defined only up to an integer multiple of $\;2\pi\;$, so the above definition in fact gives us infinite possibilities.

Another fact, way more advanced, is that $\;z=0\;$ is an acute problem here, and every time we "go around" the origin we add (substract) a multiple of $\;2\pi\;$ to the number's argument, so in order to have a more or less nice logarithm function we must choose a complex "branch" for the function, and in this case this means to choose a chunck of the complex plane that we're going to "throw away" so that in the remaining domain the function's well and nicely defined. The main branch here is the non-negative real axis $\;\Bbb R_-:=\{z\in\Bbb C\;;\;\text{Re}\,z\le 0\}\;$ , and in $\;\Bbb C\setminus \Bbb R_-\;$ the function is well defined, continuous and all.

Please do observe then that $\;\text{Log}\,y:=\log|y|\;$ , for any real $\;y\;$

There are problems when you try to $\log$ a negative value, and you will not be able to escape them. Look at the example when $x=e$. The problem is that $\ln y$ is defined as the real number $a$ for which $e^a=y$. As you may know, there is no real number $a$ for which $e^a$ would be negative, meaning that on the reals, $\ln y$ does not exist for negative indices.

If you look at complex values of $z$ and see which values $e^z$ can reach, things get complicated. Because $$e^{a+bi} = e^a(\cos b + i\sin b),$$ you can see that any number, except for $0$, can be reached with this function. However, because $$e^{a+(b+2\pi) i} =e^a(\cos (b+2\pi) + i \sin(b+2\pi))=e^a(\cos b + i\sin b) = e^{a+bi},$$ you now face a different problem. Unlike the real exponent function, this new complex function is not injective. This means it iw hard to define a function $\log$ to work on the whole convex plain. For example, if you are solving the equation $$e^z=y$$ where $y<0$ is a real number, then one solution is $\ln|y| + i\pi,$ but $\ln |y| + 3\pi i$, $\ln|y| + 5\pi i$ and $\ln |y| + (2k+1)\pi i$ in general all solve the same equation.


Multiple irrational roots of irreducible polynomial over Q Is it possible for irreducible polynomial $f \in Q[x]$ to have irrational multiple root?

I guess not, so I supposed to assume that my polynomial has factor $(x - a)^k$ in some extention, where $a$ is irrational, $k>1$ and come to contradiction, but I don't know what  could it be.

No.  The field $\mathbb{Q}$ is a perfect field: no irreducible polynomial over it has multiple roots.

To see it, recall that a polynomial $P$ has multiple roots if and only if it has a non-constant common factor with its formal derivative $P'$.  Now, since $P$ is irreducible, its only factors are itself and constant polynomials.  Since $P'$ has degree smaller than that of $P$, the only factors it can share with $P$ are constant polynomials.

No, let $G$ be the Galois group of $Q[X]/P$, $(X-a)^k$ divides $P$ implies that  $(\Pi_{g\in G}(X-g(a)))^k$ divides $P$ which implies that $P$ is not irreducible if $k>1$ since $\Pi_{g\in G}(X-g(a))\in Q[X]$.


Prove that a bipartite graph has a unique bipartition if and only if it is connected Prove that a bipartite graph has a unique bipartition if and only if it is connected.

I am completely stuck on this question. I assume it is a proof that involves a mixing of definitions but I am not sure.

If the graph is connected and bipartite it has a unique bipartition, to see this take a spanning tree, pick a root vertex. Then the parity of the distance from each vertex to the root will define the two sets.

If the graph is not connected and is bipartite each component must be bipartite, therefore you can find a bipartition for each component. So suppose the components are $C_1,C_2\dots C_n$ and each component has parts $A_1,A_2\dots A_n$ and $B_1,B_2\dots B_n$. Then you can take any set of the form $D_1,D_2,\dots D_n$ where $D_i$ is ethier $B_i$ or $C_i$ will give you a bipartition.

let's assume that the graph is connected.let's say that the graph is partitioned into two sets of vertices x and y such that every edge lies between one vertex of x and one vertex from y. I will try to prove by contradiction.So let's assume that there are two different bipartite partitions 'p1' and 'p2' of a graph.then we have at least one pair of vertices (a,b) such that in p1 they are in different sets(one in x and one in y),and in p2 they are in the same set(both of them in either x or y). Here we have two cases

case(1): There is an edge between a and b .then p2 is not a bipartite partition since a and b are in the same set and they have an edge between them.hence the bipartite partition is unique.

case(2): there is no edge between 'a' and b.here we try to obtain p2 from p1 by simply shifting the vertices from x to y.let's say that b was in set y and a was in x in the partition p1. First of all, 'a' was shifted from x to y.then all the vertices in y which are adjacent to 'a' must be moved to x.This way we keep moving the vertices,since the graph is connected b must have had an edge with some vertex 'v' in x in p1.But as we move the vertices we must have moved 'v' from x to y.Now v and b both lie in set y.so b must be moved to x.that is every such vertex as b is moved ,during this transition ,which implies that the partition p2 is same as p1.


If $\tan 9\theta = 3/4$, then find the value of $3\csc 3\theta - 4\sec 3\theta$. If $\tan9\theta=\dfrac{3}{4}$, where $0<\theta<\dfrac{\pi}{18}$, then find the value of $3\csc 3\theta - 4\sec 3\theta$.

My approach:-
$$\begin{align*} \tan9\theta &=\frac{3}{4} \\[6pt] \implies \theta & = \frac{37^{\circ}}{3} \end{align*}$$
By using this, we get value of  $(3\csc3\theta - 4\sec3\theta) =9.95$ by using calculator.

I want know if there's any way to solve this problem without calculator.

Set $3\theta=t$

$\tan3t=\dfrac34\implies\dfrac{\sin3t}3=\dfrac{\cos3t}4=\dfrac1k$

$\implies(k\sin3t)^2+(k\cos3t)^2=3^2+4^2\implies k=5$ as $0<t<\dfrac\pi6$

$$F(t)=\dfrac3{\sin t}-\dfrac4{\cos t}=\dfrac{k\sin3t}{\sin t}-\dfrac{k\cos3t}{\cos t}$$

Method$\#1:$

As $\sin t\cos t\ne0,$ using $\sin3t,\cos3t$ formula,

$$F(t)=k(3-4\sin^2t)-k(4\cos^2t-3)=k(3+3-4)$$

Method$\#2:$

$$F(t)=\dfrac{k\sin(3t-t)}{\dfrac{\sin2t}2}=2k$$ as  $\sin t\cos t\ne0$


How would I solve this exponential = linear function? Would I solve for $ab^x = kx+m$ ?  I am lost to how to solve it

This equation is what I need help on: $7\cdot 8^x +3 = 9x+2$

Im puzzled to how to solve this.

$\require{begingroup} \begingroup$
$\def\e{\mathrm{e}}\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$

\begin{align}
ab^x &= kx+m
\tag{1}\label{1}
\end{align}

There is absolutely no need for any numerical methods in this case,
since the Lambert $\W$ function was specifically invented
to express a solution(s) to such kind of equations,
as the symbol $\sqrt{\ }$ was invented
to represent a solution $x=\pm\sqrt a$ for
equation $x^2=a$, and the most suitable numerical methods
were well-studied already,
and were implemented in software.

The equation \eqref{1} just needs to be transformed
to the form $u\exp(u)=v$ to apply  $\W(u\exp(u))=\W(v)$
to find $u=\W(v)$.

\begin{align}
\frac ak\,\exp(x\ln(b))\,\ln(b)
&=
x\,\ln(b)+\frac mk\,\ln(b)
\tag{2}\label{2}
,\\
\frac ak\,\exp\left(x\ln(b)+\frac mk\,\ln(b)-\frac mk\,\ln(b)\right)\,\ln(b)
&=
x\,\ln(b)+\frac mk\,\ln(b)
\tag{3}\label{3}
,\\
\frac ak\,\exp\left(x\ln(b)+\frac mk\,\ln(b)\right)\exp\left(-\frac mk\,\ln(b)\right)\,\ln(b)
&=
x\,\ln(b)+\frac mk\,\ln(b)
\tag{4}\label{4}
.
\end{align}

Let $\left(x+\frac mk\right)\,\ln(b)=y$ for now, so \eqref{4} becomes

\begin{align}
\frac ak\,\exp(y)\exp\left(-\frac mk\,\ln(b)\right)\,\ln(b)
&=
y
\tag{5}\label{5}
,\\
\frac ak\,\exp\left(-\frac mk\,\ln(b)\right)\,\ln(b)
&=
y\exp(-y)
\tag{6}\label{6}
,
\end{align}

\begin{align}
-y\exp(-y)
&=
-\frac ak\,\exp\left(-\frac mk\,\ln(b)\right)\,\ln(b)
\tag{7}\label{7}
,\\
-y\exp(-y)
&=
-\frac{a\,\ln(b)}{k\,\exp\left(\frac mk\,\ln(b)\right)}
=
-\frac{a\,\ln(b)}{k\,b^{m/k}}
\tag{8}\label{8}
,\\
\W(-y\exp(-y))
&=
\W\left(
-\frac{a\,\ln(b)}{k\,b^{m/k}}
\right)
\tag{9}\label{9}
,\\
\tag{10}\label{10}
-y
&=
\W\left(
-\frac{a\,\ln(b)}{k\,b^{m/k}}
\right)
,\\
\left(x+\frac mk\right)\,\ln(b)
&=
-\W\left(
-\frac{a\,\ln(b)}{k\,b^{m/k}}
\right)
\tag{11}\label{11}
,\\
x
&=
-
\frac1{\ln(b)}
\W\left(
-\frac{a\,\ln(b)}{k\,b^{m/k}}
\right)
-\frac mk
\tag{12}\label{12}
.
\end{align}

At this point we have the solution for $x$
in terms of the Lambert $\W$ function.

The last and a very important step is to find out,
is there any real solutions, and how many of them.
And there is a well-known condition, based on the value of the operand of
the $\W(z)$ function, in this case it is
$z=-\frac{a\,\ln(b)}{k\,b^{m/k}}$, so

\begin{align}
\begin{cases}
z<-\tfrac1\e &\Rightarrow \text{no real solutions}
,\\
z\ge0 &\Rightarrow \text{one real solution},
x
=
-\frac1{\ln(b)}\Wp(z)-\frac mk
,\\
-\tfrac1\e < z<0 &\Rightarrow \text{two real solutions},
x_0=-\frac1{\ln(b)}\Wp(z)-\frac mk
,\quad
x_1=-\frac1{\ln(b)}\Wm(z)-\frac mk
,\\
z=-\tfrac1\e &\Rightarrow \text{in this case}
\Wp(-\tfrac1\e)=\Wm(-\tfrac1\e)=-1, \text{ so }
x=\frac1{\ln(b)}-\frac mk
.
\end{cases}
\end{align}

For $a=7,\, b=8,\,k=9,\, m=-1$ we have the argument of $\W(z)$

\begin{align}
z&=-\frac{a\,\ln(b)}{k\,b^{m/k}}
=-\frac79\,\ln(8)\,8^{1/9}\approx -2.03772502141
<-\tfrac1\e
,
\end{align}

so indeed, in this case there are no real solutions.

$\endgroup$

Yes indeed, plotting the linear function and the exponential function (or making a spreadsheet table) will show you clearly that there is no point of intersection.  And if there were a point of intersection for such a pair of functions (with different coefficients), then you would generally need to solve for it numerically.

Ob the other hand, one can take the derivatives of both functions:

d/dx (7 * 8^x + 3) = 7*ln(8)*8^x

d/dx (9x +2) = 9

The exponential function has a slope that is always positive (in fact, the second derivative is also positive which means its curvature is always upward).

It can be seen that the exponential function has a horizontal asymptote of 3 (as x approaches negative infinity), where the slope is extremely small (but positive).  As the curve passes the y-axis, the slope equals 14.56 (=7*ln(8)) and then continues to grow (exponentially).  By setting the two derivatives equal to each other (i.e., =9) and solving, we can find the point at which the curves are the closest (vertically):

x= ( ln(9)-ln[7*ln8] ) / ln(8)  =  -0.2312

We can plug that x value into each equation and see that the exponential curve is 7.408963 above the linear equation.  At that point, the exponential had been inching closer to the straight line (its slope to the left was less than 9), but after this point (moving to the right), the exponential curve will grow faster than a slope of 9 and will climb away from the straight line.


Prove $\exp(x+y) = \exp(x) \exp(y)$ for $\exp(x) = \sum_{n=0}^\infty \frac {x^n}{n!}$ I am trying to prove $\exp(x+y) = \exp(x) \exp(y)$.

I may use that $$\exp(x) = \sum_{n=0}^\infty \frac {x^n}{n!}$$
I further know how to multiply two power series in one point, i.e. if $f(x) = \sum_{n=0}^\infty c_n(x-a)^n$ and $g(x) = \sum_{k=0}^\infty d_n(x-a)^n$ then
$$
f(x)g(x) = \sum_{n=0}^\infty e_n(x-a)^n
$$
with
$$
e_n = \sum_{m=0}^n c_md_{n-m}
$$

Assume the function $\frac{exp(x+y)}{exp(x)}$, where $y$ is some constant. Differentiate with respect to $x$ and find that the derivative is zero. Therefore $\frac{exp(x+y)}{exp(x)}=c\in\mathbb{R}$. From the fact that $exp(0)=1$ we get $exp(y)=c$ which proves what was asked.


Find coordinates of two vector product given each vector in coordinate format $\vec{a} = (-1,0,2), \vec{b} = (3,-1,1), \vec{c} = (2,1,2) $

I need to find coordinates of vectors, as well as modulus of the following:

1) $\vec{a}$ x $\vec{b}$

2) $\vec{b}$ x $\vec{c}$

I am not sure I not understand how to find above.

Should I sum up coordinates to get $\times$? And afterwards take squares to find modulus?

Using formulas given in link by @Rory Daulton

Thus
$\vec{a}$ x $\vec{b} = (0+2)i + (6+1)j + (1+0)k = 2i + 7j + 1k $

If you do not remember formula, it could be easier to remember determinant form:

from where determinant can be calculated using two different formulas:

or

Thus $\vec{a}$ x $\vec{b} = (0\times{i} + 6\times{j} + 1\times{k}) - (-2\times{i} - 1\times{j} + 0\times{k})=2{i} + 7{j} + 1{k}$

Modulus can be found as $\sqrt{2\times2 + 7\times7 + 1\times1}=\sqrt{54}$

The very same idea is for $\vec{b}$ x $\vec{c}$, finding
$\vec{a}$ x $\vec{b} = -3i-4j+5k$ and modulus = $\sqrt{42}$


How can I prove $x \cos\alpha + y \sin\alpha =p$ using vectors? A straight line satisfies the following condition.

$p>0$ is the length of the perpendicular on the line from the origin
$\alpha$ is the angle made by this perpendicular with the positive $x$-axis.

Vectorically derive that the equation of this straight line is $$x \cos\alpha + y \sin\alpha =p$$

With some ideas I have made a figure. But not sure that the figure is correct. Please help me with this and the further derivation.

Let $P$ denote the point on the line that is closest to the origin.

Then $\vec{OP}=\begin{pmatrix} p\cos \alpha \\ p \sin \alpha\end{pmatrix}$

Let $R$ be an arbitrary point on the line, let $\vec{OR}=\begin{pmatrix} x \\ y\end{pmatrix}$

$\vec{PR}$ and $\vec{OP}$ are perpendicular, hence their inner product (or dot product) should give us $0$.

$$\vec{PR}.\vec{OP}=0$$
$$(\vec{OR}-\vec{OP}).\vec{OP}=0$$
$$\vec{OR}.\vec{OP}=\vec{OP}.\vec{OP}$$

$$xp\cos (\alpha)+yp\sin(\alpha)=p^2$$
$$x\cos(\alpha)+y\sin(\alpha)=p$$

Firstly, you haven't marked the angle $\alpha$ in your diagram though it is implicit from your question.

Once you mark $\alpha$ express $\cos(\alpha)$ and $\sin(\alpha)$ in terms of the distances in the figure.

Hint: $$\cos(\alpha) = \frac{PO}{OA}$$

You should get your proof.


If σ is a cycle of length n, then σ^r is also a cycle if and only if n and r are relatively prime If σ=(1.2.3), σσσ= identity permutation, which is cyclic

in this case n=3 and r=3

but their gcd is not 1.

I don't understand why  -> this direction of theorem is true

By relabelling, we may suppose $\sigma = (1,2,\ldots, n)$.  Thus $\sigma(i) \equiv i+1 \mod n$.  Then $\sigma^r(i) \equiv i+r \mod n$.  This is not an $n$-cycle iff there is some $i$, $1 \le i \le n$, and some $k$, $1 \le k \le n-1$, such that $\sigma^{rk}(i) = i$.  That
says $i + r k \equiv i \mod n$, i.e. $r k \equiv 0 \mod n$.
If $g = \gcd(r,n) > 1$, this will be true for $k = n/g$, but if $g = 1$, $r k \equiv 0 \mod n$ implies $k \equiv 0 \mod n$.

After working through the problem, I think you are right that $\gcd(n,r)=1$ for $\sigma^r$ to be a cycle.

The example you provided isn't a counter example because the identity isn't a cycle.

Instead, it is a product of cycles (each containing one element of $A$). So our condition holds true.

Because $\gcd(n,r)=3$ in the example you described, we don't get a cycle, as the theorem predicts.


The polynomial $a z^n+z+1$ has at least one root in $|z| \leq 2$ I am trying to solve this problem, but I don't have any idea.

The problem is:

Prove that for arbitrary $a \in \mathbb{C}$ and $n \geqslant 2$, polynomial $P(z) = a z^n+z+1$ has at least one root inside the disc $S = \{ z \in \mathbb{C} \mid |z| \leqslant 2 \}$.

Does anyone have idea? Thanks in advance.

The result follows immediately if $a=0$, so suppose $a \neq 0$.

The product of the zeros of the polynomial

$$
P(z) = az^n + z + 1 = a \left(z^n + \frac{1}{a} z + \frac{1}{a} \right)
$$

is $(-1)^n/a$. If all $n$ zeros of $P$ satisfy $|z| > 2$ then this product must be greater than $2^n$ in absolute value.  In other words, this would imply that

$$
\frac{1}{|a|} > 2^n,
$$

which is equivalent to $|a| < 2^{-n}$. Consequently,

  If $|a| \geq 2^{-n}$ then $P$ must have a zero in the disk $|z| \leq 2$.

This is part of what was asked. We need to use a different argument to handle the values of $a$ satisfying $|a| < 2^{-n}$.

Suppose that $|a| < 2^{-n}$.  Then on the circle $|z| = 2$ we have

$$
|az^n| = |a|2^n < 1
$$

and

$$
|z+1| \geq |z| - |1| = 1,
$$

so that $|az^n| < |z+1|$. By Rouché's theorem we conclude that the polynomial

$$
az^n+z+1 = P(z)
$$

has as many zeros inside $|z| = 2$ as does the polynomial $z+1$, which is one. Restated,

  If $|a| < 2^{-n}$ then $P$ has exactly one zero in the disk $|z| < 2$.

Combining these conclusions yields the desired result, that

  For arbitrary $a \in \mathbb C$ and $n \in \mathbb N$ with $n \geq 2$, the polynomial $P(z) = az^n + z + 1$ has at least one zero in the disk $|z| \leq 2$.

It's worth noting that if $a=(-2)^{-n}$ then $P$ has a zero at $z=-2$ and no zeros inside $|z| < 2$, so this bound is sharp.

Take f(z)=1 and g(z)=az^n+z such that |f(z)|<|g(z)| inside S. Hence by Rouche's theorem f(z)+g(z) and g(z) has same number of zeroes (including multiplicity) inside S. As g(z) has a zero at z=0 which is always inside S, hence the result.


Eigenvalue of Block matrix: Adjacency of complete bipartite Graph Let $A\in \{0,1\}^{mn \times mn}$ be the adjacency matrix of a complete bipartite graph with $m$ and $n$ vertices each, i.e. let $A$ be the matrix consisting of two blocks $A_1\in \{0,1\}^{m \times m}$ with zeros on the diagonal and 1 elsewhere, and $A_2 \in \{0,1\}^{n \times n}$ likewise.

I know that the eigenvalues of $A_1$ and $A_2$ are given $m-1, -1, \dotsc, -1$ and $n-1, -1, \dotsc, -1$, respectively.

It seems as if there were an easy way to combine these to the eigenvalues of $A$, and I think that the eigenvalues of $A$ should be $\sqrt{mn}, 0, \dotsc, 0, - \sqrt{mn}$, but I am not sure how to get to this.

More general: How can one compute the eigenvalues of a block matrix given the eigenvalues of the blocks.

Thank you very much for your help.

Here is what I could find
"The eigenvalues of a disconnected graph
are just the eigenvalues of its components and
their multiplicities are sums of the corresponding
multiplicities in each component."
Hope this helps


PDF of $Z^2$ where $Z$ is uniform $[-1,1]$ (a) Let $Z$ be uniform $[-1,1]$. What is the PDF of $Z^2$.

I would like to fix my previous statement. Using the textbook for uniform $[0,1]$ the PDF of $Z^2$
Say $Y=Z^2$.

The CDF
$$P(Z^2\le y) = P(-\sqrt {y} \le Z \le \sqrt{y})= F_Y(\sqrt{y})-F_Y(\sqrt{-y})$$
Differentiate to obtain the PDF
$$\frac{1}{2\sqrt y}f_Y(\sqrt y)+\frac{1}{2\sqrt y}f_Y(\sqrt{-y})$$ This is correct if $Z$ ~ $[0,1]$. Does the problem change greatly due to the new interval?

\begin{align}
P(Z^2 \leq y) &= P(-\sqrt{y} \leq Z \leq \sqrt{y}) \\
&= F_Z(\sqrt{y}) - F_Z(-\sqrt{y})
\end{align}

Differentiating

$$f_{Z^2}(y) = \frac{1}{2\sqrt{y}}f_Z(\sqrt{y}) + \frac{1}{2\sqrt{y}}f_Z(-\sqrt{y}) = \frac{1}{\sqrt{y}}f_Z(\sqrt{y})$$

I will leave the evaluation of $f_Z(\sqrt{y})$ as an exercise.


Show that $f(x) = x\sin(1/x)$ is Differentiable everywhere where $x\ne0$. I have to show that  $f(x)= x\sin(1/x)$ is continuous everywhere differentiable everywhere where $x\ne 0$.

I can show the continuous property, and how it is not differentiable when $x=0$, but how would I go to prove that it is differentiable for all $x$, such that $x\ne 0$.

Trying to put it into the definition of the derivative and simplifying does not work (I could not get a proper answer.)

Any hints on how to proceed would be greatly appreciated.

do you know that $\sin$, $x$ and $1/x$ are continuously differentiable everywhere (but $0$ in the last case)? if so just use the fact that a composition and a product of $C^1$ functions is $C^1$

edit: to be clear - $C^1$ means continuously differentiable

$$\text{Suppose the limit did exist then use the below to get a contradiction.}$$

$$\text{Limit Rule:} \ \lim x_n \to X \ \text{and} \ \lim y_n \to Y \Rightarrow \lim x_n y_n = XY$$


Where did the bracket notation for vectors come from and is using parentheses to represent vectors valid? In class I learned that a vector can be represented using <>, a column vector, and unit vector notation. However, one of the old math books that I study from notes that a vector in unit vector form can be represented with parantheses:

"Let $n$ be a positive integer. An ordered n-tuple of real numbers $$(a_1,a_2,...,a_n)$$ is called a vector in n-space. The set of all such vectors, $a_i$ real, under operations to be explained, is called Euclidean n-space, $E^n$."

This book is Calculus And Analytic Geometry Fourth Edition (1968) by George B. Thomas, Jr.

Is this parentheses notation still used? If so, is it uncommon?

This is the standard notation in quantum mechanics. It's great, because it allows to incorporate operators as well.

Copied from Wikipedia

  Bra–ket notation was effectively established in 1939 by Paul
  Dirac1[2] and is thus also known as the Dirac notation.


To find the smallest integer with $n$ distinct divisors For example, if $n=20$, how can I find the smallest integer which has exactly $20$ distinct divisors?  Can someone give me some hints?

If $m=p_1^{a_1}p_2^{a_2}\ldots p_k^{a_k}$, we have number of divisors to be
$$(1+a_1)(1+a_2)\ldots(1+a_k)$$
For this to be $20$, we want
$$(1+a_1)(1+a_2)\ldots(1+a_k) = 20 = 2 \cdot 2 \cdot 5$$
This means $m$ can have at most $3$ distinct prime divisors.

$3$ prime divisors. The smallest possible $m$ is $2^4 \cdot 3 \cdot 5 = 240$.
$2$ prime divisors. The smallest possible $m$ is of the form $2^a \cdot 3^b$. This means we need $(1+a)(1+b) = 20$, where $a\geq b$.

$a=9,b=1$, gives $m=2^9 \cdot 3 = 1536$.
$a=4,b=3$, gives $m=2^4\cdot3^3 = 432$.

$1$ prime divisor. The smalles possible $m$ is of the form $2^{19} = 524288$.

Hence, the smallest $m$ is $240$.


Ladyman's alternative justification for path induction in Homotopy Type Theory I'm confused by the main point of this paper: http://philsci-archive.pitt.edu/11079/1/Identity_in_HTT_public.pdf

The authors ask whether HoTT is a valid foundation for all of mathematics in a certain strong sense. They identify the justification of path induction as crucial point. Then, it is shown that path induction follows from the uniqueness principle for identity types (UIP) and substitution of equal expressions ("transport" in HoTT).

Two Questions:

1) Why does the question arise at all? Why is it path induction and not induction on any other type that needs further justification? And more important
2) How does the result help? I thought that the point of HoTT is that we dont want UIP because "higher dimensional equalities" are interesting and more diverse.

Confused. Thanks for tipps.

1) I can't speak for Ladyman and Presnell here, but I think the notion of identity is a philosophically poignant case study for Hott. The identity type's justification and presentation is perhaps the oddest, from a classically philosophical point of view, if we use the homotopy interpretation. The centrality of identity in HoTT and philosophy might speak to the authors' choice to focus on it.

2) And they don't want UIP in its usual sense of Uniqueness of Identity Proofs. They want something like it that doesn't ruin intensionality but can also be philosophically justified, using their standards for that justification. They're trying to show that contractability and transport are justified in a philosophically safe way since the combination of these two is equivalent to path induction. Once they justify these two principles -- they use different names -- they justify path induction.

That's how I understand it, at least.


Range of $\sqrt{x-1}$ Problem:

  Find the range of $f(x)=\sqrt{x-1}$

Now the problem I face is this: is the range $[0,\infty),$ or is it $(-\infty,\infty)$? $$$$I had learnt that $\sqrt{x^2} = \pm x$. However, on the Net, I read that $\sqrt{x^2}=|x|$ ie the output of a square root function is positive. If the output of a square root is always positive then the Range of  $\sqrt{x-1}$ is obviously $[0, \infty)$ $$$$

I would be very grateful for any help in clearing this doubt. Thanks very, very much in advance!

$\sqrt{z}$ is a function(Maps a point to a single point). So $\sqrt{z^2}$can only take a single value. By convention, $\sqrt{z}\ge0$. So,$\sqrt{z^2}=|z|$. So to answer your question, range will be $[0,\infty)$

for square root function to be defined x-1>=0,or x>=1 there for the range of function is [0,+inf)


Prove that $f(x)=(x^2-2)(x^2-3)(x^2-6)$ has a root in $\mathbb{F}_p$ for every $p$ prime number This is the solution that my book gives ($\mathbb{F}_p$ is the finite field with $p$ elements):

We can assume that $p\neq 2,3$ because $0$ is a root of $f(x)$ in $\mathbb{F}_2$ and $\mathbb{F}_3$. Let $Q$ be the "squares subgroup" (?) of index $2$ in $\mathbb{F}_p^*$. If $2$ and $3$ are not in $Q$ (this means that $(x^2-2)(x^2-3)$ doesn't have roots in $\mathbb{F}_p$), then $2 \cdot 3 \in Q$ (?), and then $x^2-6$ has a root in $\mathbb{F}_p$.

Honestly, I don't understand this solution at all: what is the "squares subgroup" of $\mathbb{F}_p^*$? And why $2 \cdot 3$ must be in this group?

By the squares subgroup, they mean the subgroup $$Q = \{a \in \Bbb F_p^* \mid a = b^2 \text{ for some } b \in \Bbb F_p^*\}.$$
In other words, $Q$ is the image of the homomorphism $\varphi : \Bbb F_p^* \to \Bbb F_p^*.$
Since $\ker \varphi = \{1, -1\}$ and $-1 \neq 1$ (why?), we see that $G$ has index $2$ in $\Bbb F_p^*$.
Now, if $2, 3, 6 \notin Q$, then this means that the cosets $2Q, 3Q, 6Q$ are all not equal to $Q$. Since $Q$ has index $2$ in $\Bbb F_p$, this means that $2Q = 3Q = 6Q$.
But then, $6Q = 3Q \implies 6 \cdot 3^{-1} = 2 \in Q$, a contradiction.
Thus, one of $2, 3, 6$ must be in $Q$. In turn, the given polynomial has a root in $\Bbb F_p$ (in fact, in $\Bbb F_p^*$, if $p \neq 2, 3$).

If 2 is a quadratic residue modulo p, then first factor has a root.
If 3 is a quadratic residue modulo p, then second factor has a root.
If 2 and 3 are quadratic nonresidues modulo p, then 6 is a quadratic residue (because Legendre symbol is multiplicative), therefore third factor has a root.


Numerical approximation of the modified Bessel function $I_0$ with radical argument for integration purposes I have to numerically calculate the following definite integral

$$\int_{\alpha}^{\beta}I_0(a\sqrt{1-x^2})dx$$

for different values of $\alpha$ and $\beta$, where $a$ has a value of, say, $30$. I'm not using quadrature rules, for the sake of calculation speed. Conversely, I'm trying to approximate the integrand so to obtain a closed form primitive.

According to

J.M. Blair, "Rational Chebyshev approximations for the modified Bessel functions I_0(x) and I_1(x)", Math. Comput., vol. 28, n. 126, pp. 581-583, Apr. 1974.

when the argument of $I_0$ is less than $15$, then $I_0$ can be approximated as a rational function involving only even powers of $x$. In this case, the presence of $\sqrt{1-x^2}$ introduces no difficulty, being $I_0(a\sqrt{1-x^2})$ still a rational function, so that the integration can be done in closed form for the interval $(\alpha,\sqrt{1-(15/a)^2})$.

The problem arises for the interval $(\sqrt{1-(15/a)^2},\beta)$. In this case, according to Blair's paper,

$$I_0(a\sqrt{1-x^2})\simeq \frac{e^{-a\sqrt{1-x^2}}}{(1-x^2)^{1/4}}p(a\sqrt{1-x^2})$$

where $p$ is a Chebyshev polynomial. In this case, closed form integration is not possible anymore (myself and Mathematica are not able to find a closed form primitive). I have also tried with Mathematica to find a rational function approximation for the range $(\sqrt{1-(15/a)^2},\beta)$. Unfortunately, I was not successful to achieve a satisfactory solution (of the order of machine double precision accuracy), perhaps due to my bad skills in using Mathematica.

Is anyone aware of any other approximation of $I_0(a\sqrt{1-x^2})$ in $(\sqrt{1-(15/a)^2},\beta)$ (or even of a better usage of Mathematica) which could be integrated in a closed form?

There is a nice expression for the modified Bessel function of the first kind of order zero:

$$I_0(z)=\frac1{\pi}\int_0^\pi \exp(z\cos\,u)\mathrm du$$

The nice thing about this expression is that one can use the simple-minded trapezoidal rule or midpoint rule to evaluate the integral numerically, with good accuracy; for an explanation of why this is so, see e.g. this article or this article.

As an aside, the substitution $x=\sin t$ should improve the numerical behavior of your integral.

In fact the best approach should be first expand the bessel function and then integrate it term by term.

$\int_\alpha^\beta I_0(a\sqrt{1-x^2})~dx$

$=\int_\alpha^\beta\sum\limits_{n=0}^\infty\dfrac{a^{2n}(1-x^2)^n}{4^n(n!)^2}dx$

$=\int_\alpha^\beta\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{a^{2n}C_k^n(-1)^kx^{2k}}{4^n(n!)^2}dx$

$=\int_\alpha^\beta\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^ka^{2n}x^{2k}}{4^nn!k!(n-k)!}dx$

$=\left[\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^ka^{2n}x^{2k+1}}{4^nn!k!(n-k)!(2k+1)}\right]_\alpha^\beta$

$=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^ka^{2n}(\beta^{2k+1}-\alpha^{2k+1})}{4^nn!k!(n-k)!(2k+1)}$


Dirac Delta function at a point From my understanding of the Dirac Delta function, it is infinitely thin and has a value of infinity at only a particular point. I also learned that
$$\int_{-\infty}^{\infty} \delta(x-a) dx = 1$$
What if we have a case where we are faced with the integral:
$$\int_{-\infty}^{a} \delta(x-a) dx$$
How does the fact that the function is infinitely thin at a affect the result?

Coming from a measure-theoretic point of view, one can write the following:

$\int_{-\infty}^{+\infty}\delta(x-a)dx = \int_{\mathbb R} 1 d \mu_a(x)$, where $\mu_a$ denotes the point measure with mass $1$ at $a$ and $0$ everywhere else.

Now, if we want to integrate over an interval $I$, we integrate the characteristic function $\chi_I(x)$ over $\mathbb R$:

$$\int_I \delta(x-a)dx= \int_{\mathbb R}\chi_I(x)d\mu_a(x).$$

And here comes a important distinction which we dont need to make when using the Riemann integral: If $a$ is a border point of the interval $I$, do we consider the open interval $I_1 =(-\infty,a)$, or the interval $I_2=(-\infty,a]$ ? Because the mass of $\mu_a$ lies at $a$, this is an important distinction:

$$\int_{(-\infty,a)} \delta(x-a)dx= \int_{\mathbb R}\chi_{(-\infty,a)}(x)d\mu_a(x) = 0 $$

while

$$\int_{(-\infty,a]} \delta(x-a)dx= \int_{\mathbb R}\chi_{(-\infty,a]}(x)d\mu_a(x) = 1.$$

So we have to be very careful with what we mean by $\int_{-\infty}^a\delta(x-a)dx.$

Summing up: The fact that it's 'infinitely thin' translates to the measure having a point mass, and this means that the inclusion or the exclusion of single points will change the value of the integral.

The Dirac Delta function is a symmetrical function.Thus $$\int_{-\infty}^{a} \delta(x-a) dx = \int_{a}^{+\infty} \delta(x-a) dx = \frac{1}{2}$$  Furthermore $$\int_{a^-}^{a^+}\delta(x-a) dx = 1$$ And $$\int_{a^-}^{a}\delta(x-a) dx = \int_{a}^{a^+}\delta(x-a) dx = \frac{1}{2}$$


Law Of Cosines Confusion I am trying to understand finding the joint angles to a robot arm and i am reading this tutorial on  a website and either i am really dense or it is wrong..
https://appliedgo.net/roboticarm/

The 5th picture from the top where the author starts to label angles for law of cosines, he previously states that Angle A1 is split into D1 and D2
and then he applies law of cosines but it looks more like Pythagorean theorem he is saying that angle D1^2 = X^2 + Y^2 which i dont understand ? He has two unknowns , side C and angle c, so how does he solve for law of cosines?

enter image description here

The figure is nonsense. As you say, the Pythagorean theorem involves side lengths, not angles. The correct way to compute D1 is via arctangent (as the author does do later in the code).

The pythagorean theorem is just a generalization of the Law of Cosines.

As per your given link, the author first uses the pythagorean theorem to find out the value of $D1$. He then uses this value (by substituting it into the formula for the law of cosines) to find $D2$ by considering the second triangle (which is not right-angled).

He then reuses the Law of Cosines but considering another angle now $A2$ to find out its cosine. And Hence. he deduces the value of the angle itself.

In short, he uses the law of cosines two times. One to find out the side $D2$ and one time to find out $A2$


square root of two is 1.41..., can we evaluate infinite(th) root of 2? I was just thinking about square roots. and then a thought struck me.
since square root of $2 = 2^{1/2} = 1.414..$
      cube root of $2 = 2^{1/3} = 1.259$
then  $2^{1/\infty} = 2^{0} = 1$
is this true that whatever number we take, $n^{1/\infty}$ root will be 1?
please correct me if i am wrong.

We tend to avoid treating $\infty$ as a number, because we lose a lot of nice rules about numbers when we do so.

What we do talk about is "limits." So we can look at the sequence of numbers:

$$2^{1/2},2^{1/3},2^{1/4},\dots$$

and we say that this sequence "has limit $1$" or that it "converges to $1$." Most students don't learn explicitly about limits until they get to calculus.

Most likely, the first place where you implicitly encountered a limit (without it being called such) is in repeated decimals.

For example, we write: $$\frac{1}{3}=0.3333\dots$$

This doesn't mean that $\frac{1}{3}=0.3$, nor does it mean that $\frac{1}{3}=0.33$. It means, rather, that $\frac{1}{3}$ is the "limit" of the sequence of numbers $$0.3,0.33,0.333,0.3333,\dots,$$

It took mathematicians a remarkably long time to formalize this notion of limit - it wasn't until the 19th century that they did so, even though they were implicitly using limits for much longer.

Yes, infiniteth root of $2$ is $1$.

$$\lim_{x\to\pm\infty} \sqrt[x]{2}=1$$


Show that $v\mapsto \|v\|^p$ for $1\le p<\infty$ is strictly convex I want to show that $f(v)=\|v\|^p$ for $1\leq p<\infty$ is strictly convex.

In the simplest case when $p=2$, we have:
$$\left\|\frac{v_1+v_2}{2}\right\|^2=\left(\frac{v_1+v_2}{2},\frac{v_1+v_2}{2}\right)=$$
$$\frac{\|v_1^2\|}{4}+\frac{\|v_2^2\|}{4}+\frac{1}{2}(v_1,v_2)$$
$$\leq \frac{\|v_1^2\|}{4}+\frac{\|v_2^2\|}{4}+ \frac{\|v_1^2\|}{4}+\frac{\|v_2^2\|}{4}$$
(by Cauchy inequality). That is convex, but not strictly convex - where did I make a mistake?

Please help me. Thanks.

The $p=2$ case:

If $f(\frac12(u+v))=\frac12(f(u)+f(v))$ then $\frac14\|u\|^2+\frac14\|v\|^2+\frac12\langle u,v\rangle=\frac12\|u\|^2+\frac12\|v\|^2$ by the computation in your post, hence $\|u\|^2+\|v\|^2-2\langle u,v\rangle=0$, that is, $\|u-v\|^2=0$. Thus, for every $u\ne v$, $f(\frac12(u+v))\lt\frac12(f(u)+f(v))$, QED.

The way that you show a function is convex is to draw a line between any two values; the values should always be on the same side.

It relies on a definition of convexity, where a line that strikes it twice is only inside (ie a specified $<$ or $>$) between these two striking points, and outside (the opposite inequality) everywhere else.

So, for example, one might find $f(a)=p_a$, and $f(b)=p_b$, then $f(ax+by)<x\cdot f(a)+y\cdot f(y)$, where $x+y=1$.

So, one deconstructs the function as $f(p+dx) < fp(p) + d.s$, where $s=f(p+x)-f(p)$ while $d<1$, then the function is convex.


Prove that $\int_a^c f(x)\ dx = \int_a^b f(x)\ dx +\int_b^c f(x)\ dx$ Suppose $a,b,c$ are constants, $a < b < c$. The interval $[a,c]$ is in the domain of $f$. Prove that

$$\int_a^c f(x)\ dx = \int_a^b f(x)\ dx+\int_b^c f(x)\ dx$$

This statement(property) is often stated in textbooks directly without proof. However, I encounter an assignment question which asks me to prove this statement using mathematical argument but not using a graph. I totally don't have any idea on how to prove the statement.

It depends on the definition of integral.

The easiest case would be if the integral is defined as the antiderivative.
Then the result follows from the antiderivative at b canceling out.

Riemann and Lebesgue I'll leave to others.

I would consider the integral as approximated by a Riemann sum...

$$\int_a^b f(x)\ dx\approx \sum_a^b f(x)\ \Delta x$$

By definition, a sum is additive so...

$$\sum_a^b y+\sum_b^c y=\sum_a^c y$$

Ergo, by taking the limit $\Delta x\to0$...

$$\int_a^c f(x)\ dx = \int_a^b f(x)\ dx+\int_b^c f(x)\ dx$$


Square root equation I have the equation $\sqrt{(7-x)} - \sqrt {(x+13)} = 2 $ The square root should be expanded so it is square root of $7-x$ - square root of $x+13 = 2$.
When i square both sides i get:
$7-x - x-13 = 4 $

then i clean on the LHS so i get
$-2x-6 = 4$ which leads to $(-2x-6)^2 = 4^2$ and after working that out i get:

$4x^2 + 24 x + 36 = 16$.

Next step will then be $4x^2 +24x +20 = 0$

Using the pq formula i get $x^2 = -6/2$ +- $\sqrt {6 \over 2}^2 -5$

And after that i get $x^2 = -6/2 +- 2$ which leads to $x^2 = -1$ or $x^2= -5$.

The answer according to the textbook is $-9$. Do you know how to solve this ?

Rembember that $(a-b)^2 = a^2 + b^2 -2ab$ which isn't $a^2 - b^2$.  So $\sqrt{7-x} - \sqrt{x-13}$ squared is
$$
(7-x) + (x-13) - 2\sqrt{(7-x)(x-13)}
$$

Now you can isolate the square root term and square again.


Projection of the intersection between a paraboloid and a sphere I am trying to find the projection in the xy-plane from the intersection between a sphere with radius sqrt(3) and the paraboloide z=1/2(x^2+y^2).

I tried to insert z^2=1/4(x^2+y^2) in the equation for the sphere x^2+y^2+x^2=3, but then I get a equation with x^4 and y^4. From plotting it implicitly, it seems like the answer should be x^2+y^2=2. How can I reduce it to that form?

Thank you.

The paraboloid

$$
z = \frac 12(x^2+y^2)
$$

and the sphere

$$
x^2+y^2+z^2=3
$$

intersect at

$$
z = \frac 12(3-z^2)
$$

or at the feasible $z = 1$ hence the intersection is the circle

$$
x ^2+y^2=2
$$

Given, 2z=x^2 + y^2

Substitute this in the equation of sphere

Solve the quadratic in z

You will get z=1, (and imaginary root -3)

This means x-y plane that is shifted towards positive z axis by a unit distance and so the projection on z=1 will be the projection on x-y plane (z=0)

Substitute z=1 in the equation of the sphere.


How to compute "weight"? Assume I have a list of 100 used cars, which the following information for each: year, make, model, mileage, selling price. How could I figure out how much does the year, make, model, and mileage affect the cars selling price?

Is there a program that'll do this? If someone can explain the math to me - or at least point me in the right direction, maybe I can write a simple program to figure this out myself.

Have considered using regression analysis?


Is every submodule of a free module of finite rank over a PID a direct summand? Suppose $F$ is a free module of rank $n$ over a PID, with $N$ a submodule. Is $N$ always a direct summand of $F$?

I think the answer is yes, $N$ is also free of rank $m\leq n$ since we are working over a PID, and then $N\simeq R^m$, thus $F/N\simeq R^{n-m}$ which is a free $R$-module, so $N$ is a direct summand of $F$.

I'm just a little unsure since I can't find a reference for this online, and it seems like it should be a commonly documented fact if it is indeed true.

No. Let $R$ be a principal ideal domain that is not a field, and let $p\in R$ be a prime element. Then the ideal $I=pR$ cannot be a summand of $R$. Indeed, if there were an $R$-module isomorphism $R\cong I\oplus M$ for some $R$-module $M$, then we would have an injection $R/I\cong M\hookrightarrow R$, which is impossible as the source is a non-zero torsion $R$-module.

Your error is in concluding that the quotient of two free modules is free. This is not so, as the above example shows.

Suppose $F$ is a free module of rank $n$ over a PID, with $N$ a
  submodule. Is $N$ always a direct summand of $F$?

Yes this is true. First,  $N$ is a free module, as it is a submodule of a free module over a PID. Now let $\{x_1, x_2, \ldots, x_n\}$ be a basis for $F$, and let $\{y_1, y_2, \ldots, y_k\}$ be a basis $N$. Since we are working over a PID, we know $k \le n$. Then we can define an $R$-linear map $\phi:F\rightarrow N$ where
$$\phi(x_i) = y_i$$ for $1 \le i \le k$, and
$$\phi(x_j) = 0$$
for all $k \le j \le n$. Then $\phi$ is surjective. Then consider the short exact sequence
$$0 \rightarrow \ker(\phi) \xrightarrow{\iota} F \xrightarrow{\phi} N \rightarrow 0,$$
where $\iota:\ker(\phi)\rightarrow F$ is the inclusion map. Then since $N$ is free and hence projective, the above exact sequence splits. Then
$$F \cong \ker(\phi)\oplus N,$$
hence $N$ appears as a direct summand.


Group of order 7.11.17 is cyclic Let $G$ be a group of order $7.11.17$. Show that $G$ is cyclic.

I tried to find a solution using Sylow theorems but I got stuck, here it goes:

We know that $$n_7 \equiv 1 (7) \space, n_7|11.17 \implies n_7=1,$$$$n_{11} \equiv 1 (11) \space, n_{11}|7.17 \implies n_{11}=1,$$$$n_{17} \equiv 1 (17) \space, n_{17}|7.11 \implies n_{17}=1$$

If we call $H,M,N$ to the unique 7-Sylow, 11-Sylow and 17-Sylow subgroups respectively, then $H,M,N \lhd G$. We have $H \cong \mathbb Z_7,M \cong \mathbb Z_{11}$ and $N \cong \mathbb Z_{17}$. Each of these groups is cyclic so $Z_7 \times Z_{11} \times Z_{17}$ is cyclic as well. I don't know how could I conclude from here that $G$ is cyclic, I've tried with semidirect products but I didn't arrive to anything concrete. Any suggestions would be appreciated.

Since $H$, $M$ and $N$ are the unique Sylow subgroups (of their respective orders) of $G$, then they are normal, so in particular $MN$ is normal. Then $G$ is the semidirect product $H\rtimes_f MN$ under the action of conjugation. But there are only $6$ automorphisms of $H$ (i.e. $\operatorname{Aut}(H)$ has order $6$), and $MN$ has order $11\cdot 17$, which is coprime with $6$, so the only homomorphism $MN\to\operatorname{Aut}(H)$ is the trivial one, that is $G=H\times MN$. Similarly, $MN$ is a semidirect product $M\rtimes_g N$ under some action $g$ which has to be trivial since the order of $\operatorname{Aut}(M)$ is 16 and is coprime with the order of $N$, which is 11. Therefore $G\simeq H\times M\times N$.

Well, certainly

$$\begin{cases}G=HMN\\{}\\|G|=|H||M||N|\end{cases}\stackrel{\text{why? Care here!}}\implies G\cong H\times M\times N$$


What are the applications of finite calculus I'm reading through Concrete Mathematics [Graham, Knuth, Patashnik; 2nd edition], and in the section regarding Summation, they have a sub-section entitled "Finite and Infinite Calculus". In this section they introduce to the reader the concept of finite calculus, the discrete analog of the traditional infinite calculus.

Throughout the text, they use the following notations for use in finite calculus (I'm not sure if this is standard notation, so I'd be grateful for any clarification):

$$\Delta f(x)\equiv f(x+1)-f(x)$$

Which is described as the "finite analog of the derivative in which we restrict ourselves to positive integer values of h".

Along with the following notation as the analog for the anti-derivative, and definite integral:

$$g(x)=\Delta f(x) \iff \sum{g(x) \:\delta x}=f(x)+C\\\sum_{a}^{b}{\Delta f(x)\:\delta x}=\left.f(x)\right|_{a}^{b}=f(b)-f(a)$$

It then goes on to introduce various identities relating to these interesting operators, but never describes any application for them, aside from satisfying one's mathematical curiosity, and for shortening a few proofs later in the book (although even these seem to be somewhat contrived applications).

So my question is as follows:

  What applications are there for finite calculus, in fields such as mathematics, computer science, physics, etc.?

Thanks in advance!

As I didn't really get a satisfactory answer (although thanks to user35071 for his link, and to those of you that commented), I have decided to answer my own question as best as I can (if I've missed anything important, please let me know in the comments):

Calculating closed-form summations from known expressions.
Approximation of infinite-calculus (forward-difference methods, numerical solutions to PDEs and ODEs, etc.)

It is often the case that we are trying to express the summation of some expression $f(x)$ in closed form, and this can be tricky, using standard-techniques.

For instance, computation of $\sum{x^{2}\:\delta x}$ can be done by using two known formulae:

$$x^{n}=\sum_{k}{\left\{n \atop k\right\}x^{\underline{k}}} \text{ and } \sum{x^{\underline{m}}\:\delta x}=\frac{x^{\underline{m+1}}}{(m+1)}\tag{1}$$

Where $\left\{n\atop k\right\}$ is read "$n$ subset $k$", and is a Stirling number of the second kind. And $x^{\underline{k}}=x(x-1)\cdots(x-k+1)$ is the falling-factorial function (note this is also written as the Pochhammer symbol: $(x)_{k}$).

Expanding $x^{2}$ in terms of falling factorials, we get:

$$x^{2}=x^{\underline{2}}+x^{\underline{1}}\implies\sum{x^{2}\:\delta x}=\sum{x^{\underline{2}}\:\delta x}+\sum{x^{\underline{1}}\:\delta x}$$

Using our known formulae in $(1)$, we get:

$$\sum{x^{2}\:\delta x}=\frac{x^{\underline{3}}}{3}+\frac{x^{\underline{2}}}{2}=\frac{x(x-1)(x-2)}{3}+\frac{x(x-1)}{2}=\frac{x(x-1)(2x-1)}{6}$$

Which corresponds to our standard closed form for $\sum_{k=1}^{n}{k^{2}}$.

If we don't restrict ourselves to integral finite-differences, then we can also use it to numerically approximate derivatives and integrals of continuous functions. There are several types of difference methods often used in numerical approximation to calculus,

$$\Delta_{h}[f](x)=f(x+h)-f(x) \tag{2}$$
$$\nabla_{h}[f](x)=f(x-h)-f(x) \tag{3}$$
$$\delta_{h}[f](x)=f\left(x+\frac{h}{2}\right)-f\left(x-\frac{h}{2}\right)\tag{4}$$

Where $(2)$ is called the forward-difference, $(3)$ is called the backwards-difference, and $(4)$ is called the central-difference. These can be used to approximate the derivative using the following formulae:

$$f'(x)\approx\frac{\Delta_{h}[f](x)}{h}\approx\frac{\nabla_{h}[f](x)}{h}\approx\frac{\delta_{h}[f](x)}{h} \tag{5}$$

Which is often used when a non-analytic evaluation of a derivative is required.

Discrete calculus (a preferable term I think) has the same applications "continuous" calculus has: free fall, planetary motion, population growth, spring oscillation, heat transfer, wave propagation, Newtonian physics... It's what calculus has just before you take that limit (http://calculus123.com/wiki/Discrete_calculus).