How do I prove a combinatorial statement about the change-making problem when using the greedy algorithm? Let $D$ be set of denominations and $m$ the largest element of $D$. We say that $c$ is a counterexample if the greedy algorithm gives an answer different from the optimal one.

Now, apparently, if for a given set $D$, the greedy algorithm does not return an optimal solution, then the smallest $c$ will be smaller than $2m-1$.

Is this really true? How can we prove it? If not, is there a relatively small range to look for the smallest counterexample?

The paper "Optimal Bounds for the Change-Making Problem" (by Kozen and Zaks, TCS 1994) gives a bound of $x < c_M + c_{m-1}$ where $x$ is the counter example and $c_m$ and $c_{m-1}$ are the largest and second largest coin. They claim the bound is optimal. (I just browsed the paper, and I did not take the time to understand whether this automatically means you cannout expres it in the largest coin value)

Jeffrey Shallit (famous for his paper "What this country needs is an 18c piece") in 1998 posted a bibliography on the subject: mathforum. Shallit adds "... it is a problem in which it is notoriously
easy to make wrong conclusions -- some of them have even been published".

Good luck with your investigations.

I recently came up with 1 solution that seemed to show if the given 2 conditions are satisfied, the greedy algorithm would yield the optimal solution.

a) The G.C.D (All the denominations except 1) = 2nd Smallest denomination.

b) The sum of any 2 consecutive denominations must be lesser than the 3rd consecutive denomination.

For eg. $c_2 + c_3 < c_4$.

(Where $c_1 = 1; c_2, c_3, c_4$ are coin denominations in ascending order).

I understand this is not a complete solution. However, I believe that if these 2 conditions are met, the greedy algorithm will yield the optimal solution.


Galois group. $K$-automorphisms take adjoined roots to other roots of minimal polynomial or take roots of $f$ to other roots of $f$ Say $f\in \Bbb Q[x]=K$ is irreducible, and $L/K$ is a field extension where $L=\Bbb Q(\alpha,\beta)$ where $\alpha,\beta$ are not in $\Bbb Q$ i.e. we need to adjoin both for $L$ to be the splitting field of $f$ since $\alpha,\beta$ are both roots of $f$. The minimal polynomials of $\alpha,\beta$ however aren't $f$.

Do 1) $K$-automorphisms of $L$ take roots of the minimal polynomial of $\alpha$ to other roots of the minimal polynomial of $\alpha$ and roots of the minimal polynomial of $\beta$ to other roots of the minimal polynomial of $\beta$

Or 2) Take roots of $f$ to other roots of $f$?

Or 3) These are somehow the same?

The first makes sense to me, but the second seems to be what we do in the examples. Thanks

I just mean that if we had $L=K(\alpha,\beta)$ is the splitting field of $f$, and we want to find $K$-automorphisms of $L$, in order to find elements of $\rm{Gal}(L/K)$, do we determine $\sigma\in\rm{Gal}(L/K)$ by seeing where it maps $\alpha,\beta$ in which we know these can only go to either, other roots of their respective minimal polynomials, or other roots of $f$.(Hopefully that makes sense)

Both are true. In general, let $g(x) = a_0 + a_1x + \cdots + a_nx^n$ be a polynomial with coefficients $a_i \in K$, $y \in L$ be a root of $g$, and $\phi$ be a $K$-automorphism of $L$. Then
$$0 = \phi(g(y)) = a_0 + a_1\phi(y) + \cdots + a_n\phi(y) = g(\phi(y)).$$
Thus $\phi$ permutes the roots of $g$.


Expected Value of a Probability Density Function with Absolute Value I'm given a probability density function $f(x) = c |(x^2 - 1)|$ for $-2 \leq x \leq 3.$ I found $c$ by by integrating from -2 to 3 and setting this equal to 1 and got $c = \frac{3}{28}$, so $f(x) = \frac{3}{28} |(x^2 - 1)|$. However, this was tedious because I couldn't think of any other way to do it besides splitting the function into three parts (two positive chunks and one negative chunk), due to the absolute value.

Now I'm asked to find E(X), the expected value. (I did what my textbook told me and simply multiplied each by $x$.) I split it into three integrals again.

$$E(X) = \int\limits_{-2}^{-1}{x\frac{3}{28}(x^2 - 1)}\,dx + \int\limits_{-1}^{1}{x\frac{-3}{28}(x^2 - 1)}\,dx + \int\limits_{1}^{3}{x\frac{3}{28}(x^2 - 1)}\,dx$$

Now it seems like I'm finding the expected value of three separate components, so I thought I might have to average them. But my calculator says

$$\int\limits_{-2}^{3}{x\frac{3}{28}|{(x^2 - 1)}|}\,dx = \frac{165}{112}$$

which is what I got when I did the addition out above.

But I think this must be wrong, since the function $\frac{3}{28}|(x^2 - 1)|$ doesn't even attain that value over the interval, and my understanding is that expected value is like the average of the probability density function.

Is my methodology wrong?

You're finding the average value (not the average probability), and the value ranges from $-2$ to $3$, so $165/112$ looks like a possible average value (especially since the pdf is weighted toward the high numbers).

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
 \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
 \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,}
 \newcommand{\dd}{{\rm d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}}
 \newcommand{\expo}[1]{\,{\rm e}^{#1}\,}
 \newcommand{\fermi}{\,{\rm f}}
 \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,}
 \newcommand{\half}{{1 \over 2}}
 \newcommand{\ic}{{\rm i}}
 \newcommand{\iff}{\Longleftrightarrow}
 \newcommand{\imp}{\Longrightarrow}
 \newcommand{\Li}[1]{\,{\rm Li}_{#1}}
 \newcommand{\norm}[1]{\left\vert\left\vert\, #1\,\right\vert\right\vert}
 \newcommand{\pars}[1]{\left(\, #1 \,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\pp}{{\cal P}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,}
 \newcommand{\sech}{\,{\rm sech}}
 \newcommand{\sgn}{\,{\rm sgn}}
 \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}}
 \newcommand{\ul}[1]{\underline{#1}}
 \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
If we do not want 'to split' the integral, it's easier to use some properties related to the Dirac Delta Function $\ds{\delta}$ such as
$$
\quad\totald{\verts{x}}{x} = \sgn\pars{x}\,,\quad
\totald{\sgn\pars{x}}{x} = 2\delta\pars{x}\quad\mbox{and}\quad
\delta\pars{\fermi\pars{x}}=\sum_{i}{\delta\pars{x - x_{i}} \over \verts{\fermi'\pars{x_{i}}}}
$$
where $\ds{\fermi\pars{x_{i}} = 0}$. We recommend to any reader to check carefully the above link.

\begin{align}&\color{#66f}{\large\int_{-2}^{3}\verts{x^{2} - 1}\,\dd x}
=\overbrace{\left.\vphantom{\LARGE A}x\verts{x^{2} - 1}\ \right\vert_{-2}^{\phantom{-}3}}^{\ds{=\ \dsc{30}}}\,\,\,\,\,\,\,
-\,\,\,\int_{-2}^{3}x\sgn\pars{x^{2} - 1}2x\,\dd x
\\[5mm]&=30 -\ \overbrace{\left.\vphantom{\LARGE A}{2 \over 3}\,x^{3}\sgn\pars{x^{2} - 1}\ \right\vert_{-2}^{\phantom{-}3}}^{\ds{=\ \dsc{70 \over 3}}}\,\,\,\,\,\, +\,\,\,
\int_{-2}^{3}{2 \over 3}\,x^{3}\bracks{2\,\delta\pars{x^{2} - 1}2x}\,\dd x
\\[5mm]&={20 \over 3}
+{8 \over 3}\ \overbrace{\int_{-3}^{2}
\bracks{{\delta\pars{x + 1} \over 2} + {\delta\pars{x - 1} \over 2}}\,\dd x}
^{\ds{=\ \dsc{1}}}
\ =\ \color{#66f}{\large{28 \over 3}}
\end{align}


(NOT a physics question) Is electric field always asymptopic to $x^{\alpha}$ for some rational $\alpha$? In three dimensional space with origin $O$, you pick a finite number of points $P_1, P_2, \cdots, P_n$. To each point $P_i$ you assign a nonzero integer (positive or negative) $q_i$. For all other points $R$ in the plane, define the vector valued function $$\displaystyle \vec{F(R)} = \sum_{i = 1}^{n} \frac{q_i}{D(P_i, R)^2} \vec{r_i}, $$ where $D(P_i, R)$ is the Euclidean distance between $P_i, R$, and $r_i$ is a vector of unit magnitude directed from $P_i$ to $R$. Now you pick a ray $\vec{\ell}$ originating from $O$ in any direction.

Is it true that for any such configuration of such points, there always exist an rational number $\alpha$ such that $\displaystyle \lim_{x \rightarrow \infty} \| F(R_x) \| x^{\alpha}$ converges to some nonzero constant, where $R_x \in \ell$  with $D(O, R_x) = x$ and $\| F(R_x) \|$ is the magnitude of the function at $R_x?$

No. Place positive unit charges at $(\pm 1,0,0)$ and negative unit charges at $(0,\pm 1,0)$. Then the field is zero on the whole $z$-axis.

Far away from any localized distribution of point charges the value of $||F(R_x)||$ will drop off as $1/x^2$ so you want $\alpha=2.$ In particular:

$$\lim_{x\to \infty}||F(R_x)||x^2=\sum_{i=1}^nq_i$$


Why is Euler's formula for connected planar graphs different on Toruses? I was a youtube video on Euler's formula for planar graphs. https://www.youtube.com/watch?v=VvCytJvd4H0&vl=en
The creator of the video provides a very satisfying proof of why Euler's formula is V - E + F = 2 for planar graphs on a flat sheet of paper and then asks the viewers to attempt to determine why V - E + F = 0 for a planar graph on a torus. I have tried many variations of the graph of a mug (which I understand has the same topological properties as a torus) and have only been able to determine V - E + F to be equal to 1 by including the handle in my calculations.
Could someone please provide me with a proof for this?

The best thing to do is to use an abstraction for the torus to get a better intuitive handle on things. In particular, imagine taking a square and joining its left edge to its right one by bending it backwards. You glue that edge in place - giving yourself a cylinder with an open top and bottom. Then, you stretch the cylinder and bend it to mate the top edge with the bottom edge. This gives you a torus!

We can pictorially represent this process in two dimensions: a torus is really just a square in which, whenever you hit an edge, you pop out the opposite edge that you would have glued to - so you wrap around each time you reach the edge. Try playing around with that model a bit to get a feel for how it works. Note that a few strange things can happen - like, you could draw a single point in the middle, then draw two edges connecting that point to itself: one going straight up, wrapping back to the bottom, then continuing up back to the point, and one going left and wrapping back to the right. If you count, you find $1$ vertex and $2$ edges along with $1$ face (which is a square).

You can also realize that, while the result for planes is fairly robust, the result for a torus is weaker: if I took the previous diagram and removed an edge, I'd have $1$ vertex and $1$ edge, but only $1$ "face" - except that the face would be a cylinder, which is not allowed. Basically, there is a condition: the Euler formula only works if all of the regions you have are actually polygons. This isn't an issue for connected non-empty graphs in the plane, but causes trouble for toruses.

A reasonable, if somewhat informal, proof for the torus would be to nudge your graph around in this square - without changing the number of vertices, edges, and faces - so that no vertex is on the edge of the square and no edge passes through a vertex of a square. Then, you draw, as part of a graph, the edges of the "imaginary" square you were using as your torus, adding a vertex wherever these edges intersect an edge of your original graph. This gives you a procedure that turns graphs on a torus into graphs on a plane.

Exactly proving it can be difficult without deeper mathematical ideas - you end up with trouble because the usual argument of "remove an edge and see what happens" no longer works, since doing so might merge two polygons into something that's not a polygon.

A simple, if much less than formal, intuition is that, if you want to draw a graph on a torus, you could start by putting a vertex at the corner of the square -which, in your square, will appear at every corner. Then, you draw some number $k$ of vertices on each edge - noting that these appear on both of the glued edges in your graph - and you connect these vertices along the edge of the square. Then, you draw some graph inside the square, allowing for edges to pass to the edge of the square. You view this as a planar graph with $V'$ vertices and $E'$ edges and $F'$ faces and get that $V'-E'+F'=2$.

However, if you let your graph on the torus have $V$ vertices and $E$ edges and $F$ faces, you find that $V'=V+3+k$ because the corner appears $4$ times in the plane and each of the $k$ vertices on the edge appears twice times. You find that $E'=E+2+k$ because, as there are $2k+4$ total vertices on the boundary of the square viewed in the plane, connected in a cycle, there are equally many edges. Each edge is the same as the one on the opposite side of the square when viewed on the torus, so half of these - $k+2$ - were added when we view this in the plane. Finally, $F'=F+1$ because no two faces in the square are identified, but we do add an extra face for the "exterior" which wasn't there before.

If we substitute in these formulas, we get $$(V+3+k)-(E+2+k)+(F+1)=2$$
which simplifies to $V-E+F=0$.

The part of the proof that is missing is that this only covers the case where we include the boundary of the imaginary square inside of the graph itself - a true proof would need to start from an arbitrary graph and show it from there. A quick fix to this issue is to start with an arbitrary graph and then add edges and vertices - making sure not to modify the quantity $V-E+F$ in doing so - until you get to a case where this proof applies.

Euler's formula generalise to any closed orientable surface of genus $g$ (genus $\approx$ number of "handles" in your surface):
$$ V-E+F=2-2g$$

Or to any closed non-orientable surface of genus $k$
$$V-E+F=2-k$$

The wikipedia article on Euler's characteristic should give you plenty enough information.


Trigonometric equation: $2 \cot^2 \alpha - 2 \sec^2 \alpha = 1$ Can someone help me solving this equation:

$$
2 \cot^2 \alpha - 2 \sec^2 \alpha = 1
$$

I have to solve this for $\alpha$. Is there any relation I'm missing? I guess I'm going to have to bring both to the same trigonometric number?

Hint

$$\sec^2\alpha=1+\tan^2\alpha$$

so

$$\frac{1}{\tan^2\alpha}-1-\tan^2\alpha=\frac{1}{2}$$

Now use $u=\tan^2\alpha$

$$\frac{1}{u}-u-\frac{3}{2}=0 \rightarrow 2u^2+3u-2=0$$

What give us $u=-2$ or $u=1/2$.

Can you finish?

$$
\sec^2(\alpha)=1+\tan^2(\alpha)
$$
And
$$
\cot^2(\alpha)=\frac{1}{\tan^2(\alpha)}
$$
Simplifying we get a quadratic equation which can solved to get the value of $\tan^2(\alpha)$


Prove that $(x^2+1)\mathbb Z[x]$ is a prime ideal of $\mathbb Z[x]$, but not maximal Prove that $(x^2+1)\mathbb Z[x]$ is a prime ideal of $\mathbb Z[x]$, but not maximal.

I'm supposed to show this for my homework. My first thought is to show that $\mathbb Z[x]/(x^2+1)\mathbb Z[x]$ is an integral domain but not a field, but I don't know how to do that. Please don't give me the answer but if you can give me a hint I'd really appreciate it.

Update: I used the unique factorization of $\mathbb Z[x]$ and the irreducibility of $x^2+1$ in $\mathbb Z[x]$ to show that $\mathbb Z[x]/(x^2+1)\mathbb Z[x]$ is an integral domain. I'm still working on the other part.

You have the right idea: the quotient ring is a domain but isn't a field. It might be helpful to note that the quotient is isomorphic to $\mathbb{Z}[i]$, and then show that $2$ (for instance) isn't invertible.

You could define a natural homomorphism which maps $\mathbb{Z}$ to its quotient ring $\mathbb{Z}[x] / <x^2+1>$ which can be defined as $\{[a] | a\in \mathbb{Z}\}$ and we let the homomorphism be $\pi$; that is $\pi:\mathbb{Z}[x]/<x^2+1> \mapsto \mathbb{Z}, \pi([a]) = a$. It is trivial to show that is an isomorphism, since it is a bijection as well as a homomorphism.

We know that $\mathbb{Z}$ is an integral ring but not a field (because $\forall z \in \mathbb{Z} \setminus \{-1, 1\}$, we cannot find an inverse of $z$); therefore $<x^2+1>$ is a prime ideal of $\mathbb{Z}[x]$ but not maximal.


Are Elementary Functions chosen by convention? Are the elementary functions chosen by convention? What leads me to this question is that many non-elementary functions have power series representations, allowing them to be computed numerically, just as the trig and exponential functions. The sine function and the error function both usually require numerical methods as you can't just "plug in" a number and perform the calculation (I.e. power series or other infinite series would be required). So then what differentiates elementary functions from non-elementary functions?

There is a technical definition used in differential algebra.  Elementary functions are the smallest field containing complex constants and the variable $x$ and closed under the following operations:

(Algebraic extensions): if $P(X)$ is a polynomial of degree $> 1$ with elementary function coefficients, then there is an elementary function $g$ such that $P(g) = 0$.
(Exponential extensions): if $f$ is an elementary function, then there is an elementary function $g$ such that $g' = f' g$.
(Logarithmic extensions): if $f$ is an elementary function (not identically $0$), then there is an elementary function $g$ such that $g' = f'/f$.

Yes, elementary functions are indeed chosen by convention, for all the reasons you've stated. For most people, the elementary functions are the functions that their calculator in school could compute.


What is meant by a continuous-time white noise process? What is meant by a continuous-time white noise process?

In a discussion following a question a few months ago, I stated that as an engineer, I am used to thinking
of a continuous-time wide-sense-stationary white noise process
$\{X(t) \colon -\infty < t < \infty\}$
as a zero-mean process having autocorrelation function $R_X(\tau) = E[X(t)X(t+\tau)] =  \sigma^2\delta(\tau)$ where $\delta(\tau)$ is the Dirac delta or impulse, and power spectral density $S_X(f) = \sigma^2, -\infty < f < \infty$. At that time, several people with very high reputation on Math.SE
assured me
that this was an unduly restrictive notion, and that no difficulties arise
if one takes the autocorrelation function to be
$$E[X(t)X(t+\tau)] = \begin{cases}\sigma^2, & \tau = 0,\\
0, & \tau \neq 0. \end{cases}$$

What engineers like to call a white noise process is a hypothetical
beast that is never observed directly in any physical system, but
which can be used to account for
the fact that the output of a linear time-invariant system whose
input is thermal noise is well-modeled by a wide-sense-stationary
Gaussian process whose power spectral density is proportional to
$|H(f)|^2$ where $H(f)$ is the transfer function of the linear
system. Standard second-order random process theory says that
the input and output power spectral densities $S_X(f)$ nd $S_Y(f)$
are related as
$$S_Y(f) = S_X(f)|H(f)|^2.$$
Thus, pretending that thermal noise is a white Gaussian noise process in the
engineering sense and pretending that the second-order theory
extends to white noise processes (even though their variance is
not finite) allows us to get to the result that the output power
spectral density is proportional to $|H(f)|^2$.

My query about the definition of a white noise process
is occasioned by a more recent question regarding the variance of a random variable $Y$ defined as
$$Y = \int_0^T h(t)X(t)\ \mathrm dt$$
where $\{X(t)\}$ is a white Gaussian noise process.
The answer given by Nate Eldredge
leads to
$$\operatorname{var}(Y) = \sigma^2 \int_0^T |h(t)|^2\ \mathrm dt$$
(as I pointed out in a comment on the answer) if the autocorrelation
function is taken to be $R_X(\tau) = \sigma^2\delta(\tau)$
(the engineering definition).  However, the OP on that question
specified $R_X(0) = \sigma^2$, not $\sigma^2\delta(\tau)$,
that is, the definition accepted by mathematicians.  For
this autocorrelation function, the variance is
$$\int_0^T \int_0^T E[X(t)X(s)]h(t)h(s)\mathrm dt\mathrm ds = 0$$
since the integrand is nonzero only on a set of measure $0$.

So, what is the variance of the random variable $Y$? and what
do readers of Math.SE understand by the phrase white noise process?

Perhaps this question should be converted to a Community wiki?

This is a bit late, but I see that the main points in this question have not been completely addressed. I'll set
\begin{equation}
\sigma = 1
\end{equation}
for this answer.

The definition of white noise may be context-dependent: How you define it depends on what you want to do with it. There's nothing inherently wrong with saying that white noise (indexed by a set $T$) is just the process of iid standard normal random variables indexed by $T$, i.e. $E[X(t)X(s)] = \begin{cases} 1 & t = s \\ 0 & t \neq s \end{cases}.$ However, as cardinal noted here, Example 1.2.5 of Kallianpur's text shows that this process is not measurable (as a function of $(t, \omega)$). This is why, as Did commented above, $Y$ is undefined (with this definition of $X$). Thus, this definition of white noise is not appropriate for defining objects like $Y$.

Rather, you want $X$ to have covariance given by the Dirac delta. But the $\delta$ function is not a function but rather a measure and the best context for understanding it is the theory of distributions (or generalized functions---these are not to be confused with "probability distributions"). Likewise, the appropriate context for white noise is the theory of random distributions.

Let's warm up with a heuristic explanation: We'll think of white noise as the "derivative" of Brownian motion: "$dB_t/dt = X_t$". So ignoring rigor for a moment, we could write
\begin{equation}
\int_0^T h(t) X(t) dt = \int_0^T h(t) \frac{dB_t}{dt} dt = \int_0^T h(t) dB_t.
\end{equation}

The reason this isn't rigorous is that Brownian motion is nowhere differentiable. However, the theory of distributions allows us to "differentiate" non-differentiable functions. First of all, a distribution is a linear functional (linear map taking values in the real numbers) on a space of "test functions" (usually smooth functions of compact support). A continuous function $F$ can be viewed as a distribution via the pairing
\begin{equation}
(F, f) = \int_0^\infty F(t) f(t) dt.
\end{equation}
The distributional derivative of $F$ is the distribution $F'$ whose pairing with a test function $f$ is defined by
\begin{equation}
(F', f) = -(F, f').
\end{equation}

Thinking of Brownian motion as a random function, we can define white noise $X$ as its distributional derivative. Thus, $X$ is a random distribution whose pairing with a test function $f$ is the random variable
\begin{equation}
(X, f) = -(B, f') = -\int_0^\infty B(t) f'(t) dt.
\end{equation}
By stochastic integration by parts,
\begin{equation}
(X, f) = \int_0^\infty f(t) dB_t;
\end{equation}
this is the Itô integral of $f$ with respect to $B$.

Now a well-known fact in stochastic calculus is that $M_T = \int_0^T f(t) dB_t$ is a martingale starting at $M_0 = 0$, so $E (X, f) = 0$. Moreover, by the Itô isometry,
\begin{equation}
\mathrm{Var}((X, f)) = E (X, f)^2 = \int_0^\infty f(t)^2 dt.
\end{equation}
It can also be verified that $(X, f)$ is Gaussian.

My main point is that a more appropriate definition of $Y$ might be
\begin{equation}
Y = \int_0^T h(t) dB_t.
\end{equation}

As a last note, because of the way $X$ was defined above, $X_t$ is not defined but $(X, f)$ is. That is, $X$ is a stochastic process but whose index set is given by $T = \{ \text{test functions} \}$ rather than $T = [0, \infty)$. Moreover, again by the Itô isometry,
\begin{equation}
E (X, f) (X, g) = \int_0^\infty f(t) g(t) dt.
\end{equation}
Abandoning rigor again, this becomes
\begin{equation}
E (X, f) (X, g) = \int_0^\infty \int_0^\infty f(s) \delta(s - t) g(t) ds dt
\end{equation}
and it is in this sense that the covariance of $X$ is the Dirac delta.

Edit: Note that we could leave the definition of $(X, f)$ in terms of the ordinary integral and do all the above calculations using Fubini's theorem and (ordinary) integration by parts (it's just a bit messier).

Actually X is the digital signal and Y is the average analog signal power generated in the time-domain. The variance shows how deviation goes if signal Y goes for a long time.

Moreover, you should learn more about the delta function (your function becomes one if x = 0 ) . The engineering and mathematical aspects to explain the autocorrelation function are also correct. There is no contracdiction between them.


find delta, maybe by congruent triangles? Hey there: I think this is a rather short and easy question for you. Can anyone either way please give me a hint? Would be very lovely!
I found out about almost all angles in this triangle. In my solution sheet it says that $\delta$ should be $10^\circ$. But how can I explain it or prove it? I don't have a clue how to find $\beta$ or $\delta$. Can anyone please give me a hint?

Hint: Angle subtended by a CHORD on the circumference is equal. (Given the fact that $w$ is the Angular bisector).

Notice here: $ \angle EBC=\angle EDC=70^0$. Now you can go on.

Given:
A
C
‾
AC
  and
B
D
‾
BD
  bisect each other.

Prove:
B
C
‾
≅
A
D
‾
BC
 ≅
AD
 .

Note: quadrilateral properties are not permitted in this proof.
Step    Statement   Reason
1
A
C
‾
AC
  and
B
D
‾
BD
  bisect each other
Given
try
Type of Statement
A
B
C
D
E
Note:
A
C
‾
AC
  and
B
D
‾
BD
  are segments.

You must answer all questions above in order to submit.


Question concerning prime ideals of $\mathbb{C}[x,y]$ I know that $(0)$, $(x-a,y-b)$ for $(a,b)\in\mathbb{C}^2$ and $(f(x,y))$ for $f(x,y)$ irreducible in $\mathbb{C}[x,y]$ are all prime ideals in $\mathbb{C}[x,y]$.
What I'd like to understand is why they are the only prime ideals. In particular, I'd like to know why the following outline of a proof is valid (which comes from Vakil's algebraic geometry notes): Let $P$ be a prime ideal that is not principal. "Show you can find $f(x,y),g(x,y)\in P$ with no common factor. By considering the Euclidean algorithm in the Euclidean domain $\mathbb{C}(x)[y]$, we can find a nonzero $h(x)\in (f(x,y),g(x,y))\subset P$."
I have two questions.

Why can we find such $f(x,y),g(x,y)$ without a common factor? If we were to pick to distinct generators of $P$, I don't see why they must be coprime. Further, I don't think we can just factor out a common factor either.
Why does the Euclidean algorithm imply we can find such a nonzero $h(x)$? Might this not still be of the form $h(x,y)$ since the Euclidean algorithm only guarantees the existence of $q(x,y)$ and $h(x,y)$ with $f(x,y)=g(x,y)q(x,y)+h(x,y)$ with the $y$ degree of $h(x,y)$ less than that of $g(x,y)$? Does this somehow follow from $f(x,y)$ and $g(x,y)$ having no common factors?

We know that $P$ can be generated by finitely many generators (because $\mathbb C[x, y]$ is Noetherian), so suppose that $f_1, \dots, f_n$ is a minimal set of generators for $P$. Since $P$ is not principal, $n \geq 2$.
Now recall that $\mathbb C[x, y]$ is a unique factorisation domain. So suppose that $h$ is a greatest common divisor for $f_1$ and $f_2$. Then $f_1 = hg_1$ and $f_2 = hg_2$, where $g_1, g_2$ have no non-trivial common factor.
Now $h$ can't be in $P$, otherwise $P$ would be generated by $h, f_3, \dots, f_n$, contradicting my assumption that the generators $f_1, f_2, \dots, f_n$ were minimal.
But $f_1 = hg_1 \in P$ and $P$ is prime, so either $h\in P$ or $g_1 \in P$. Since $h \notin P$, we have $g_1 \in P$. By a similar argument, $g_2 \in P$ too.
Thus we have constructed elements $g_1, g_2 \in P$ that have no non-trivial common factor.
For the second part, see this answer.

Ad 1: Pick $f$ of minimal (total) degree (but non-zero, of course). Then $P\ne(f)$, so there exists $g\notin (f)$ and again we pick it of minimal degree. What could you say about the degree of  a common factor?


Probability of landing on a particular point in an infinite 1D random walk $50\%$ of the time you walk right one unit, and $n$ units to the left o.w. The probability in question is ever landing one unit to the right of where you start at (as your number of moves goes to infinity).

I've read somewhere that the answer is (not sure if I remember correctly):
$$P=\frac{1}{2}+\frac{1}{2}P^{n+1}$$

but I cant figure out why. The first term is self-explanatory (half the time you reach your goal right-then-and-there), but I can't find an intuitive explanation for the second (recursive) term - the coefficient of the second term makes sense (half of the other time something else happens), but I can't explain the second factor and its power. Please explain what the interpretation is.

If you start by going left $n$ units, then you have to go right $n + 1$ times.

You might find the idea of Markov Chains useful to help solve your problem.


Mathematical ideas that took long to define rigorously It often happens in mathematics that the answer to a problem is "known" long before anybody knows how to prove it.  (Some examples of contemporary interest are among the Millennium Prize problems: E.g. Yang-Mills existence is widely believed to be true based on ideas from physics, and the Riemann hypothesis is widely believed to be true because it would be an awful shame if it wasn't.  Another good example is Schramm–Loewner evolution, where again the answer was anticipated by ideas from physics.)
More rare are the instances where an abstract mathematical "idea" floats around for many years before even a rigorous definition or interpretation can be developed to describe the idea.  An example of this is umbral calculus, where a mysterious technique for proving properties of certain sequences existed for over a century before anybody understood why the technique worked, in a rigorous way.
I find these instances of mathematical ideas without rigorous interpretation fascinating, because they seem to often lead to the development of radically new branches of mathematics$^1$.  What are further examples of this type?
I am mainly interested in historical examples, but contemporary ones (i.e. ideas which have yet to be rigorously formulated) are also welcome.

Footnote: I have some specific examples in mind that I will share as an answer, if nobody else does.

The notion of probability has been in use since the middle ages or maybe before. But it took quite a while to formalize the probability theory and giving it a rigorous basis in the midst of 20th century. According to wikipedia:

  There have been at least two successful attempts to formalize probability, namely the Kolmogorov formulation and the Cox formulation. In Kolmogorov's formulation, sets are interpreted as events and probability itself as a measure on a class of sets. In Cox's theorem, probability is taken as a primitive (that is, not further analyzed) and the emphasis is on constructing a consistent assignment of probability values to propositions. In both cases, the laws of probability are the same, except for technical details.

  There are other methods for quantifying uncertainty, such as the Dempster–Shafer theory or possibility theory, but those are essentially different and not compatible with the laws of probability as usually understood.

The Egyptians and the Babylonians knew a lot of mathematical facts (e.g. the Pythagorean theorem, the quadratic formula, the volumes of prisms and pyramids, etc.) which were later proved by the Greeks.

Also, Pascal and Fermat used mathematical induction and the well ordering of the naturals (in the form of infinite descent) 250 years before Peano formalized the axioms defining the natural numbers.


Confused about the $\pm$ sign? I have multiple questions about the $\pm$ sign, since it seems to confuse me in general...

Question 1:

Say I have $15=\pm(a+x)$, Can I use the distributive property so it becomes $15=\pm a \pm x$? Or does that mean I went from 2 to 4 solutions?

I get very confused when I encounter this sign in equations I need to simplify. Whenever I need to deal with these kind of situations I tend to just turn it into two equations, $15= a+x$ and $15=-a-x$. But if I were to do that with bigger equations that still need to be simplified it means I'm wasting alot of time since I'm doing twice the labor...

Question 2:

Does the $\pm$ sign only make sense in equations, or can they be used in normal expressions as well (e.g. $\pm y + 3$). But then I wonder, in what scenario would you do something like this?

Question 3:

$\sqrt {x^2}+37=y+40$

Say I were to simplify $\sqrt{x^2}$ in that equation, I don't know where I should put the $\pm$ sign. Where would I put it? The extra terms 37 and 40 confuse me...

If the $\pm$ sign is confusing you, get rid of it.  If you have $$15 = \pm(a+x)$$ you can turn that into two equations:  $$15 = a+x\\15 = -(a+x)$$ and then deal with the two equations separately, one at a time.  That is exactly the meaning of the $\pm$ sign.

The reason you're confused is because the notation is confusing!  The expression $\pm a \pm b$ is actually ambiguous: in some contexts it means four values, and in other contexts it means two. Sometimes there is a convention that the two $\pm$ signs must represent the same sign; sometimes there isn't.

In your third example, I'd suggest that you write $z = \sqrt{x^2}$ and turn the equation into $z+37 = y + 40$.  Then solve as usual.  When you get to the end, you have $z$ and $y$.  Then you can conclude that $x$ could be either $z$ or $-z$.

what if variable x = +1 and y = -1, since it is a variable, maybe 4x could be 4+1= 5, not multiplication in this state? y= -1, 2y = 2-1 = 1.
4x -2y=3.
4x-2y=3
_____=3
           _

4-2y=4-2y
x = 3
    ______
     4-2y

3             = +1

4-2y

3
-   = +1
3

x=+1 , y=-1


Is there a rational way to conceptualize an irrational number? This is a request for help, not an attempt to challenge anything.

Since $\pi$ is irrational, this tells me that it's impossible to express the distance around a circle in terms of the distance accross.

That boggles my mind, but maybe it should not.

I think crazy thoughts like: "this means that a the path of a circle around a unit lenth line segment has a non existant length".

Is there a way to accept that the number is irrational and not break from reason?

That $\pi$ is irrational does not mean that it's impossible to express the distance around a circle in terms of the distance across.  It only means it can't be expressed as a ratio of integers.

BTW here you can read some proofs that $\pi$ is irrational.

It is quite easy to prove that $\sqrt{2}$ is irrational.  It's even easier to prove that $\log_2 3$ is irrational.  But it's hard to prove $\pi$ is irrational.

Here is a related question that may be of interest, where you will read about how to prove some numbers are irrational.

Irrational numbers can and do correspond to length

Yes. An irrational number is a number that cannot be expressed as a ratio between two integers. It does exist as a length. For example, $\sqrt{2}$ is famously irrational, but it is extremely easy to show it as a length: just draw the diagonal of a square of side $1$.

Conceptualisation of irrational numbers

The whole rational vs irrational business started in ancient Greece. The Pythagoreans (the members of a secret group of mathematicians) believed that the whole universe could be described in terms of whole numbers: $1,2,3,4,...$
They were doing geometry, and they realised that some lengths were not integer multiples of other ones (more than $2$ times as long, but not quite $3$ times).

That's when they invented ratios: relationships between numbers. For example, $2$ is to $3$ just like $4$ is to $6$, in the sense that if you were to draw a rectangle with sides $2$ and $3$ and, next to it, one with sides $4$ and $6$, they would look the same (try it!). We express that relationship, by saying:
\begin{equation}
\frac{2}{3} = \frac{4}{6}
\end{equation}

Since they have faith in the integers, they believed that any length could be understood this way, two lengths coming from the simplest shapes: the circumference of the circle of diameter $1$ and the diagonal of the square of side $1$.

Now you know that it turns out that both $\pi$ and $\sqrt{2}$ are irrationals. So what does it mean?  It just means that it cannot be expressed as a ratio!

Check out this video for a brilliant exploration of all of this!


Proving that if $AB=A$ and $BA=B$, then both matrices are idempotent Let $A, B$ be two matrices such that $AB=A$ and $BA=B$, how do I show that $A\cdot A=A$ and $B\cdot B=B$?

Steps I took:

Let $A= \left[\begin{array}{rr}
    a & b \\
    c & d \\
   \end{array}\right]$ and let $B= \left[\begin{array}{rr}
    w & x \\
    y & z \\
   \end{array}\right]$
$\left[\begin{array}{rr}
    a & b \\
    c & d \\
   \end{array}\right] \cdot \left[\begin{array}{rr}
    w & x \\
    y & z \\
   \end{array}\right] = \left[\begin{array}{rr}
    aw+by & ax+bz \\
    cw+dy & cx+dz \\
   \end{array}\right] $ (Which is also equal to A)
$\left[\begin{array}{rr}
    w & x \\
    y & z \\
   \end{array}\right] \cdot \left[\begin{array}{rr}
    a & b \\
    c & d \\
   \end{array}\right] = \left[\begin{array}{rr}
    wa+xc & wb+xd \\
    ya+zc & yb+zd \\
   \end{array}\right] $ (Which is also equal to B)

At this point I am stuck. I don't know how to proceed and I imagine that I started off on the wrong track to begin with. I'd like a hint to guide me in the right direction.

My proof after consultation with answerers below:

Proof:

1) Since $AB=A$, we can say that: $(AB)A=AA$ which is equal to $A^2$

2) Then, (by associativity of matrix multiplication), we can say that $A(BA) = AB$ (since $BA=B$)

3) Then, $AB=A$ (since $AB=A$ was given)

4) Therefore, $AA$ is equal to $A$

5) Since $BA=B$, we can also say that: $(BA)B=BB$ which is equal to $B^2$

6) Then, (by associativity of matrix multiplication) we can say that $B(AB)=BA$ (since $AB=A$)

7) Then, $BA=B$

8) Therefore, $BB$ is equal to $B$

9) Thus, $AA$ is equal to $A$ and $BB$ is equal to $B$

Q.E.D.

Notice that $AA = (AB)(AB) = A(BA)B = ABB = AB = A$.

Since $AB=A$ and $BA=B$, to show that $A$ is an idempotent we have to show that $A=A^2$
Therefore $A^2=(AB)^2 = A^2B^{ } = AB = A(BA) = AA = A^2 = A$. Proven


natural language proof assistant I was wondering whether there has been any attempt to create a proof assistant that you write in it, in english,

I mean you write your proof the usual way in TeX(maybe use a 'simpler english') then instead of sending it to a journal to have it verified you use the proof assistant to have it verified for you.

There are programming languages like inform7 in which you program in english. I think what is needed is a set of macro's to turn the tex into, lets say, something that Coq can verify. Is there any such thing out there?

Do you think if this happens casual mathematicians will use it? or are there deeper problems that people don't use them now??

The proof-checker CalcCheck takes input via $\TeX{}$ in the form of formulas and
accompanying English hints/justifications.
Given the input file, the system will output that the proof is valid at all steps or indicate which steps are poorly justified.
To the best of my knowledge, it currently recognizes most theorems of first order logic and set theory ---based on the great text ``A Logical Approach to Discrete Math.''
If I recall correctly, the back-end is in Haskell.

Main system site is at http://calccheck.mcmaster.ca/.
Manual: http://calccheck.mcmaster.ca/CalcCheckDoc/

On a final note, this system has been used in first-year logic courses to assist students in proof-writing. It is helpful to have a system check one's proof when in-doubt.

Edit The above was 2013, now as of 2017, it now supports

creation of logical theories via named modules in the style of the Agda language
unicode input directly via latex-style bindings
"code completion" for theorem names and definitions
coloured and somewhat helpful error messages.

Moreover, the system now no-longer needs to be installed as an application but can be used directly via a browser.
It has been successfully used in-place of paper-and-pencil examinations at the university level with over 200 students in the 2017 fall term alone.
Unfortunately, it is not open source.

I believe the RISC (Research Institute for Symbolic Computation) group at Castle Hagenberg in Austria have been involved in such things for many years.

See:  www.theorema.org

Based on the OP's comments below, it appears that they are not seeking an automated 'theorem prover', but an automated 'proof verifier'. If so, there is already a thread on this at:

How do proof verifiers work?


why $C_{2}[a,b]$ is an inner product space under the given inner product, while $C[a,b]$ is not? I could not see the problem here How can I prove if this is an inner product? as the calculation of the inner product is 1 with me, and also why $C_{2}[0,1]$ is an inner product space with the inner product defined by
$<f,g>=∫_{0}^{1}f(x)\bar g(x)dx$

, what is the difference?

The linear space $C[a,b]$ of continuous functions on $[a,b]$ is an inner product space under $\langle f,g\rangle = \int_{0}^{1}f(t)\overline{g(t)}dt$. The only property in question is whether $\langle f,f\rangle =0$ iff $f=0$. But this property follows easily from the Fundamental Theorem of Calculus. For suppose
$$
   \int_{0}^{1}|f(t)|^2dt = 0.
$$
Then $\int_{0}^{x}|f(t)|^2dt=0$ for all $0 \le x \le 1$ because
$$
 0 \le \int_{0}^{x}|f(t)|^2dt \le \int_{0}^{x}|f(t)|^2dt + \int_{x}^{1}|f(t)|^2dt = \int_{0}^{1}|(f)|^2dt = 0.
$$
$|f(t)|^2$ is continuous because $f$ is. Hence, $0=\frac{d}{dx}\int_{0}^{x}|f(t)|^2dt=|f(x)|^2$ for all $x$.

The linear space $C_2[0,1]$ is also an inner product space under the same inner product.

Note that neither space is a Hilbert space; the completion of either space is $L^2[0,1]$.

The normed space is an inner product space if and only if the parallelogram law holds. This is Jordan-von Neumann Theorem (1937). Check whether or not the parallelogram law holds in $C[0,1]$.


real-analysis Limit property question $\text{Here is a question that I have been given, }$
$\lim_{x\to a}f(x)=A \iff $for each $c\in\mathbb R,\lim_{x\to a-c}f(x+c)=A$.

For this question the proof follows:

$\text{ Proof: Assume } \lim_{\theta\to a}f(x)=A;
 \
\\\forall\epsilon\gt0,\exists\delta\gt0,\forall{\theta}\in\mathbb R, 0\lt | \theta-a| \lt\delta\implies\ |f(x)-A|\lt\epsilon.\\ \text{Given }\epsilon\gt0, \\\text{If$\;$ }     0\lt\; |\theta-(a-c)| \lt\ \delta, \text{choose$\;$ } \theta=x-c,$
$
 $\begin{equation}
\begin{split}
\text{Then$\;$}     0\lt\; |\theta-(a-c)| \lt\ \delta,  & = |\theta-(a-c)| \lt\ \delta \\
& =|(x-c)-(a-c)| \lt\ \delta \\
& =|x-a| \lt\ \delta\\
& \implies\ |f(x)-A|\lt\epsilon\\&\implies\lim_{\theta\to a-c}f(x+c)=A
\end{split}
\end{equation}

$\text{$\therefore$ }$
$\lim_{x\to a}f(x)=A \iff $for each $c\in\mathbb R,\lim_{x\to a-c}f(x+c)=A$.

Prove or disprove: Now this one I have tried similar method's with no success.I can't figure what $\theta$ should be in this example.

$\lim_{x\to a}f(x)=A \iff $for each $c\neq0,\lim_{x\to a/c}f(cx)=A$.

You can easily solve your problem by applying the theorem stated by Pedro Tamaroff in this thread. Roughly speaking, you change variable: $cx=u$, and remark that $u \to a$ whenever $x \to a/c$. All this seems rather obvious, but it isn't.


The Diophantine equation $ax+by = b+c$ is solvable in integer $x , y$ iff $ax+by =c$ is solvable. Let $a,b,c \in \Bbb Z$. Show that the Diophantine equation $ax+by = b+c$ is solvable in integer $x , y$ iff $ax+by =c$ is solvable.

We know that a Diophantine equation $ax+by =c$ is solvable iff $d|c$ where $d=gcd(a,b)$.
From here how can we do the above proof??

$(x,y)=(m,n)$ satisfies $ax+by=c$ iff $(x,y)=(m,n+1)$ satisfies $ax+by=b+c$.
Thus $\exists (x,y)(ax+by=c)\iff\exists (x,y)(ax+by=b+c)$.

As you have said we use the following theorem :

A Diophantine equation $ax+by =c$ is solvable iff $d|c$ where $d=gcd(a,b)$.

$\Rightarrow$ Let us assume that $ax+by =b+c$ is solvable in integer $x,y$. Thus $d|b+c$ where $d=gcd(a,b)$ and since $d|b$ so $d|c$ thus $ax+by =c$ is solvable in integer $x,y$ using the above theorem.

$\Leftarrow$ can be done similarly!!


Find number of real roots of the polynomial $x^3+7x^2+6x+5$. I want to find the number of real roots of the polynomial $x^3+7x^2+6x+5$.
Using Descartes rule, this polynomial has either only one real root or 3 real roots (all are negetive). How will we conclude one answer without doing some long process?

I note that this is "close" to $$(x+5)(x+1)(x+1)=x^3+7x^2+11x+5$$ which has a repeated root at $-1$, and another root at $-5$. The repeated root at $-1$ is a local minimum, considering the general shape of a cubic with positive leading coefficient.

So you have $$(x+5)(x+1)(x+1)-5x$$

Adding that $-5x$ is going to push the local minimum upward, since $-5x$ is positive near $-1$. The doubled root will be perturbed into two non-real complex conjugate roots. And only the perturbed root near $-5$ will still be real.

If you actually want the roots of a cubic, you can use the rational root test to find one of the roots. After you have one of the roots you can easily do a long division to find the resulting second degree polynomial from there you just apply the quadratic formula. This will yield the exact roots, but if you want to just know if you have three real or one real and two complex that is a different scenario.


Why does factoring eliminate a hole in the limit? $$\lim _{x\rightarrow 5}\frac{x^2-25}{x-5}  = \lim_{x\rightarrow 5} (x+5)$$

I understand that to evaluate a limit that has a zero ("hole") in the denominator we have to factor and cancel terms, and that the original limit is equal to the new and simplified limit. I understand how to do this procedurally, but I'd like to know why this works. I've only been told the methodology of expanding the $x^2-25$ into $(x-5)(x+5)$, but I don't just want to understand the methodology which my teacher tells me to "just memorize", I really want to know what's going on. I've read about factoring in abstract algebra, and about irreducible polynomials (just an example...), and I'd like to get a bigger picture of the abstract algebra in order to see why we factor the limit and why the simplified is equal to the original if it's missing the $(x-5)$, which has been cancelled. I don't want to just memorize things, I would really like to understand, but I've been told that this is "just how we do it" and that I should "practice to just memorize the procedure."
I really want to understand this in abstract algebra terms, please elaborate. Thank you very much.

First, and by definition, when dealing with

$$\lim_{x\to x_0}f(x)$$

we must assume $\,f\,$ is defined in some neighborhood of $\,x_0\,$ except , perhaps, on $\,x_0\,$ itself, and from here that in the process of taking the limit we have the right and the duty to assume $\,x\,$ approaches $\,x_0\,$ in any possible way but it is never equal to it.

Thus, and since in our case we always have $\,x\ne x_0=5\,$ during the limit process , we can algebraically cancel for the whole process.

$$\frac{x^2-25}{x-5}=\frac{(x+5)\color{red}{(x-5)}}{\color{red}{x-5}}=x+5\xrightarrow[x\to 5]{}10$$

The above process shows that the original function behaves exactly as the straight line $\,y=x+5\,$ except at the point $\,x=5\,$ , where there exists "a hole", as you mention.

In limits the value of x is always tending and not exact

x→a means the value of x tends to a. Similarly  x→5 means tending to 5. So basically the value of x is not 5, but a little more or less than 5.

Hence you cannot take the denominator 0 in any case as the denominator will be, again, tending to zero. That is why we try to factorize the numerator first in order to check if there are any terms common both on the numerator and the denominator.


Show that $\lim_{p→∞} ||x||_p = ||x||_∞$ For any $x ∈ \mathbb{R}^n$ and $p ≥ 1$, define $$||x||_p = \left(\sum_{i=1}^n|x_i|^p\right)^{\frac1p},||x||_∞ = \underset{1≤i≤n}{\max}|x_i| $$
  Show that $$ \underset{p→∞}{\lim} ||x||_p = ||x||_∞$$

I understand this in theory, that if $x_i < x_j$ then as they are both raised to the power of ∞ then $x_i$ will become arbitrarily small relative to $x_j$, and so their sum to the power of $\frac1∞$ is $\approx x_j$. But I don't know how to go about proving this formally.

HINT: Note that
$$\max_i \left\vert x_i \right\vert^p \leq \sum_{i=1}^n \left\vert x_i \right\vert^p \leq n \max_i \left\vert x_i \right\vert^p$$

Case 1: Assume without loss of generality, that $x_1$ is your maximum and assume that $|x_1| > |x_i|\ \forall i=2,..,n$. Then we get $$\left(\sum_{i=1}^n|x_i|^p\right)^{\frac1p} = \sqrt[p]{|x_1|^p\cdot(1+ \frac{|x_2|^p}{|x_1|^p}+...+\frac{|x_n|^p}{|x_1|^p}) } \to \sqrt[p]{|x_1|^p} =||x||_∞ \ \mathrm{for }\ p\to\infty , $$ because $$\frac{|x_i|^p}{|x_1|^p} \to 0 \ \mathrm{for }\ p\to\infty\forall i=2,..,n,$$this follows from the convergence of the geometric series.

Case 2: Assume w.o. loss of generality that $x_1$ is your maximum and that $|x_1| \geq |x_i|\ \forall i=2,..,n$. Equality be the case $q$ times. Then:
$$\left(\sum_{i=1}^n|x_i|^p\right)^{\frac1p} = \sqrt[p]{|x_1|^p\cdot q\cdot(1+ \frac{|x_2|^p}{|x_1|^p}+...+\frac{|x_{n-q}|^p}{|x_1|^p}) } = q^{\frac{1}{p}}\sqrt[p]{|x_1|^p\cdot(1+ \frac{|x_2|^p}{|x_1|^p}+...+\frac{|x_{n-q}|^p}{|x_1|^p}) } \to \sqrt[p]{|x_1|^p} =||x||_∞ \ \mathrm{for }\ p\to\infty, $$because $q^{\frac{1}{p}}\to1 \mathrm{for }\ p\to\infty.$

This completes the proof.


Question about induction to infinity with regard to Bolzano's philosophy I'm a philosophy and mathematics student, and I'm writing a paper on a proof put forward by Bolzano that if we can know one thing to be true, then we can know infinite truths. Put simply, he states that if we know A to be true then the statement "A is true" is true, and the statement "'A is true', is true" is true, and so on. (There are a lot of problems with this but for now lets just focus on the maths of it)
Obviously this is a proof by induction. Russell writes about the inductive numbers as a subset of the natural numbers, and how we can't use induction to infinity, because each number the statement is true for must have a predecessor and successor, which are indefinable for ∞. Can someone explain this further, and possibly more correctly for me? Thanks!

Mathematical induction per se cannot prove the existence of an infinite set, like the set $\mathbb N$ of natural numbers.

Usually, induction is assumed as an axiom "defining the essence" of our "intuitive" undertsanding of the succession of natural numbers.

The principle of mathematical induction is stated, in formal way, as :

  if $P(0)$ holds and, for any $n$, if $P(n)$ holds, then $P(n+1)$ holds, then $P(n)$ holds for any $n$.

The key-point is the "operation" $+1$ which produce, for any number $n$, its successor $n+1$.

The weak-point of a proof like Bolzano's one (at least in his simplest form) is that induction is not enough to ensure infinity.

For clerness, call $s(x)$ the successor function and consider a simple "universe" $U$ with only two objects : $U = \{ 0,1 \}$ and define the successor as follows :

  $s(0)=1$ and $s(1)=0$.

Assume now a property $P$ such that : $P(0)$ holds and that : for any $n$, if $P(n)$ holds, then also $P(s(n))$ holds.

Mathematical induction licenses us to conclude that $P(n)$ holds for any $n$, i.e. that both $P(0)$ and $P(1)$ hold.

But our "universe" $U$ is not infinite.

There are many ways of saying the same thing. Saying it in many ways does not express a new truth each time. So, yes, the number of ways of saying something is without theoretical bound; but, no, this "count" does not represent more than one truth.

Saying "$A$ is true", rather than saying what $A$ represents---a possibly long and complex statement---is a linguistic convention: a short sentence replaces a long one. In ordinary speech, rather than saying "$A$ is true", we would more likely say "it is true", "that is true", "what XY says is true", and so on, as long as the reference is clear from the context; we do not say just "it", "that", etc., because these are not sentences.


Why is intersection of two independent set probability a multiplication process? Why is the probability of intersection of two independent sets $A$ and $B$, a multiplication of their respective probabilities i.e. Why is
$$\mathbb{P}(A \cap B) = \mathbb{P}(A) \cdot \mathbb{P}(B)?$$

this question is about the intuition behind the definition of independence of sets in a probability space

The fact that the probability of the intersection of independent events $A$ and $B$ is the product of their probabilities is actually the definition of independent events.

before I give the proof---lemme me mention about conditional probability---.(..if you already know skip this para) its represented by
P(E/F) it means-->Probability of event E given Event F has already occurred. Now if F has occurred the sample space of even E will 'reduce' or get limited to outcomes of F.
 eg: While tossing a die... 1,2,3,4,5,6 is the sample space. now if event of getting an even number has already occurred then find the probability of getting 2---- since the number on the die is surely an even no: the sample space reduces to 2,4,6 ...so
P(E/F)=p(E[intersection)F)/P(F) ;
{we take intersection....cuz we need only what is common in both the events ..because event F surely has occurred)

Now here is the proof.... for independent events E and F ....whether event E has occurred or not --it doesn't depend on event F So p(E/F)=p(e[intersn]F)/p(F)=p(E)
===>>
 p(E[intersc]F)=p(E)*p(F) for independant events

it doesn't simply come from the definition ...see there's the proof!!


What is the result of $\lim\limits_{x \to 0}(1/x - 1/\sin x)$? Find the limit:

$$\lim_{x \rightarrow 0}\left(\frac1x - \frac1{\sin x}\right)$$

I am not able to find it because I don't know how to prove or disprove $0$ is the answer.

For fun, and because of the pre-calculus tag, we give a proof without calculus. It turns out that there is a geometric argument that $|x-\sin x|$ is less than a constant times $|x^3|$ for $x$ near $0$.

I will need some help from you, to draw the missing picture.  We have
$$\frac{1}{x}-\frac{1}{\sin x}=\frac{\sin x-x}{x\sin x}.$$
Let
$$f(x)=\frac{x-\sin x}{x\sin x}$$
(the change of sign is for convenience). We will show that $\lim\limits_{x\to 0}\,f(x)=0.$

We are interested in the behaviour of $f(x)$ when $x$ is close (but not equal) to $0$.  Note that $f(-x)=-f(x)$. So we will be finished if we can show that $f(x)$ approaches $0$ as $x$ approaches $0$ through positive values.

Let $x$ be small positive. Draw $\triangle OPQ$ as follows.  The base of the triangle is $OP$, and has length $1$.  The triangle is right-angled at $P$.  Finally, $Q$ is such that $\angle QOP =x$.

Draw the circular sector with centre $O$, radius $1$, and going from $P$ to a point on $OQ$.  So the sector has angle $x$.

Note that the circular sector is contained in $\triangle OPQ$.  The circular sector has area $(1/2)x$, and $\triangle OPQ$ has area $(1/2)\tan x$.  Thus the geometry gives us the inequality
$$\frac{x}{2}<\frac{\tan x}{2}.$$
Since $x>\sin x$, we get the estimates
$$0<x-\sin x< \tan x-\sin x.$$
The right-hand side only involves trigonometric functions, so is easier to deal with than $x-\sin x$:
$$\tan x-\sin x=\sin x\left(\frac{1-\cos x}{\cos x}\right)=\sin x\left(\frac{1-\cos^2 x}{\cos x(1+\cos x)}\right)=\frac{\sin^3 x}{\cos x(1+\cos x)}.$$
We conclude that
$$0 <\frac{x-\sin x}{x\sin x}<\frac{\sin^2 x}{x\cos x(1+\cos x)}.$$
Since $\sin x<x$, we find that
$$0 <\frac{x-\sin x}{x\sin x}<\frac{\sin x}{\cos x(1+\cos x)},$$
and it is clear that $\dfrac{\sin x}{\cos x(1+\cos x)}$ approaches $0$ as $x$ approaches $0$ through positive values.

Comment: In this problem, there is no virtue in avoiding the calculus. The Taylor expansion is the natural  approach.

I did it that way:$$\lim_{x\to0} \left(\frac{1}{x} - \frac{1}{\sin x}\right) =\lim_{x\to0} \left(\frac{1}{x} - \frac{1}{\frac{\sin x}{x}*{x}}\right) $$
because $\lim_{x\to0} \frac{\sin x}{x} = 1$ then
$$ \lim_{x\to0} \left(\frac{1}{x} - \frac{1}{\frac{\sin x}{x}*{x}}\right) \\= \lim_{x\to0} \left(\frac{\frac{\sin x}{x}* x - x}{\frac{\sin x}{x}*x}\right)\\= \lim_{x\to0} \left(\frac{x(\frac{\sin x}{x}* 1 - 1)}{\frac{\sin x}{x}*x}\right) \\=\lim_{x\to0} \left(\frac{\frac{\sin x}{x} - 1}{\frac{\sin x}{x}}\right) =[\frac{0}{1}] = 0 $$


If $xa=xb$ then $a=b$ We just defined the axioms of a group in our lecture notes on algebra, but silently assumed that the properties of the $=$ relation are known. A few lemmas after the definition of a group we prove that (if $G$ is a group):

$\forall x \in G, \forall a,b \in G: \ xa=xb \implies a=b$

by multiplying both sides by $x^{-1}$.
Now I wonder why we are allowed to do so. Is it a general property of equivalence relationships that if $a=b,c=d \implies ac=bd$ , or is it a group property?

Now I wonder why we are allowed to do so. Is it a general property of equivalence relationships that if $a=b,c=d \Rightarrow ac=bd$ , or is it a group property?

It is property of binary operations. Binary operation on a set $G$ is defined as a function $G\times G\to G$.

This means that the result $ac$ is uniquely determined by $a$ and $c$. Which is just rephrasing of the implication you wrote: $a=b \land c=d \Rightarrow ac=bd$.

(For functions we have $x=y$ $\Rightarrow$ $f(x)=f(y)$.)

Be: $ g * h = g * k$ ; then there is a $s$ in $ G $, so that: $g*s = e $;

$s * (g * h) = (s*g) * h = e * h = h;  $

$s * (g * k) = (s*g) * k = e * k = k;$

And this gets you:  $ h = k; $

You can show in very similar manner that $ h = k $, if $ h * g = k * g; $


Find $\alpha(t,a)$, integrable w.r.t. to t, that bounds $\vert e^{-t}t^{a} \frac{t^h - 1}{h} \vert$ ; $a,t > 0$, for all $\vert h \vert < h_0$. I think I have found an answer to this which I have given at the end but is there a better answer?

I am trying to prove differentiability of gamma function $\Gamma(z)$ for $Re(z) > 1$ using dominated convergence theorem.
So for differentiability, $$ \lim_{h\to 0} \frac{\Gamma(z+h) - \Gamma(z)}{h} $$ must converge. To prove this, I want to be able to take the limit inside the integral, i.e. $$ \lim_{h\to 0} \int^{\infty}_0 e^{-t}t^{z-1} \frac{t^h - 1}{h} dt =  \int^{\infty}_0 \lim_{h\to 0}  e^{-t}t^{z-1} \frac{t^h - 1}{h} dt $$
which is possible if the integrand is bounded by an $\mathbf{integrable}$ function $\alpha(t,z)$, i.e.
$$ \Bigg | e^{-t}t^{z-1} \frac{t^h - 1}{h}  \Bigg | < \alpha(t,z)$$ for all $\vert h \vert < h_0$ (for some real positive $h_0$) and $$ \int^{\infty}_0 \alpha dt < \infty $$
This is equivalent to $$ \Bigg | e^{-t}t^{a} \frac{t^h - 1}{h}  \Bigg | < \alpha(t,a)$$ for real $a>0$.

So finding such a function $\alpha$ is the goal.

Edit: The answer that I had given before was wrong. I was treating h as a real no. when actually it can be complex.

Here is another solution that I found:

$$ \Bigg | \frac{t^h - 1}{h} \Bigg | = \Bigg |\frac{e^{h\log t} - 1}{h} \Bigg | $$
$$ = \Bigg | \frac{h\log t + \frac{(h\log t)^2}{2!} +  \frac{(h\log t)^3}{3!}+ \cdots}{h} \Bigg | $$
$$ \leq \vert \log t (1 + \frac{(h\log t)^1}{2!} +\frac{(h\log t)^2}{3!} + \cdots)\vert $$
 $$= \vert \log t \vert (1+  \frac{(\vert h\log t \vert)^1}{2!} + \frac{(\vert h\log t \vert)^2}{3!} + \cdots) $$
$$ \leq \vert \log t \vert (1+ \vert h \log t \vert + \frac{(\vert h\log t \vert)^2}{2!} + \cdots) $$
$$= \vert \log t \vert e^{\vert h\log t \vert } $$

Thus required function is
$$ \alpha(t) = e^{-t}t^{a} \vert \log t \vert e^{\vert h\log t \vert }$$
And it can be shown that it's integral exists and is finite.

There cannot be an intergable function $g$ on $(0,\infty)$ such that $|\frac {t^{h}-1} h | \leq g(t)$ for all $t>0$ and all $h$ with $|h|<h_0$. For, it you take limit as $h \to 0$ you will get $|\log (t)| \leq g(t)$ for all $t$ and $|\log (t)|$ is not integrable.

Hint : Usually to show the derivability of the gamma function, you use theorems of dominated convergence. You first derive the function under the integral and show that it can be dominated.


Variant of Picard-Lindelöf theorem Question

Let $I=[0,a]$ and define the norm $||f||_{\lambda}=\sup_I |e^{-\lambda x}f(x)|$ for $f\in C(I)$. Let $\phi:\;\mathbb{R}^2\to\mathbb{R}$ satify $|\phi(x,u)-\phi(y,v)|\leq\rho |u-v|$ for all $x,y,u,v\in\mathbb{R}$ and some $\rho >0$. Define $\tau:\;f\mapsto \int_0^x \phi(t,f(t))\;dt$

I need to find a $\lambda$ such that $\tau$ is a contraction under the norm $||\cdot||_{\lambda}$

Thoughts

I am not too sure how to do this; my first line of thought was:

$$\begin{aligned}||\tau (f)-\tau (g)||_{\lambda} &=\sup_I\Big| e^{-\lambda x} \int_0^x \phi(t,f(t))-\phi(t,g(t))\;dt\Big| \\ &\leq \sup_I e^{-\lambda x} \int_0^x |\phi(t,f(t))-\phi(t,g(t))|\;dt\\ &\leq\sup_I \rho e^{-\lambda x} \int_0^x |f(t)-g(t)|\;dt \end{aligned}$$

But I can't see how to get $\cdots \leq \alpha\sup_I |e^{-\lambda x}(f(x)-g(x))|$ for some $\alpha<1 $ and some $\lambda$ from this. Any help would be appreciated.

In general you can prove that if $w : I := [a,b] \longmapsto \mathbb{R}$ is strictly positive and continuos, it defines a norm equivalent to $\|f\|_{\infty}$ on $C^{0}(I)$, defined as $\|f\|_{w} := \sup\limits_{t \in I} \left\lbrace w(t) |f(t)|\right\rbrace$

Defining $w(t) := e^{-2L |t-t_{0}|}$ we define the norm $\|f\|_{w} := \sup\limits_{t \in I} \left\lbrace  e^{-2L |t-t_{0}|} |f(t)|\right\rbrace$

Let's prove that with this choice, $T$ is a contraction on $I = [t_{0}-\delta,t_{0}+\delta]$ :

$$e^{-2L|t-t_{0}|} |\int_{t_{0}}^{t}[f(s,v(s))-f(s,u(s))]ds| \leq e^{-2L|t-t_{0}|} |\int_{t_{0}}^{t}|f(s,v(s))-f(s,u(s))|ds| \leq$$

$$e^{-2L|t-t_{0}|} |\int_{t_{0}}^{t} L |v(s)-u(s)|ds| \leq Le^{-2L|t-t_{0}|} |\int_{t_{0}}^{t} \|v-u\|_{w}e^{2L|s-t_{0}|} ds|$$

$$ = \frac{\|v-u\|_{w}}{2}(1-e^{-2L|t-t_{0}|}) \leq \frac{1}{2}\|v-u\|_{w}$$

So, taking the supremum for $t \in I$ we get exactly

$$\|T(v)(t)-T(u)(t)\|_{w} \leq \frac{1}{2}\|v-u\|_{w}$$

In other words $T$ is a contraction $\hspace{0.2cm} \Box.$

In you case $I$ becomes $[0,a]$, $T$ becomes $\tau$, $L$ becomes $\rho$ and $\lambda$ is the constant chosen in the definition of the weighted norm.

Well the supremum can split under multiplication, therefore if $f$ and $g$ were bounded in $I$, then you could call the supremum for the difference between $f$ and $g$ out of the integral, because the supremum of that difference remains constant in $I$ and the supremum and after done that you could join both and constrain the difference of $I$ on the delta and choose the delta you need.

Maybe not the cleanest but I do not see something wrong, but still, a person who can make mistakes, If I made some, please feel free to do all the corrections needed...


Is there an analytical solution to this differential equation? I am not very knowledgeable in the field so I apologize in advance if this question might look naive.
But is there an analytical solution to a differential equation that looks like this :

$$\frac {\mathrm{d}y}{\mathrm{d}t} = a\cdot v(t) - \mathrm{b}\cdot y $$

My issue here is the function $v(t)$ and how I should deal with it.

Thanks

Yes,there is:

$$y(t)=ce^{-bt}+ae^{-bt}\int_1^t e^{bx} \nu(x)\,dx$$

where $c$ is a constant.

If I understand correctly, your equation is of the from
$$ \frac{\mathrm{d}y}{\mathrm{d}y} = -by(x) + h(x) $$

A solution then is available (http://eqworld.ipmnet.ru/en/solutions/ode/ode0103.pdf, just setting $g(x) = 1, f_1(x) = -b , f_0(x)=h(x) $)

$$ y = Ce^{-bx} + e^{-bx} \int e^{-bx} h(x)\mathrm{d}x $$


How often is $a^b$ rational when $a,b>0$ are irrational. This question was motivated by the proof of the existence of irrational numbers $a,b$ such that $a^b$ is rational. I am wondering how many such numbers there are.

Let $\mathbb I = \mathbb R_+ \setminus \mathbb Q$ ( I'm choosing positive numbers to avoid complex numbers). Now consider the following set $P= \{ (x,y) \in \mathbb I \times \mathbb I \, \, | \, \, x^y \in \mathbb Q \}$. Now is it true that $\overline{P} = (\mathbb R_+ \cup \{0\})\times (\mathbb R_+ \cup \{0\})$ ?

Let $q > 0; q \in \mathbb Q$.  Let $a$ be irrational such that $a^r \ne q$ for any $r \in \mathbb Q$.  (For instance let $a$ be transcendental; or if $q = 2$ and $a = \sqrt 3$)  Then $b = \log_a q$ is irrational.

Then $a^b = a^{\log_a q} = q$.

There are a huge number of these and easy to find.

Second part.  Let $S = \{(x,\log_x q)| x \text{ transcendental}; x > 0; q \in \mathbb Q^+\} \subset P$.

Technically $S = \cup_{x\in \mathbb T}\{x\} \times \{\log_x q| q \in Q\}$

It's obvious $\{\text {positive transcendental}\}$ is dense in $\mathbb R^{\ge 0}$.  (Their complement, the algebraics, are countable for one thing).  Likewise for any $x > 0; x\ne 1$ and any positive $w < z$ we can find rational $q$ so that $q$ is between $x^w$ and $x^w$ which are not equal.  So $w < \log_x q < z$. And so for any positive real $x; x \ne 1$, in particular for any positive transcendental $x$, $\{\log_x q| q \in \mathbb Q+\}$ is dense in $\mathbb R^{\ge 0}$.

So $\overline S = \overline{\cup_{x\in \mathbb T}\{x\} \times \overline{\{\log_x q| q \in Q^+\}}}=\overline{\cup_{x \in \mathbb T}\{x\} \times \mathbb R^{\ge 0}} = \mathbb R^{\ge 0} \times \mathbb R^{\ge 0}$

So $(\mathbb R^{\ge 0})^2\ = \overline{S} \subset \overline{P} \subset (\mathbb R^{\ge 0})^2$

Take any open interval, map it forward by raising to a power $b\in\mathbb I$, then the image open set must contain some rational number $q/p$ with $p$ prime.


$f (f(x))=-x$ - there's no such function Prove that there's no continuous function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $f(f(x))=-x $.

Any hints?

Hint: $f \circ f \circ f=f^{-1}$ so the function is bijective, since the inverse exists. Every continuous bijection $\mathbb R \to \mathbb R$ is monotone.

How about $f(x)=ix$. Then $f(ix)=i(ix)=i^2x=-x$.


Application of Bayes Theorem I am reading Nate Silver's book "The Signal and the Noise" and have a question about Bayes Theorem. I've created my own example and am trying to wrap my mind around the conclusion.

Let's say, before any information, I think there is a 5% chance humans have caused global warming.

Then, I hear information that scientists think there is a 99% chance that humans have caused global warming.

I also know that the probability that the 99% claim is wrong is 10%.

Using the Bayes Theorem calculation, the result is a 34% chance that humans cause global warming.

Here is the calculation:

X = initial probability of humans causing global warming = 5%

Y = probability of humans causing global warming, given scientist evidence = 99%

Z = probability of humans not causing global warming, given scientist evidence = 10%

The formula presented in the book (page 247) is:

Revised probability (given the new information) = XY / (XY + Z(1-X))

Revised probability (given the new information) = 34%

My intuition says that, after this new knowledge, the chances that humans have caused global warming is instead (10% * 1%) + (90% * 99%) or 90%.

This would be based on the fact that theres a 10% they're wrong and 90% chance they're right.

What is wrong about my application of the theorem or understanding of the theorem that causes this mental roadblock?

Thanks.

I think the problem is that $Z$ and $Y$ are likely not correct. I may be getting parts of your example wrong, so let me explicate all my steps. There are two possible interpretations I came up with:

Define the following Bernoulli RVs:

$A$: Humans cause global warming
$B$: Scientists are correct

First interpretation: We have the following data:

$P(A) = 5\%$, the initial probability that humans cause GW
$P(B) = 90\%$, the probability that scientists are correct
$P(A\mid B) = 99\%$, the probability that humans cause GW, given that scientists are correct

We can now calculate $P(B \mid A)$ with Bayes' Theorem; the probability that scientists are correct given that humans cause GW. This isn't what we wanted.

Alternatively: We have the following data:

$P(A) = 5\%$, the initial probability that humans cause GW
$P(B\mid A) = 90 \%$, the probability that scientists are correct given that humans cause GW
$P(A \mid B) = 99 \%$, the probability that humans cause GW given that scientists are correct.

Bayes' Theorem gives us a method to calculate $P(B)$, the probability that scientists are correct.

In neither of these interpretations, we have obtained the desired $P(A \mid B)$. This is due to the fact that what we are given (some scientists playing oracle) does not relate in any way to humans causing GW or not. This is because they only say "There is a chance of $99\%$ that humans cause GW" which doesn't relate to the RVs $A$ and $B$, but rather to:

$C$: The RV $A$ is Bernoulli distributed with $p = 0.99$

As an example of information that would relate $A$ to $B$ (yielding $P(B \mid A)$ and making Bayes' applicable), consider:

"The scientists' measurements occur with a probability of $99\%$ if humans cause GW."

However, this is rather strange (since it would mean that the measurements almost cannot vary if humans cause GW - this would likely mean that the probability of obtaining those measurements would be small, which is contradicted by the probability of $90\%$ that the scientists are correct).

In conclusion, your example doesn't lend itself for Bayes' Theorem unless more information is given or some information is altered.

I think that the problem is that you are adding all the information about GW as a single factor, while each individual journal entry added to the scientific literature advanced the case a bit further.  Let's imagine that there have been 3000 published journal entries advancing the idea of anthropogenic global warming.  We know from the Ioannidis article cited in the same that 2/3rds of them will be wrong, and lets imagine that each of them has at least a 75% probability of humans causing global warming given the evidence presented in the article.  I am choosing 75% because by definition an article advancing a theory must exhibit a probability greater than 50% lest it would not be counted among those advancing the theory.  Since 51% is little better than flipping a coin, then I can assume that the journals would be unlikely to unmistakenly publish an article with such a low probability as less than 75%. Of course they could do so mistakenly, but those mistakes are among those accounted in Ioannidis's 67%.  Now your values are as such:

X = initial probability of humans causing global warming = 5%

Y = probability of humans causing global warming, given scientist evidence = 75%

Z = probability of humans not causing global warming, given scientist evidence = 67%

The formula presented in the book (page 247) is:

Revised probability (given the new information) = XY / (XY + Z(1-X))

Revised probability (given the new information) = 5.5%

And that's after the first journal entry.  After the second it is up to 6.1% by the 100th positive article you're up to 99.98% likelihood. By the 3000th it may as well be 100% because Excel does not even handle that many 9’s.

This model, however, does not account for the journal articles that deny the cause of AGW in those cases we would see different math:  in this case I assigned the Y value to 25% for the reverse logic of articles that support AGW... by definition they have to be <50%, and in order to not be counted in Ioannidis's 67% it must be lower still.
This raises a new problem: when do the AGW refuting articles get published.  For simplicity sake I spaced the negative articles every 10 and re-ran the model. It takes longer to get to 99.999%, but we still get there.  We reach five nines by the next iteration, the 1710th.   It drops  back below again several times, but the last time is on the 1820th.

Last issue is my 75% assumption… let’s say it is 70% instead of 75%?  This changes the math significantly as Bayes now takes us down to near zero instead of up to near certainty.  But then 75% was my floor, not the full range of how much each journal article may advance or refute AGW.  I think we can safely assume a mean advance of >75%


Is there a faster way to add/subtract fractions then having to draw a factor tree each time? Do you really have to draw a factor tree and work with primes every time you encounter adding or subtracting fractions?

Not this way - LCM(8,15)...

15: 15, 30, 45, 50, 65, *80* --
8: 8, 16, 24, 32, 40, 48, 56, 64, 72, *80* --

This makes adding and subtracting fractions quite a lot of work.

What is the most efficient and effective practice in regards to dealing with adding or subtracting fractions? Is there a faster way to add or subtract fractions? I heard of the "Butterfly Method" but it involves a lot of rules. The factor tree seemed easier. I came here to see if determining the least common denominator of two fractions can be done even more efficiently.

We don't need the least common multiple to add fractions.

But if you want the least common multiple (lcm) of $x$ and $y$, where $x$ and $y$ are BIG, first use the Euclidean Algorithm to find the greatest common divisor $\gcd(x,y)$ efficiently. Then use the fact that $\operatorname{lcm}(x,y)=\frac{xy}{\gcd(x,y)}$.

For very large numbers, this is far more efficient than factoring using the best currently known algorithms. But for smallish familiar numbers, factoring works well.

Google or Youtube search: The fast way to adding and subtracting fractions.
https://www.youtube.com/watch?v=lKiR4BBt6TE . Number 3 has like factors. Cancel them out yields your result.


Let $f : \Bbb R \rightarrow \Bbb R$ be a func such that $p>0$, that $f(x+p) = f(x)$ for all $x \in \Bbb R$ . Show that $f$ has an absolute max and min Problem: Let $f : \Bbb R \rightarrow \Bbb R$ be a contiunous function such that for some real number $p>0$, $f(x+p) = f(x)$ for all $x \in \Bbb R$. Show that $f$ has an absolute max and min.

Thoughts:
By rolle's theorem, I know that between $f(x+p)$ and $f(x)$ there has to be a local minimum if $f$ is differentiable on this open interval, but I am outright confused by the precise statement of the question, and in fact I have included an image which may be a counter-example if the question has not been stated properly(sorry for the crudeness of the image) assuming the function continues in this manner infinitely

Edit: As per a comment, since $f$ is not shown to be differentiable on this interval, then Rolle's theorem does not apply.

Also see my answer for a response to my initial confusion.

Recall that if a real-valued function $g(x)$ is continuous on a closed and bounded interval $[a,b]$, then it has a global maximum and minimum on $[a,b]$.

And since $f(x)$ is periodic with period $p$, it follows that $f$ is determined by its values on $[0,p]$. Therefore the global maximum for $f$ on $[0,p]$ is a global maximum for $f(x)$ on $\mathbb{R}$.

I feel I should add an interesting fact that was the basis of my confusion. Take any $x \in \Bbb R$. Then for some $p > 0$, $f(x+p) = f(x)$, and $f((x+p) + p) = f(x+p) = f(x)$ and so continuing on in this way $f(x+np) = f(x)$ for all $n \in \Bbb N$.


Solve $\frac{dy}{dx} - \frac{dx}{dy} = \frac{y}{x} - \frac{x}{y}$ Solve:
$$\frac{dy}{dx} - \frac{dx}{dy} = \frac{y}{x} - \frac{x}{y}$$

What I have done till now:
$$\left(\frac{dy}{dx}\right)^2  -1= \frac{dy}{dx}\left(\frac{y}{x} - \frac{x}{y}\right)$$
$$xy\left(\frac{dy}{dx}\right)^2  - xy= \frac{dy}{dx}\left({y^2} - {x^2}\right)$$
Now I am stuck.

EDIT: I have been through a course on ODEs (More like a course which added tools to our toolbox to solve ODEs) but it has been a long time since then. Everytime someone helps me with an answer it is an "Oh Yes! I forgot about that" moment. Is there any resource (Succinct PDF preferably) which can help me revise the tools used for solving ODEs? This looked good but I want more examples and methods.

A hint: If $u$ and $v$ are real numbers $\ne0$ then
$$v-{1\over v}=u-{1\over u}$$
holds iff either $v=u$ or $v=-{\displaystyle{1\over u}}$.

The differential equation reads

$$f\left(\frac{\partial y}{\partial x}\right)=f\left(\frac{y}{x}\right)$$
and
$$\frac{\partial y}{\partial x}=\frac{y}{x}$$
has solution
$$y(x)=c\ x.$$

The kernel of the right hand side of the equation is a factor of $\pm 1$, which gets sucked up by $c$ already.


Checking for convergence/divergence of the nth root. I've been working on this problem for a while, but I can't really get it.
I get it, but I don't actually get it.

The question is to find whether or not this series converges:
$\displaystyle\sum_{n=1}^\infty(2 ^{1/n} - 1) $

I am almost certain that it diverges, but the way I did it proved convergence, and looking into it online didn't give results I could understand, but generally agreed that it was divergent.

Let $f(x) = 2^{1/x} - 1$

The way I did it was using a comparison test of a p series $1/x^{1.1}$. Since I know $1/x^{1.1}$ is greater than $f(x)$ (this is apparently wrong, as a source online claims that $2^{1/x} - 1 \ge 1/n$) and because $p>1, 1/x^{1.1}$ converges, which would therefore, by the limit comparison test, mean that $f(x)$ must also converge since it's less than $1/x^{1.1}$.

I am at a disagreement with $1/x$ being less than or equal to $f(x)$ because when I graphed it, it was greater. $1/x^{1.1}$ was also greater when graphed.

If anyone can shed light on how $1/x$ is less than or equal to $f(x)$, that'd be great, since then by that comparison if $1/x\ge f(x)$, $f(x)$ would also diverge.

Notice that

$$2^{1/n}-1=\exp\left(\frac1n\ln2\right)-1\sim_\infty\frac{\ln2}{n}$$
 so the series is divergent by comparison with the harmonic series $\sum\frac1n$.

Show the terms don't even approach zero (they approach 1) as n goes to infinity.  Then I think it's just usually called the divergence test that is applied.


Big theta notation of harmonic series I want to prove that big theta notation of the harmonic series is $\Theta(\log n)$. I want to work with integral to show that.

I attempted this:

$$\ln(n)=\int^n_1  \frac{dx}x \le \sum _{k=1} ^n  \frac1k \le 1 + \int^n_2 \frac{dx}x = 1 + \ln(n)$$
This approach was not demanded, because I have not proven that $\Theta(\log n)$ is a tight bound for the harmonic series.

How can I show this, and how to overcome this obstacle? Please help.

Thanks.

If we extend the definition of $=$ and $\le$ to $\Theta$s in the obvious way, you have shown that

$$\begin{align}
\Theta \ln (n) &= \Theta \int^n_1  \frac{dx}x \\
&\le \Theta \sum _{k=1} ^n  \frac1k \\
&\le \Theta (1 + \int^n_2 \frac{dx}x) \\
&= \Theta (1 + \ln(n)) \\
\text {which, since } \Theta (1) \lt \Theta (\ln (n))\text {, }\\
&= \Theta \ln (n) \\
\end{align}$$


Proof of $e^x (\ln x+\frac{1}{x})>\ln 8$ Prove that
$$e^x \left(\ln x+\frac{1}{x}\right)>\ln 8$$

I found that the minimum of $e^x \left(\ln x+\frac{1}{x}\right)$ is close to $\ln 8$, then how do we prove that it's greater than $\ln 8$?

$f(x)=e^x \left(\dfrac{1}{x}+\log x\right)$

$f'(x)=\dfrac{e^x}{x^2} \left(x^2 \log x+2 x-1\right)$

$f'(x)=0\to x^2 \log x+2 x-1=0\to x\approx 0.59$

Taylor polynomial at $x=\dfrac{1}{2}$

$f(x)=\sqrt e \left[x^2 \left(5 -\dfrac{1}{8}  \log 16\right)+x \left(-5 -\dfrac{1}{8}  \log 16\right)+\dfrac{13 }{4}-\dfrac{1}{8}  \log 32\right]+O(x^3)$

$f(x)\approx 7.6722 x^2-8.81501 x+4.64409$ in a neighbourhood of $x=\frac12$

To estimate the error we need the third derivative

$f^{(3)}(x)=\dfrac{e^x}{x^4} \left(x^4 \log (x)+4 x^3-6 x^2+8 x-6\right)$

On the interval $[0.4,0.6]$ we have $|f^{(3)}(x)|\le 36.0234$

thus the error is $R_3(x)\le \dfrac{ |f^{(3)}(x)| \cdot \left|\,x-\dfrac{1}{2}\right|}{3!}\approx 0.006$

Now as
$7.6722 x^2-8.81501 x+4.64409>\log 8;\;\forall x\in\mathbb{R}$

we can conclude that $f(x)>\log 8$ for any $x\in\mathbb{R}$

Hope this helps

Edit

A graph can explain better. Remember that the minimum is at $x\approx 0.59$


Probability of attending cinema and theater. Problem:
It is researched that 60% of people in city goes to cinema on daily basis, 40% of people goes to theater on daily basis. It is also known that 20% simultaneously goes to both theater and cinema.

What is

Probability that chosen person does not attend both.
Probability that chosen person attends theater OR cinema.
Probability that person, which attends cinema will
attend theater too.
Probability that person, which attends theater will
attend cinema too.
Probability that person, which doesn't attend theater
will attend cinema.
Probability that person, which doesn't attend cinema
will attend theater.

So far I have got:

A-person attends cinema
B-person attends theater

1)$P(\overline{A}\cap\overline{B})=P(\overline{A})\cdot P(\overline{B})=0.4\cdot 0.6=0.24$

2)$P(A \cup B)=P(A)+P(B)-P(A \cap A)=0.6+0.4-0.24=0.76$

I have no clear understanding of how to solve rest of questions.

For example,  in 3rd questions I guess that I need to find $P(A \cap B)$, which gives exactly the same as 4th questions $P(B \cap A)$, which seems unlikely for me.

And I guess that 5th and 6th would be:

5) $P(\overline{B}\cap A)=P(\overline{B})\cdot P(A)=0.6\cdot
   0.6=0.36$

6) $P(\overline{A}\cap B)=P(\overline{A})\cdot P(B)=0.4\cdot 0.4=0.16$

Is there a something I am not getting here?

First a note to your solution of item (2), it should be $P(A)+P(B)-P(A\cap B)$. Regarding item (3)-(4), you are in fact not looking for the joint probability, but the conditional probability i.e. $P(B|A)$ and $P(A|B)$ respectively, where the conditional probability is defined as usual:
\begin{equation}
P(X|Y) = \frac{P(X\cap Y)}{P(Y)}.
\end{equation}
The solutions to items (5)-(6) are found using the conditional probability as well. That is, by $P(A|\bar{B})$ and $P(B|\bar{A})$ respectively.

Hope that helps.

On your answer (1), you are implying that the two events are independent which does not seem to be the case.
I would have use the De Morgan's Law to solve it:
$$ \Pr \big( ( A \cup B ) ^ { \mathrm c } \big) = 1 - \Pr ( A \cup B ) = 0.2 $$
$$ \Pr \big( ( A \cup B ) ^ { \mathrm c } \big) = \Pr ( A ^ { \mathrm c } \cap B ^ { \mathrm c } ) $$


Lottery with ordered draw - check my proof Let's assume that the order in which the numbers are drawn shall be considered.
6 numbers will be drawn out of 30 numbers and each number can only be drawn once.

Calculate the probability that your lottery ticket has exactly 3 numbers which also appear on the winning ticket.

My approach:

Let be $\Omega:=\{(\omega_1, ..., \omega_6) \vert \omega_i \in \{1, ..., 30\}\}$ and $\vert \Omega \vert = { 30! \over 24!}$.

Let $B:=\{\omega \in \Omega \vert \omega \text{ contains 3 winning numbers}\}$. $\color{red}{\text{How do I rigorously write this down?}}$

Let $B^c$:= complement of $B$

Now I am not sure how to rigorously define $A$ the set of positive outcomes. It should contain three elements of $B$ and three elements of $B^c$ considering all possible permutations ot them.
$\color{red}{A:=(\{\omega_1, ...,\omega_6,)\vert \omega_i \in ~??\} }$.

Then the cardinality of $A$ should be: $\vert A\vert = {6! \over 3!}\cdot { 24! \over 3!}\cdot {6 \choose 3}$. Hence, $P(A)=\frac{{6! \over 3!}\cdot { 24! \over 3!}\cdot {6 \choose 3}}{30! \over 24!}\approx 0.068$

Is this correct? Any comments are appreciated.

What we are interested in is whether exactly three of the six numbers on the lottery ticket match the winning numbers drawn in the lottery.

If there are exactly three matches, then the lottery ticket must contain exactly three of the six winning numbers and three of the other $24$ numbers.  Since six of the thirty numbers are winning numbers, the probability that this occurs is
$$\Pr(\text{exactly three matches}) = \frac{\dbinom{6}{3}\dbinom{24}{3}}{\dbinom{30}{6}}$$
which agrees with the answer you obtained.

A favorable outcome is any set of six numbers containing exactly three winning and three losing numbers.  If you wish to take the order of selection into account, favorable outcomes consist of ordered sextuples with three winning and three losing numbers.


Smallest area that can be formed by 3 non intersecting diagonal of a hexagon We have a regular hexagon: a polygon with six sides, in which all internal angles have 120 degrees and length is the 5 cm of each side. We are going to draw three non-intersecting diagonals in some way. These will divide the hexagon into four triangles. We will then compute their areas. Find the smallest of those areas.

Let $ABCDEF$ be our hexagon and $O$ be a circumcircle of the hexagon.

Thus, the radius of the circle is equal to $5$ and $\Delta AFE\cong\Delta AOE,$ which gives $$S_{\Delta AFE}=\frac{1}{2}S_{\Delta AED}$$ and since $\Delta AFE\cong \Delta ABC$ and $\Delta AED\cong ACD,$ we see that $S_{\Delta AFE}$ is a need area.

Id est, $$S_{\Delta AFE}=\frac{5\cdot5\cdot\sin120^{\circ}}{2}=\frac{25\sqrt3}{4}$$ gives the answer.


Show that $a(XY)=X(aY)$ where $a$ is an element of field and $X$ and $Y$ are $m\times n$ and $n\times p$ matrices Show that $a(XY)=X(aY)$ where $a$ is an element of field and $X$ and $Y$ are $m\times n$ and $n\times p$  matrices.

By using the notation to designate the coefficient of the $i$th row and $j$th column of a matrix $A$: $(A)_{i,j}$, we have

$$(a(XY))_{i,j}=a\cdot (XY)_{i,j}.$$

On the other hand,

\begin{align}
(X(aY))_{i,j} & =\sum_{k=1}^{n} (X)_{i,k} (aY)_{k,j}
\\ & =\sum_{k=1}^{n} (X)_{i,k}\  a\cdot(Y)_{k,j}
\\ & = a\cdot\sum_{k=1}^{n} (X)_{i,k} (Y)_{k,j} = a\cdot(XY)_{i,j}.
\end{align}

Matrix multiplication $\pi:(X,Y)\mapsto XY$ is linear in each slot. By definition $X(aY) = \pi(X,aY)$, which by linearity in the second slot equals $a \pi (X,Y)  = a (X Y)$. Hence $a(X Y) = X(aY)$.


Determining stability of ODE I'm working on a prey-predator model. I'm using the following system of differential equations for it:
\begin{align} x'&=-a_1x+a_2xy+a_3xz\\
y'&=b_1y-b_2xy\\
z'&=c_1z-c_2xz
\end{align}
Where $a_i, b_i, c_i >0$.

One of the stationary points is $P=(\frac{b_1}{b_2},\frac{a_1}{a_2},0)$.

Question:
How can I determine the stability of this point $P$?

Attempt:

First I wrote the equation as:
\begin{align} \frac{\mathrm d \underline{v}}{\mathrm d t}=\underline{F}(\underline{v}), \hspace{10pt} \underline{v}=(x,y,z)\end{align}
I looked at the linearized ODE, and I found:
\begin{align} \frac{\mathrm d \underline{\hat{v}}}{\mathrm t}=
\begin{pmatrix}
0 & \dfrac{a_2b_1}{b_2} & \dfrac{a_3b_1}{b_2}\\
-\dfrac{a_1b_2}{a_2} & 0 & 0 \\
0 & 0 &c_1-\dfrac{c_2b_1}{b_2}\\
\end{pmatrix}
\underline{\hat{v}}
\end{align}
The eigenvalues are $c_1-\dfrac{c_2b_1}{b_2}, \pm i \sqrt[]{a_1b_1}$. My problem is the pair with zero real part. I have learned a theorem that only says something when all eigenvalues have negative real part (then it's stable) or at least one has positive eigenvalue (then it's unstable). My other approach was Lyapunov's theorem. I found (with a little bit puzzling) the following Lyapunov function:
\begin{align}V(x,y,z) = a_2y+ b_2x  -a_1\left(1+\ln (y)-\ln\left(\frac{a_1}{a_2}\right)\right)-b_1\left(1+\ln(x)-\ln\left(\frac{b_1}{b_2}\right)\right)  \end{align}
Differentiating it brought me eventually to this:
\begin{align} V'=a_3b_2xz-a_3b_1z\end{align}
Now the only thing that I can say is that $V'$ is both positive and negative in every neighborhood of $P$, so the theorem doesn't say something about that case.

What I also thought about is to Taylor approximate the function $F$ till the second order term, like it can be done with one-dimensional ODE if the first derivative is equal to zero. But then here I get a matrix in a matrix (a tensor?) which is pretty vague. I can not visualise what is going on there, but if someone can explain how to do it that way, then you are welcome. I have read many articles about this problem on internet. I couldn't find something useful/understandable. Some are talking about manifolds, but I have not learned that yet. What I understand is that it is something like a solution curve.

I really couldn't sleep well because of this problem. Your help is appreciated! Thanks in advance.

Update
I have realised that the function $V$ that I have used missed an imortant requirment, it was $0$ in all points $(\frac{b_1}{b_2},\frac{a_1}{a_2},z)$, so that was not a Lyapunov function. My new approach was to pick:
\begin{align} V(x,y,z)=\left(x-\frac{b_1}{b_2}\right)^2+\left( y-\frac{a_1}{a_2}\right)^2+z^2\end{align}
Okay, this one is Lyapunov for sure. I differentiate w.r.t. $t$ and get:
\begin{align} V'=(-a_1x+a_2xy+a_3xz)\left(x-\frac{b_1}{b_2}\right) + (b_1y-b_2xy)\left( y-\frac{a_1}{a_2}\right) +z(c_1z-c_2xz)
\end{align}
This is zero in $(x,y,z)=P$ but that is also a saddle point, which means that is both negative and positive in every neighborhood of $P$. I also tried some Lyapunov functions of the form:
\begin{align} V(x,y,z)=f(x-\frac{b_1}{b_2})+g(y-\frac{a_1}{a_2})+h(z)\end{align}
where $f,g,z$ are all even functions that has minimum in zero. Examples that I have tried are: $-\cos(x), -e^{-x^2}$ and they also made $V'$ have a saddle point in $P$.

I also transformed everything to cilindar coordinates, but the equation that I have got was not so beautiful. I believe there is a way to prove the stability of this point.

Can you guys help me? I really don't know what to do after this.

Heh, it's actually simpler than expected. The system is stable but
not asymptotically stable. Here is a rough sketch of a proof:

First, for simplicity, set
\begin{align*}
\beta &= \frac{b_1}{b_2}\\
\alpha&= \frac{a_1}{a_2}\\
\end{align*}
and then center your system at the equilibrium point $P =
\left(\frac{b_1}{b_2}, \frac{a_1}{a_2}, 0 \right) = \left(\beta,
\alpha, 0 \right)$ which means shift the variables as follows
\begin{align*}
x &\mapsto x + \frac{b_1}{b_2}\\
y &\mapsto y + \frac{a_1}{a_2}\\
z &\mapsto z
\end{align*}
The system becomes
\begin{align*}
\dot{x} &= \beta a_2  \, y  + \beta a_3 \, z + a_2
\, xy +  a_3 \, xz\\
\dot{y} &=  - \alpha b_2\, x - b_2 \, xy\\
\dot{z} &=  \left(c_1 - \beta c_2\right) z -c_2 \, xz
\end{align*}
which could also be reorganized as
\begin{align*}
\dot{x} &= \beta a_2  \, y   + a_2
\, xy + a_3(\beta + x) z\\
\dot{y} &=  - \alpha b_2\, x - b_2 \, xy\\
\dot{z} &=  \left(c_1 - \beta c_2\right) z -c_2 \, xz
\end{align*}
Now observe that the $x,y$-coordinate plane, also given by the
equation $z=0$, is invariant under the flow of the system, meaning
that the vector field is tangent to $z=0$. You can even check
directly that if you take a solution $\big(x(t), y(t) \big)$ of
the 2D system
\begin{align*}
\dot{x} &= \beta a_2  \, y  + a_2
\, xy \\
\dot{y} &=  - \alpha b_2\, x - b_2 \, xy\\
\end{align*}
then $\big(x(t), \, y(t), \, 0 \big)$ is a solution of the
original system
\begin{align*}
\dot{x} &= \beta a_2  \, y   + a_2
\, xy + a_3(\beta + x) z\\
\dot{y} &=  - \alpha b_2\, x - b_2 \, xy\\
\dot{z} &=  \left(c_1 - \beta c_2\right) z -c_2 \, xz
\end{align*}
All of this yields that the plane $z=0$ is a center manifold of
your system and the restriction of the system to that invariant
manifold is
\begin{align*}
\dot{x} &= \beta a_2  \, y  + a_2
\, xy \\
\dot{y} &=  - \alpha b_2\, x - b_2 \, xy\\
\end{align*} As you already know, the latter is a conservative system
(i.e. integrable, has a first integral) Thus, all trajectories on
$z=0$ are ovals (closed orbits, cycles) and consequently, all
orbits close enough to the invariant plane $z=0$ and in a
neighborhood of the equilibrium point $P$, will tend to approach
it more and more, as long as, the restriction $c_1 <
\frac{b_1}{b_2} c_2$ is satisfied. Thus, they will follow the
cyclic closed orbits on the invariant manifold. Hence, the
equilibrium point is stable, but not asymptotically.

Observe that whenever $c_1 > \frac{b_1}{b_2} c_2$ the equilibrium
$P$ is automatically unstable.

Here is the rigorous proof. The following theorem is a direct corollary from the theory of center manifolds (see for example these slides). One can also prove it, although it requires some techniques from analysis. Notice that in this case, the center manifold is known explicitly and is $z=0$.

Theorem. There exist two small open neighborhood $U, \, V \, \in \, \mathbb{R}^2$ of the point $0 = (0,0,0)$ and a smooth change of variables
$$H \, : \, U \, \to \, V$$
$$H \, : \, (u,v,w) \, \mapsto \, (x,y,z)$$ of the form
\begin{align*}
u &= x + h_1(x,y,z)\\
v &= y + h_2(x,y,z)\\
w &= z
\end{align*}
where the functions $h_1$ and $h_2$ are smooth of order at least two, so that $H$ transforms the system of ODEs
\begin{align}
\dot{u} &= \beta a_2  \, v   + a_2
\, uv\\
\dot{v} &=  - \alpha b_2\, u - b_2 \, uv \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ (1)\\
\dot{w} &=  \left(c_1 - \beta c_2\right) w - g(u,v)w
\end{align}
into the original system of ODEs
\begin{align}
\dot{x} &= \beta a_2  \, y   + a_2
\, xy + a_3(\beta + x) z \\
\dot{y} &=  - \alpha b_2\, x - b_2 \, xy \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\ (2)\\
\dot{z} &=  \left(c_1 - \beta c_2\right) z -c_2 \, xz
\end{align}

Since system (1) is definitely stable, but not asymptotically stable, system (2) is also stable but not asymptotically because (2) is obtained from (1) by the change of variables $H$ and any change of variables preserves stability (which is a topological property).

System (1) is stable because the first two equations are separate from the third. If you pick their first integral, let's call it $J(u,v)$, one can check directly that the function $J(u,v)$ is also a first integral of (1). Then, it is easy to prove the stability of (1). On each level surface $J(u,v) = const$, a general solution $(u(t), v(t), w(t))$ close to $w=0$, which starts from a point on the surface $J(u,v) = const$, stays on the surface $J(u,v) = const$ and because of the coefficient $c_1 - \beta c_2 < 0$ it converges to the solution $(u(t), v(t), 0)$, simply because $\lim_{t \to \infty} \, w(t) = 0$. However,
$(u(t), v(t), 0)$ is actually a closed cycle (a periodic solution). Therefore system (1) is stable, but not asymptotically, because   $(u(t), v(t), 0)$ does not converge to the origin.

I also transformed everything to cilindar coordinates, but the equation that I have got was not so beautiful. I believe there is a way to prove the stability of this point.

Easiest way to go for your system is indeed transform it into the cylindrical coordinates. To see the final picture, you should set $y=\rho \cos \theta, z=\rho\sin \theta$ and keep $x$ as is. If you assume that the parameter values are generic, then it turns out that there is no non-trivial equilibrium in this system. Which immediately means that the omega-limit sets are on the boundary of positive orthant. Moreover, the analysis of the system in cylindrical coordinates should show that one of the preys should always go extinct.


$| \dfrac{-z+\sqrt{z^2-4}}{2}|\le 1$ Suppose $z=u+iv$, with $v>0$.

$$ m(z)=\dfrac{-z+\sqrt{z^2-4}}{2}$$ satisfies the equation $$m(z)+1/m(z)+z=0 $$ from which we found that $$|m(z)|=|m(z)+z|^{-1} $$

Take the branch of the square root so that the imaginary part of $m(z)$ is postive.

($m(z) $ here is the Stieljes Transformation of the Semicircle Law)

How do we show $$|m(z)|\le 1 $$ or equivalently $$|m(z)+z|\ge 1 $$

As pointed out by Conrad in the comment,$$|m(z)(z+m(z))|=1 $$If we can show $m(z)$ is the smaller of the two in norm, then we are done.

Noting that $$v>0,\qquad m(z)=\dfrac{-z+\sqrt{z^2-4}}{2},\qquad m(z)+z= \dfrac{z+\sqrt{z^2-4}}{2}$$

we have $$Im[m(z)+z] >Im[m(z)] $$ but how about the real part?

as shown in the first few lines of the accepted solution, by the constrains we have on the imaginary part of $z$ and $m(z)$. we have that the real and imaginary part of $$z$$ and $$\sqrt{z^2-4}$$ are of the of the sign. With this we can conclude that $$|m(z)+z|\ge |m(z)| $$

Let $z=a+ib,\sqrt{z^2-r^2}=c+id$ where $a,b,c,d$ are real and $r>0$

$$r^2=z^2-(z^2-r^2)=(a+ib)^2-(c+id)^2=a^2+d^2-b^2-c^2+2i(ab-cd)$$

Equating the imaginary parts, $ab-cd=0\implies\dfrac ac=\dfrac db=k$(say)

$\implies a=ck,d=bk$

Equating the real parts,  $$r^2=a^2+d^2-b^2-c^2=(b^2+c^2)(k^2-1)$$

$\implies k^2-1=\dfrac{r^2}{b^2+c^2}>0$

$\implies$ either $k>1$ or $k<-1$

$$|-z+\sqrt{z^2-r^2}|^2=|c+id-(a+ib)|^2=(c-a)^2+(d-b)^2=(b^2+c^2)(1-k)^2$$

$$=\dfrac{r^2(1-k)^2}{(k^2-1)}=\dfrac{r^2(k-1)}{k+1}$$ which will be $\le r^2$

$\iff\dfrac{k-1}{k+1}\le1\iff0\ge\dfrac{k-1}{k+1}-1=-\dfrac2{k+1}$
$\iff 0\le\dfrac2{k+1}\iff k+1>0\iff k>-1$

So, the proposition will be nullified if $k<-1$

Let's write $m_1,m_2$ the 2 roots of the equation $m^2+zm+1=0$ where $\Im m_1 \ge 0$ so $m_1=m(z)$ in the post notation.

We know that $m_1=re^{i\theta}, 0 \le \theta \le \pi$, so $\sin \theta \ge 0$ while then $m_2=\frac{1}{r}e^{-i\theta}$, so $0>-\Im z =\Im (m_1+m_2)=(r-\frac{1}{r})\sin \theta$ hence $r-\frac{1}{r} \le 0$ and we are done!


Shorter method to find the orthocentre of this triangle? The given vertices are (0,0) (5,-1) (-2,3).

My approach::

I assumed the orthocentre of the triangle as (a,b).
Now,
I used the fact that the line through each vertex and orthocentre is perpendicular to the opposite side.

So,

m (AH).m (BC) = -1

m (BH).m (AC) = -1

From this, I got two equations and I solved them to get the orthocentre.

However,
THIS METHOD IS VERY LENGTHY.
Can someone please suggests me easier or shorter method to find it?

Thanks.

I don't think there is a much better way to do that. One particular case, that is very convenient, is when the origin lies on the circumcenter, on that case because of the Euler's Line we get:

$$\text{orthocenter}=A'+B'+C'$$

But, in order to use that result you have to make a translation to bring the origin to the circumcenter and use that translation to find the new vertex $A',B',C'$.


Urn model with replacement where each ball has a different number Suppose we have $N = 10$ balls where each ball has its own number. Thus, $C = \{0,1,2,3,4,5,6,7,8,9\}$. We draw $n = 4$ times and each time we put the ball back.

What is the probability to draw 4 balls where each ball has a different number?

The only solution I can come up with is:

If we draw a ball the first time there are $10$ allowed numbers. The next time we draw there are $9$ allowed numbers and so on. Thus, $10 \cdot 9 \cdot 8 \cdot 7 = 5040$ draws that are allowed. There are $10^4 = 10000$ possible draws.

Now, we get $\frac {5040}{10000} = 0.504$.

Is there a different solution?

Since nobody in the comments could really help I came up with a different solution myself.

We can see this as an urn model with replacements and unordered samples.

Now:

$\Omega = \{ (k_0, k_1,\dots,k_9) \in \mathbb{Z}_+^{10} \colon \sum_{i = 0}^9 k_i = 4 \}$

and

$\mathbb{P} = \operatorname{Multin}( 10, (\frac{1}{10})_{c = 0}^{10})$.

Further, we have the event $A := \{ \omega \in \Omega_1 \colon \omega \text{ has 4 different digits} \}$.

We calculate:

$\mathbb{P}(A)\\  = \mathbb{P}(\{(1,1,1,1,0,0,0,0,0,0)\}) \cdot |A| \\ = \mathbb{P}(\{(1,1,1,1,0,0,0,0,0,0)\}) \cdot {10 \choose 4} \\
= {{4}\choose{(1,1,1,1)}}\cdot \frac{1}{10^4}  \cdot {10 \choose 4} \\ = 0.504$.

We can do this because ${{4}\choose{(1,1,1,1)}} = {{4}\choose{(1,1,1,1,0,0,0,0,0,0)}}$.

Actually, we could also come to the same solution using an urn model with replacement and ordered samples (using $\Omega = \{0,1,\dots,9\}^4$ with $\mathbb{P} = \operatorname{Uniform}(\Omega)$).