What *can* Euclid prove? It is well-known that Euclid's axioms for geometry are not up to modern standards of rigor: in particular, there are a lot of times when he used "obvious" facts about the geometric objects which were neither stated in, nor derivable from, the axioms.

For example, in his very first proposition, Book I Proposition I, he sets out to construct an equilateral triangle. Seems easy enough, right? He gives a construction involving two circles drawn at the ends of a line segment which is to be one side of the triangle and uses this to produce the remaining two sides. Trouble is, he has no axioms which guarantee that the circles will intersect, as they must to produce the third vertex of the triangle! Another, even more subtle, problem is that when you draw the segments to the third vertex, how do you know what you form is in fact a triangle? You need another axiom for that -- specifically, the problem is you don't know the sides don't meet somewhere else than the new point before they get there. This trouble was actually known even back in old Greece: Zeno of Sidon pointed this one out.

So that makes me wonder: just what can you prove from Euclid's axioms, definitions, etc. without resorting to any hidden, "obvious" assumptions? If you can't so much as even get something as basic as whether the points you are attempting to construct by crossing lines and circles really exist, it seems the answer is "not a whole heck of a lot!" Or am I wrong?

You can see :

Ian Mueller, Philosophy of Mathematics and Deductive Structure in Euclid's Elements (1981 - Dover reprint).

It is a detailed analysis of Euclid's Elements logical structure. In particular, Ch.1.2 : Book I of the Elements , is devoted to the reconstruction of Book I, with graphical maps depicting the logical dependencies between Propositions.

You can use them, to check what Prop depends from the "flawed" ones : I mean those Prop of Book I which have some "hidden assumptions" asking for additional axioms not stated by Euclid.

Nothing because  the  geometry is axiomatic
and that
Definition 1 : A point is what the game is drawn for null .

Therefore, according to 1 a point has : neither length nor width.
Definition 2 : A line is a length without a width.

But a null part added to itself will always remain identical to itself, namely null ..
in fact,
there is no nul draw
the length of a line is the set of length parts of that line ...


Pipe-fitting problem 3D I have a 3D pipe-fitting problem for which I was able to write the following equations:

$$
y = \tan (a)\sqrt{x^2 + z^2}\\
z = \tan (b)\sqrt{x^2 + y^2}\\
y = \sin (a)\sqrt{x^2 + y^2 + z^2}\\
z = \sin (b)\sqrt{x^2 + y^2 + z^2}
$$

$x > 0$ and known; $y$ and $z \ge 0$; $0 < a$, $b < 90º$

what choices for $a$ and $b$ will minimize $y + z$?
if $a$ and $b$ are constrained to $45º$, $22.5º$ or $11.25º$, what choices minimize $y + z$?

I tried rearranging and combining these equations, and I got:
$$
y = \frac{x \sin(a)}{\sqrt{cos^2(a) - sin^2(b)}}\\
z = \frac{x \sin(b)}{\sqrt{cos^2(b) - sin^2(a)}}
$$

But these equations behave strangely: as $a$ and $b$ go to $0$, $y$ and $z$ go to $0$; and in my problem, that means the pipes never meet.

Here is the physical problem: I have a vertical pipe, and a pipe aligned on the $z$-axis, separated by a fixed horizontal distance $x$.  The length of the pipes can be varied. I need to connect the ends with one pipe and $2$ fittings from $0º$ to $90º$.

There is an easy solution with one $45º$ fitting, and one $90º$ fitting, but I am trying to avoid it.

So, did I model the problem correctly?  Were my reduced equations correct?  I am also having trouble seeing which variables are independent.  Any help would be appreciated.

Please look at my blog at forexsg.wordpress.com/2014/06/19/solution-for-3d-pipe-fitting-connection-vertical-wye-tee-and-elbow-or-another-wye-tee/

I descirbed my solution there. /_a and /_b cannot go to zero at the same time. When a is increasing b is decreasing. They are inverse of each other. There are more variables than the equations can define. Two equations cannot define 5 variables. Even with two know angles you need to define at least one other distance x,y or z . So you have to fix two angles and one of the distances x y or z .


$a^2+b^2+c^2=(abc)^2-2\leq 6$. Proof or counter-example needed for $a,b,c\gt 0$ I was working on an inequality proof in which I need to use the following inequality to conclude.

  $$\forall~a,b,c\gt 0~,~a^2+b^2+c^2=(abc)^2-2\implies a^2+b^2+c^2\leq 6$$

I can't think of any way to prove this. I only need a hint since I dislike people just giving away the answers. Thanks in advance!

If this claim is false, I'd like a counter-example. Thanks again!

I found the reversed claim is true, which means the claim is false.
Put $s=x+y+z, x = a^2, y = b^2, z = c^2 \Rightarrow x+y+z = xyz-2 \leq \dfrac{(x+y+z)^3}{27} - 2 \Rightarrow s \leq \dfrac{s^3}{27}-2\Rightarrow s^3-27s-54 \geq 0 \Rightarrow (s-6)(s+3)^2 \geq 0 \Rightarrow s \geq 6.$


Let p be a prime and k a positive integer such that $a^k$mod p = a mod p for all integers a. Prove that p - 1 divides k - 1. Let p be a prime and k a positive integer such that $a^k$mod p = a mod
  p for all integers a. Prove that p - 1 divides k - 1.

I think I need to use Fermat's Little Theorem, and I can get  $a^k$mod p =  $a^p$mod p, and I think it follows that  $a^{k-1}$mod p = $a^{p-1}$mod p = 1, but from there I'm stumped.

I think it follows that $a^{k-1}$ = m$a^{p-1}$, but why couldn't m be a fraction, i.e. k - 1 divides p - 1 but not the other way around?

The OP's solution is incorrect.  This deduction is flawed:

  "Multiplying both sides by $a^{k-1}$
  $$a^{k-1}({a^{p-1}})^m=1a^{k-1} \implies a^{(k-1)+m(p-1)}=a^{k-1} \implies (k-1)+m(p-1) = k-1$$
  by the definition $x$ mod $(p-1)$".

Here is a different approach.  Let $k-1=(p-1)q+r$ for some non-negative integers $q$ and $r$ with $r<p-1$.  If $r\neq 0$, then we have by Fermat's little theorem that
$$a^{r+1}\equiv a^{r+(p-1)q+1}= a^{k}\equiv a\pmod{p}.$$
Therefore, the polynomial $x^{r+1}-x$ is of degree $r+1<p$, but it has $p$ roots in the Galois field $\operatorname{GF}(p)$.  This is impossible, so $r=0$ must be the case.

I figured out my own question. Here is is for anyone else that wants to know (maybe my classmates...)

By FML, $a^p=a$ mod $p$.

$a^p$ mod $p=a$ mod $p$

multiplying both sides by $a^{-1}$ we get:

$a^{-1}a^p=a^{-1}a$ $\implies$ $a^{p-1}=1$.

$a^{p-1}=1$ (mod $p$)

Taking each each side to the m$^{th}$ power:

$({a^{p-1}})^m=1^m \implies ({a^{p-1})}^m=1$

Multiplying both sides by $a^{k-1}$

$a^{k-1}({a^{p-1}})^m=1a^{k-1} \implies a^{(k-1)+m(p-1)}=a^{k-1} \implies (k-1)+m(p-1) = k-1$

by the definition $x$ mod $(p-1)$ iff $x=q(p-1)+r$, so

$(k-1)+m(p-1) = k-1 \implies (k-1)$ mod $(p-1)=k-1$

By hypothesis we have $a^k$=a mod p. Like in the beginning,

$a^k$=a mod p $\implies$ $a^{-1}a^k$= $a^{-1}$a mod p $\implies$ $a^{k-1}$ = 1 mod p

Substituting $(k-1)$ mod $(p-1)$ for $k-1$ we get:

$a^{(k-1) \textrm{ mod } (p-1)} = 1$ mod $p = a^0$ mod $p= a^{0\textrm{ mod }(p-1)}$mod $p$

so, $(k-1)$ mod $p = 0 $mod $p$

By definition, this means that

$(k-1)=m(p-1)$, i.e. that $p-1$ divides $k-1$. $\square$


Formal proof for $(-1) \times (-1) = 1$ Is there a formal proof for $(-1) \times (-1) = 1$? It's a fundamental formula not only in arithmetic but also in the whole of math. Is there a proof for it or is it just assumed?

We use only the usual field axioms for the real numbers. First we prove an intermediate result.

$0\times 0$

$=0\times(0+0)$

$=0\times 0+0\times 0$

Subtract $0\times 0$ from each side to get $0=0\times 0$. Now we are ready for the final kill.

$0$

$=0\times 0$

$=(1-1)\times(1-1)$

$=1\times 1+1\times (-1)+(-1)\times 1+(-1)\times (-1)$

$=1+(-1)+(-1)+(-1)\times (-1)$

$=(-1)+(-1)\times (-1)$

Add $1$ to each side to get $1=(-1)\times (-1)$.

Another possibility using Euler's identity:

$$\begin{align}
(-1)\times (-1) & = e^{\imath \pi} \times e^{\imath\pi}\\
&=e^{\imath 2\pi} = 1
\end{align}
$$


Derivative of $x\cdot|x|$ on $x=0$? $$f(x) = x |x|$$

Wolfram Alpha says is:

$$f'(x) = \frac{2x^2}{|x|}$$

and thus $f'(0)$ is indeterminate, while an HP48 says that:

$$f'(x) = |x| + x \operatorname{sgn} x,$$

which would yield $f'(0) = 0$. If I say that:

$$f(x) = -x^2$$

for $x<0$, 0 for $x=0$ and:

$$f(x) = x^2$$

for $x>0$, I kinda think that $f'(0)$ is 0; all three parts' derivatives converge to $0$ on $x=0$, and thus I think that Wolfram is wrong, but I don't really dare say that! (but I don't see any spiky bits on $f(x)$, although it seems clear to me that $f''(0)$ is indeterminate).

(excuse my poor TeX)

edit: according to the answer below, it seems that Wolfram Alpha is wrong- I've already sent them some feedback, but can anyone elaborate on that?

For $f'(0)$, let's not worry about values of $f'(x)$ near $0$. Let us instead compute directly from the definition of derivative. By definition,
$$f'(0)=\lim_{h\to 0}\frac{h|h|-0}{h}=\lim_{h\to 0}|h|=0.$$

Note that,
  $|x| =  \sqrt{x^2}$.

Now,  use the derivation rules (chain and product) and you will see that $ f'(x) = |x| +  \dfrac{x^2}{|x|} =  \dfrac{2|x|^2}{|x|} = 2|x|$.

Thus, the wolfram is not wrong. We realize that we suppose x nonzero.


Find slope of the tangent of a curve passing through fixed point I need to find the slope of the tangent of $f(x)=\sqrt{2x-10}$ that pass through point $(0,0)$.

The derivative of $f(x)$ is $1/(\sqrt{2x-10})$.

Observe that the given curve is a branch of a parabola. Since the tangent line passes through $(0,0)$. Therefore it is of the form $y=mx$. Now find the intersection of this with the given curve by solving
\begin{align*}
mx & = \sqrt{2x-10}\\
m^2x^2 & = 2x-10\\
m^2x^2-2x+10 & = 0.
\end{align*}
For this to be a tangent the line should intersect the curve at coincident points, meaning the roots of the above quadratic must be equal. Thus
$$4-40m^2 =0 \implies m=\pm \frac{1}{\sqrt{10}}.$$
Keeping in mind that this is the upper branch of the parabola. $m=\frac{1}{\sqrt{10}}$.

The tangent is given by $$mx+ b,$$ with $m$ being the derivative of $f$ in $x=a$ ($x$-value to be evaluated).


Does the equation $x^3-9=0$ have any solutions in $(\mathbb{Z}/31 \mathbb{Z})$? This is my first time "solving an equation" in a group so I feel like that may be the source of my troubles. Here's what I have so far, although I'm not sure how to progress:

  Assume $c$ is a solution to $x^3-9=0$ in $(\mathbb{Z}/31 \mathbb{Z})$.
  We then have $$c^3-9=0\implies c^3=9 \implies |c^3|=|9|$$ and since on
  the LHS we're multiplying elements in $(\mathbb{Z}/31 \mathbb{Z})$, we
  consider the order of $[9]_{31}$ in $(\mathbb{Z}/31 \mathbb{Z})^*$
  which is $15$ since $$9^{15} \equiv 1 \ \text{mod} \ 31.$$ Thus
  $|c^3|=15$.

I feel like I should be able to deduce $|c|$ from $|c^3|$ but it doesn't seem to be clear to me. Could someone assist me in figuring out how to get $|c|$?

From there, my next steps would be to ensure it divides $31$. Since $(\mathbb{Z}/31 \mathbb{Z})^*$ is abelian and cyclic thus has an element of maximal finite order of $31$, and if $|c| \leq 31$ then $|c|$ must divide $31.$

Slightly more efficiently, we get that if $c^3\equiv 9\equiv 3^2\pmod{31}$, then $c^{15}\equiv 3^{10}\pmod{31}$.

$$3^{10}\equiv (27)^3\cdot 3$$

$$\equiv (-4)^3\cdot 3\equiv -64\cdot 3$$

$$\equiv -2\cdot 3\equiv -6\pmod{31},$$

Since $\left(c^{15}\right)^2\equiv 1\pmod{31}$ for all $c\in\mathbb Z$ such that $31\not\mid c$ by Fermat's little theorem,

we get $31\mid \left(c^{15}+1\right)\left(c^{15}-1\right)$,

so $31\mid c^{15}+1$ or $31\mid c^{15}-1$ by Euclid's lemma,

so $c^{15}\equiv \pm 1\pmod{31}$, contradiction.

((In fact, $a^{\frac{p-1}{2}}\equiv \left(\frac{a}{p}\right)\pmod{p}$ for all $a\in\mathbb Z$, odd primes $p$ by Euler's criterion))

((and also, in this case, $c^{15}\equiv 1\pmod{31}$, because

if $c^{15}\equiv -1\pmod{31}$,

then $\left(3^5\right)^2\equiv -1\pmod{31}$,

contradiction, because $-1$ isn't a quadratic residue mod $31=4\cdot 7+3$ by Quadratic Reciprocity))


How can I find the surface area of a normal chicken egg? This morning, I had eggs for breakfast, and I was looking at the pieces of broken shells and thought "What is the surface area of this egg?" The problem is that I have no real idea about how to find the surface area.

I have learned formulas for circles, and I know the equation for an ellipse; however, I don't know how to apply that.

The only idea I can think of is to put an egg on a sheet of paper and trace it, and then measure the outline drawn, and then try to find an equation for that ellipse and rotate that about the $x$-axis.  Now, my problem is how I can find the equation of the ellipse from the graph, and will my tracing method really be the edge of the egg?  Also, can I use the standard surface area integral? Will I have to use some techniques to solve the integral that are not covered in the AP BC Calculus?

There has to be a better method for finding the surface area. Please, help me understand how to find the surface area of an egg; i.e., how to use my mathematical knowledge for something other than passing exams.

There is a nice equation describing the equation of an egg curve credit to Nobuo Yamamoto :
$$
(x^2+y^2)^2 = ax^3 + \frac{3a}{10}xy^2, \tag{1}
$$
where $0\leq x\leq a$, $a$ is the length of the major axis of symmetry for an egg.

In other words, we could get it by cutting a boiled egg in half and measure the distance from tip to the bottom. I just drew it in MATLAB, and the curve looks like the following for $a=1$:

I must say this curve fits pretty well with an egg. Now we have a curve, then the method of computing surface area by revolution, which is taught in Calculus II I believe, can be used to computing the surface. We just revolve the curve above around the egg's major axis of symmetry and get a surface, here is what looks like when we revolve it around the $x$-axis by degree $\pi$, we can get the lower half by revolve another $\pi$:

First we solve for $y$ in (1) when $y>0$:
$$
y = \sqrt{\frac{3ax}{20} - x^2 + x\sqrt{\frac{7ax}{10} + \frac{9a^2}{400}} },
$$
The formula of computing surface area by revolution is:
$$
S = 2\pi\int_0^a y\sqrt{1+\left(\frac{dy}{dx}\right)^2} \,dx,\tag{2}
$$
The derivative is:
\begin{align}
\frac{dy}{dx} = \frac{1}{2\sqrt{\frac{3ax}{20} - x^2 + x\sqrt{\frac{7ax}{10} + \frac{9a^2}{400}} }} \left( \frac{3a}{20}- 2x+ \sqrt{\frac{7ax}{10} + \frac{9a^2}{400}} + \frac{7ax}{20\sqrt{\frac{7ax}{10} + \frac{9a^2}{400}}}\right),
\end{align}
Plugging $dy/dx$ and $y$ into (2), then you could use your favorite tool of numerical integration to perform the computing for you(Octave, MATLAB, Mathematica, etc).

A more tweakable/numerical/experimental approach:

As Ｊ. M. suggests in the comments, the shape looks like an egg, but is a real egg being approximated nicely by this curve? I guess the answer is that "it really depends on that specific egg"!

Let's say we still want to use surface of revolution to compute the surface area.

But this time, we handle it more numerically from the very start, instead of looking for a curve to fit one thing for all.

Two assumptions:

All eggs are axial symmetric with respect to its major axis, i.e., if $x$-axis is its major axis, its surface can be obtained by revolving a curve $y= f(x)$, for $0\leq x\leq a$.
That curve $y = f(x)$ has certain smoothness: $f$ and $f'$ are continuous for $x\in (0,a)$ .

Now we want to compute the integral (2) using Simpson's rule or Trapezoidal rule, which is also taught in Calculus II in most colleges I believe. A remark is that $|f'|\to \infty$ when $x\to 0$ and $x\to a$, it would be much better if we use adaptive qudrature by putting more sample points near $0$ and $a$.

The steps are:

Boil an egg, cut it by half, hold it against a paper, use a pencil to outline its boundary (upper half is enough).
Draw the major axis, set it to be the $x$-axis, and measure its length $a$.
Choose $(n+1)$-sample points (including the end points) so that the points are equidistant to their neighbor on the curve. $n$ is chosen to be even, measure the distance to the major axis ($y$-coordinates) like the above figure.
Denote the sample point as $(x_i,y_i)$, $x_0=0$, $x_n = a$.
Approximate
$$\frac{dy}{dx}\Big|_{x_i} \approx s_i = \frac{1}{2}\left( \frac{y_{i+1} - y_i}{x_{i+1} - x_i} +  \frac{y_{i} - y_{i-1}}{x_{i} - x_{i-1}} \right).$$
For two end points:
$$
s_0 = \frac{y_1 - y_0}{x_1 - x_0},\quad \text{ and }\quad s_n = \frac{y_n- y_{n-1}}{x_n - x_{n-1}}.
$$
This step may be problematic, we can use other methods to approximate $dy/dx$: for example, approximating the curve by cubic spline using sample points $(x_i,y_i)$, but it would be beyond the content of college calculus.
Let $h = x_{i+1} - x_i = a/n$, approximate (2) by computing:
$$
\frac{2\pi h}{3}\bigg[g(x_0)+2\sum_{j=1}^{n/2-1}g(x_{2j})+
4\sum_{j=1}^{n/2}g(x_{2j-1})+g(x_n)
\bigg], \quad \text{ where } g(x_i) = y_i \sqrt{1+s_i^2}.
$$

Some results comparison:

Amzoti gave a link in his comments above that has two semi-empirical formulas:
$$
S_1 = (3.155 − 0.0136a + 0.0115b)ab, \;\text{ and }\;S_2 = \left(0.9658\frac{b}{a}+2.1378\right)ab
$$
where $a$ and $b$ are the length for major and minor axis of the real eggs. If there exists an egg's shape like (1), $a = 1$, and $b\approx 0.7242629$, the surface area computed by above formula is:
$$
S_1 \approx 2.278215  \;\text{ and }\;S_2 \approx 2.054946.
$$
Using the surface by revolution formula (2), we have:
$$
S \approx 2.042087.
$$

A few steps should allow to determine the surface of a given egg

Take a measuring cup as accurate as possible.
Pour some water in the cup, say half the cup: note the current volume $v_1$
Put the egg inside the cup and it should sink
Note the new volume $v_2$, where $v_2 - v_1$ gives the volume of the egg
consider that to be the equivalent volume of a sphere, where $$v_2 - v_1 = \frac{4}{3}\pi r^3$$ and get $r$ from that.
the surface is $4\pi r^2$


When is a halfspace a subset of another halfspace? Let's say we have the following halfspaces:
$$H_1=\{x\mid a^Tx\leq b\}$$ and $$H_2=\{x\mid\tilde{a}^Tx\leq \tilde{b}\}$$

I want to find the conditions that need to hold such that $H_1\subseteq H_2$. So these halfspaces are defined by the hyperplanes, say,
$$A_1=\{x\mid a^Tx=b\}$$ and
$$A_2=\{x\mid\tilde{a}^T x=\tilde{b}\}$$
Intuitively I understand that $H_1\subseteq H_2$ demands the hyperplanes to be parallel or equivalently the normal vectors $a$ and $\tilde{a}$ to be parallel, so:
\begin{eqnarray*}
&&A_1\parallel A_2
\\&\Leftrightarrow& a\parallel \tilde{a}
\\&\Leftrightarrow& \exists t\in \Re\text{ such that }\tilde{a}=ta
\end{eqnarray*}
The hypothesis $H_1\subseteq H_2$ also requires the $H_2$ to lie above $H_1$ which restricts us to consider $\tilde{b}\ge tb$ for positive values of $t$ ($t>0$). But how can I formally prove that these two conditions are enough and necessary?

For the general case, to appear $t> 0$, let us normalize the vectors $a, b$. Then we have,
$H_1=\{x\mid a^Tx \le b\} = \{x\mid\frac{a^T}{\|a\|} \le \frac{b}{\|a\|}\}$.
Do the same for $H_2$ and do the same as above. We will show $t =\frac{\|a\|}{\|a'\|}$.


$P=x^3 + x^2 + ax + 1$, what is $a$ when $ P$ is a rational numbers, also satisfies $ 0= x^2+2x-2$ $x^3 + x^2 + ax + 1 $

$0= x^2+2x-2$

$x= - 1 \pm \sqrt 3$

$x^3 + x^2 + ax + 1 = x^3 + 2 - 2x + ax + 1  $
 Or $x(x^2 + x) + ax + 1 = x(2-x) + ax + 1 = 2x + 2x - 2 + ax + 1 =4x+ax-1$

If i sub $x= 1 \pm \sqrt 3$ into $4x+ax-1$
 I got wrong answer. Where am i wrong?

$0=x^2+2x-2$, so $$0=(x-1)(x^2+2x-2)=x^3+x^2-4x+2$$

If we subtract this from $P=x^3+x^2+ax+1$, we get $$P=(a+4)x-1$$

As you note, $x$ is an irrational number, so $P$ is only rational if $a=-4$.


Why does the function $r = \theta$ graph a spiral? If $\theta$ denotes an angle (in radians I assume), and $r$ means the distance from the origin, then why does $r = \theta$ make a perfect spiral?  I'm not that advanced in math (only in geometry and Algebra II), and I saw a 3Blue1Brown video that involved spirals and prime numbers, and I got curious. When I graphed $r=\theta$ in Desmos, I saw the spiral and asked "Why?"

Does anyone know why this happens?

So, if we graph $r = f( \theta )$, as we often do, what does this mean? This means, for each angle $\theta$ (where $\theta$ is a real number), its distance from the origin (or radius) $r$ is given by $f(\theta)$. This is often easiest to see if you take a particular function $f$ and graph a bunch of points $(r, \theta)$ with increasingly higher values of $\theta$. Animations are also nice. Some are in this video. You can play with some demos yourself on Wolfram here.

In this sense, what does $r = \theta$ represent? Basically, for any given angle $\theta$, its distance from the origin is equal in value to $\theta$.

An angle of $\theta = 0$? The radius $r$ is $0$.
An angle of $\theta = \pi/4$? The radius $r$ is $pi/4$.
An angle of $\theta = \pi$? The radius $r$ is $\pi$.
An angle of $\theta = 2\pi$? The radius $r$ is $2\pi$.
An angle of $\theta = 10^{10}$? The radius $r$ is $10^{10}$.

All you're doing is graphing the points $(\theta,\theta)$, really (again, $r=\theta$). It's sort of like graphing $y=x$, which gives you the points $(x,x)$ -- now just with a polar flair.

On a

ω(x′,y′)=ω(ρ,φ)=a₀(ρ)+∑_{n=1}^{∞}[a_{n}(ρ)cos nφ+b_{n}(ρ)sin nφ]

si je prends

ω(x′,y′)=ω(ρ,φ)=a₀(ρ)=√(1-ρ²)

tel que

T(ζ,η)=(1/(2π))∬_{S}ω(x′,y′)exp[i(ζx′+ηy′)]dx′dy′

et

S:((x²)/(a²))+((y²)/(b²))≤1;z=0,

(ζa,ηb) = (λcosχ,λsinχ); λ∈[0,∞[,χ∈[0,2π],
(x,y) = (arcosθ,brsinθ); r∈[0,1],θ∈[0,2π],
(x′,y′) = (aρcosφ,bρsinφ); ρ∈[0,1],φ∈[0,2π],


Question on finding the range of a function involving squareroot The question is as under :

  Find the range of the real valued function : $f(x) = \dfrac{1}{\sqrt{16-x^2}}$

The answer in the book is :  $(-\infty,1/4] \cup [1/4, \infty)$.

But according to me the answer should be only $[1/4, \infty)$ because $f(x)$ cannot take negative values because if it takes negative values,
there will be two values of $f(x)$ corresponding to one value of $x$ and then $f(x)$ will not remain a function.

Then why is the answer in the book different?

Please give me the explanation. Thank you.

You are correct.

$\sqrt{16-x^2}$ only takes on positive values from $(-4,4)$, which range from $0$ to $4$.

Well, the zeros of $16-x^2$ are $\pm 4$ and the expression $16-x^2$ is non-negative if  $-4\leq x\leq 4$. Now stick it together to find the real-valued solutions of $1/\sqrt{16-x^2}$.


Two forms of $\big(A^{-1} + B^{-1}\big)^{-1}$ I am trying to calculate $\big(A^{-1} + B^{-1}\big)^{-1}$ and I find that there are two ways to get the answers:

$$\big(A^{-1} + B^{-1}\big)^{-1}
 = \big(A^{-1}(I + AB^{-1})\big)^{-1}
 = \big(A^{-1}(B+A)B^{-1}\big)^{-1} = B(A+B)^{-1}A$$

$$\big(A^{-1} + B^{-1}\big)^{-1}
 = \big(B^{-1}(BA^{-1} + I)\big)^{-1}
 = \big(B^{-1}(B + A)A^{-1}\big)^{-1}
 = A(A+B)^{-1}B$$
So my question is that: is there a way to prove that
$$B(A+B)^{-1}A = A(A+B)^{-1}B$$
without proving that they are both equal to the inverse of $(A^{-1}+B^{-1})$ ?

You only need that $A+B$ is invertible.

If $A$ and $B$ are both invertible, then as you demonstrated, it follows that $A^{-1}+B^{-1}$ is also invertible and $$B(A+B)^{-1}A = (A^{-1}+B^{-1})^{-1} =A(A+B)^{-1}B$$

If $A$ or $B$ is not invertible, then since $\sigma(A)$ and $\sigma(B)$ are finite, there exists $r> 0$ such that for all $\varepsilon \in \langle0,r\rangle$ we have that $A_\varepsilon := A-\varepsilon I$ and $B_\varepsilon := B+\varepsilon I$ are both invertible. Now from the first case it follows that
$$B_\varepsilon(A+B)^{-1}A_\varepsilon = B_\varepsilon(A_\varepsilon+B_\varepsilon)^{-1}A_\varepsilon = A_\varepsilon(A_\varepsilon+B_\varepsilon)^{-1}B_\varepsilon = A_\varepsilon(A+B)^{-1}B_\varepsilon.$$
Letting $\varepsilon \to 0$ from continuity of matrix operations we get $B(A+B)^{-1}A =A(A+B)^{-1}B$.


Do all rational numbers repeat in Fibonacci coding? In a regular, positional number system (like our decimal numbers), every rational number ends in a repeating sequence (even if it's just 0).  For example, $\frac{1}{3}$ in base 10 is $0.33333...$, in base 5 it's $0.131313...$, and in base 3 it's just $0.1$.

A less common number system uses the Fibonacci sequence as its base, so the first few digit places represent $1, 2, 3, 5, 8, 13, 21$, and so on (instead of the decimal $1, 10, 100, 1000,$ etc.).  In this system:

$$
\begin{aligned}
17_{10} &= 100101_F \\
40_{10} &= 10001001_F
\end{aligned}
$$

This can easily be extended to digits after the radix point, so $0.1_F = \frac{1}{2}_{10}$, $0.01_F = \frac{1}{3}_{10}$, $0.001_F = \frac{1}{5}_{10}$, and so on.

$$
\begin{aligned}
\frac{5}{6}_{10} &= 0.11_F \\
\frac{7}{10}_{10} &= 0.101_F
\end{aligned}
$$

Some numbers aren't as easy to write.  Greedily adding up the first places that are smaller than the remainder of the number we're trying to write:
$$
\frac{1}{4} = \frac{1}{5}+\frac{1}{21}+\frac{1}{610}+\frac{1}{1597}... = 0.001001000000101..._F \\
\frac{2}{3} = \frac{1}{2}+\frac{1}{8}+\frac{1}{34}+\frac{1}{89}... = 0.10010010100001..._F
$$

Do all rational numbers end in a repeating sequence in Fibonacci coding?

(For the specific case of $\frac{1}{4}$, see here.)

I'm not sure if this is what you meant, but integers, such as your examples 17 and 40, are also rational numbers (as they can be represented by x/1, where x is an integer), and they don't repeat in Fibonacci coding, so there are some rational numbers (such as integers) that do not repeat in Fibonacci coding


Cellular homology of projective space $\mathbb{R}P^n$ For the projective space the cell decomposition is $e_0 \cup \dots \cup e_n$ and the attaching map is $a_1 a_1 \dots  a_k a_k$ for the $k$-th cell.
So for $k \leq n$ I thought that this means that the boundary map between the cellular chain groups is multiplication by two because the boundary map of the $k$-th cell coincides with the attaching map. So I thought $im \partial_k = \langle a_1 a_1 + \dots + a_k a_k \rangle = 2 \langle a_1 + \dots + a_k \rangle \cong 2 \mathbb{Z}$.

Why is this wrong? In the example on p. 144 in Hatcher the chain complex looks like this

$$ \dots  \xrightarrow{0} \mathbb{Z} \xrightarrow{2} \mathbb{Z} \xrightarrow{0} \mathbb{Z}\xrightarrow{2} \mathbb{Z} \dots$$

How can the boundary map be the $0$ map? Many thanks for your help.

you can use of local degree and the fact degree of reflection is $(-1)^{n+1}$ .or if you use $I^n$, $\partial_k=<a_1-(-1)^{k+1}a_1+...>$ because degree of reflection is $(-1)^{n+1}$


Product of Permutations- Abstract Algebra I'm trying to find the product of the permutations: (1,2,3)(2,4,3,5)(1,3,2)
From what I've been taught in my Abstract Algebra class, the product of permutations is in fact a composition of permutations. So, in cyclic notation, I got my answer to be (1,5,3,4) (2). However, Wolfram Alpha tells me that my answer is incorrect. I must be missing something here. The way I solve these is the following: I treat each permutation as a function. So let's say our 3 functions are f,g,h. f composed with g composed with h = f(g(h)). So let's say I want to know where 3 goes to, I take f(g(h(3))= f(g(2))= f(4)= 4. And I do this for all of the other elements. What am I doing wrong here? Any help would be appreciated.

As it says on the W|A page, "the product $\sigma \tau$ represents the permutation $\tau(\sigma(\cdot))$." So they're just using the opposite, "left-to-right" notation. For instance, they get the cycle $(3)$ because $3$ is mapped to $1$, then $1$ is mapped to $1$, then $1$ is mapped to $3$.

this opposite way of expressing the composition is quite prevalent. Even JA Gallian does this way in his book on abstract algebra.


How comes that ( after Desmos graphing calculator) the " area function" of f(x)=x² gives back negative values when x<0? I'm currently trying to understand basic calculus concepts.

I asked to Desmos graphing calculator to draw the graph

(1) of f(x) = x²

(2) of the " area function" or " accumulation function" of function f, that I defined as the integral from 0 to x of f(t)dt.

What astonishes me is that the curve of the area function is below zero for x<0, while the curve of the original function is always above the X-axis.

If my " area function " represents the area under the curve of f and the X axis ( limited by x=0 and  the vertical line passsing through x), how comes that the values of the area function are negative for x < 0?

In order the " area function" to correspond to its informal concept , shouldn't we define it as

A(x)= absolute value of ( integral from 0 to x of f(t)dt ) ?

Integrals explicitly are signed areas. Your original function is always positive, and if you're integrating left-to-right, so $x_1>x_0$:
$$\int_{x_0}^{x_1}f(x) dx >0$$
but if you're integrating right-to-left, so $x_1<x_0$:
$$\int_{x_0}^{x_1}f(x) dx <0$$
The area under your original function between, say $x_0=-2$ and $x_1=0$, is $F(0)-F(-2)$, which you'll find is positive.

What you're plotting is a primitive function $F$, which has the useful property that the signed area under $f$ between $x_0$ and $x_1$ is $F(x_1)-F(x_0)$. Taking absolute values as you propose destroys this useful property.

It's normal. You have negative values for x<0
Intégration of f(x)=x^2 from a to b is always positif only if a < b


Proof of $C\left(n, k\right) = \frac{C\left(n, k-1\right) \cdot C(n - (k - 1), 1)}k$ I was wondering if anyone could explain to me why this holds true?

$$C\left(n, k\right) = \frac{C\left(n, k-1\right) \cdot C(n - (k - 1), 1)}k$$

or

$$C\left(3, 2\right) = \frac{C\left(3, 1\right) \cdot C(2, 1)}2$$

I was playing around with combinatorics, and realized that this holds true for all values that I put into it.

Here is a combinatorical explanation:

You can choose $k$ from $n$ items by

choosing $k-1$ from $n$: $\color{blue}{\binom{n}{k-1}}$ and then
choosing $1$ from the remaining $n-(k-1)$: $\color{blue}{\binom{n-(k-1)}{1}}$

But using this selection method the same selection can result in $\color{blue}{k}$ ways, since the $1$ you choose "last" can be one of the $k$. So,

$$\color{blue}{\binom{n}{k-1}\cdot\binom{n-(k-1)}{1}} = \color{blue}{k}\cdot \binom{n}{k}$$

LHS is the number of ways to choose $k$ objects from $n$ objects.
RHS is nothing but, now from $n$ objects first choose $k - 1$ and then from remaining $n-(k-1)$ objects choose 1 obeject.
but in this counting, we are repeating the same objects to objects K, so divided by Y
enter image description here

$\frac { n ! } { k ! ( n - k ) ! } \simeq \frac { 1 } { k } \frac { n ! } { ( n - 1 ) ( n - k + 1 ) ! } \frac { ( n - k + 1 ) ! } { 1 ! ( n - k ) ! }$


Let $a,b \in \mathbb{Z}$ and suppose that $a | b$ and $b |a$. Then $a=b$ or $a=-b$ Let $a,b \in \mathbb{Z}$ and suppose that $a | b$ and $b |a$. Then $a=b$ or $a=-b$

My attempt

$a|b \Rightarrow b=k_1a$ for some $k_1$

$b|a \Rightarrow a=k_2b$ for some $k_2$

By substitution $a=k_1k_2a \Rightarrow a(1-k_1k_2)=0$

So, we have two cases:-

Case $(1)$

$k_1k_2=1 \Rightarrow k_1=k_2=\pm 1$

Thus, $a=\pm b$

Case $(2)$

I am stuck here, as $a=0 \Rightarrow b=0$.

But then division by $0$ doesn't make sense. What to do?

Note

This question is not a duplicate as it explicitly asks the question for $\mathbb{Z}$ and not $\mathbb{N}$ or nonzero integers.

If $a=0$ then, since $a\mid b$ by hypothesis , we must have that $0=ak=b$ for some $k\in\mathbb{Z}$.

If you read $a | b$, then you can suppose that $a \ne 0$


Probability of $\limsup$ of a sequence of sets (Borel-Cantelli lemma) Let $(E_n)$ be a sequence of events in a probability space such that $$\lim_{n\to\infty}\mathbb P(E_n)=0.$$ I am trying to prove that if $$\sum_{n=1}^\infty\mathbb P (E_n\setminus E_{n+1}) <\infty$$ then $$\mathbb P\left(\limsup_{n\to\infty} E_n\right)=0.$$

For an example I am using the sequence of intervals $[0, 1/n]$ which goes to $0$ as $n$ increases to infinity. Then I can see how the probability of infinite sum of the intersection of the $E_n$ and the complement of $E_{n+1}$  is less than infinity and the probability of $\limsup_{n\to\infty} E_n$ equals zero.

Can anyone help me figure out how to prove this in general using any sequence? Especially to show how $\mathbb P\left(\limsup_{n\to\infty} E_n\right)=0$?

Thank you in advance for your thoughts.

An application of the Borel-Cantelli lemma gives that $\mathbb P(\limsup_n E_n\setminus E_{n+1})=0$. This means that there is a set $\Omega'$ of probability $1$ such that if $\omega\in\Omega'$, then $\omega\in E_n^c\cup E_{n+1}$ for each $n\geqslant N(\omega)$. Notice that if for some $n\geqslant N(\omega)$ we have $\omega\in E_n$, then $\omega\in E_k$ for each $k\geqslant n$ (this can be seen by induction on $k$). Hence $\limsup_n E_n\subset \liminf_n E_n$ (up to a subset of measure $0$).  The assumption $\mathbb P(E_n)\to 0$ show that $\mathbb P(\liminf_n E_n)=0$.

Note that
$$
\bigcup_{n=m}^{\infty} E_n = \bigcup_{n=m}^{\infty} E_n\setminus E_{n+1}
$$
Let $\epsilon > 0$, since $\sum P(E_n\setminus E_{n+1}) < \epsilon$, then $\exists m\in \mathbb{N}$ such that
$$
\sum_{n=m}^{\infty} P(E_n\setminus E_{n+1}) < \epsilon
$$
and so
$$
P(\bigcup_{n=m}^{\infty} E_n) < \epsilon
$$
So if $E = \limsup E_n$, then $P(E) <\epsilon$. This is true for any $\epsilon >0$, so $P(E) = 0$.


Completness and Set of Result of One Set ?!? Dear Everyone on this Wonderful Sites:

I'm so glad to participate on this site and ask the first question that mentioned on the Contest some days ago. I ran into a question that wrote this set:

$\{(p_i \vee $~ $p_{i+1}$$) $$: i \in  \mathbb{N}  \}$

and ask this question, Which of the following is False:

  -)  this is Satisfiable

  --) this is Complete

  ---) this is Decidable

  ----) set of logical result of this set is effectively enumerable.

Solution of answer sheet is (2).

We try to solve it that without any detail is mentioned in the contest. one of our TA says, It is satisfiable because the TFTF... will satisfy it.  (We couldent get it) and it is decidable because propositional logic is decidable. We read some useful website for completeness (--) and for (----), but we get stuck in it until yet. any useful hint or idea for this question will be highly appreciated.

Let $\Phi = \{ p_i \lor \neg p_{i+1}|i \in \mathbb{N}\}$

It is easy to see that all models of $\Phi$ are of the form $M_k$, where

$$M_k(p_i)= \left\{ \begin{array}{ll} T \quad i < k\\ F \quad \text{else}\end{array} \right. \quad k\in \omega \cup \{\omega\}$$

So in particular it is satisfiable.

It is not complete, since $M_0\models \neg p_0$ but $M_1\models p_0$

The set $\Phi^\models$of logical results of $\Phi$ is decidable, so in particular recursivly enumerable:

Let $\varphi(p_0,\dots,p_n)$ be a formula whose propositional variables are among $p_0 \dots p_n$.

$$\Phi \models \varphi \Leftrightarrow \{ p_i \lor \neg p_{i+1}|i \leq n\}\models \varphi \Leftrightarrow \bigwedge_{i\leq n}p_i \lor \neg p_{i+1} \models \varphi$$

The last can be checked by truth tables.

The question asks which of the following is False.  Now, if for some propositional formula $\phi$, neither $\{(p_i \vee $~ $p_{i+1}$$) $$ $$ : i \in  \mathbb{N}  \}$ $\vdash$ $\phi$ nor $\{(p_i \vee $~ $p_{i+1}$$) $$ $$ : i \in  \mathbb{N}  \}$ $\vdash$ $\lnot$ $\phi$, then $\{(p_i \vee $~ $p_{i+1}$$) $$ $$ : i \in  \mathbb{N}  \}$ is not complete.

Now, note that we completeness is defined in terms of "$\vdash$" and not in terms of "|=".  Thus, we simply declare that the propositional logic we have under study has no rules of inference.  And thus neither $\phi$ nor $\lnot$ $\phi$ is provable, and thus $\sum$ is incomplete and it is fase that $\sum$ is complete.  Or do something like set up an axiomatic propositional calculus with only conditional connectives, forbid definitions, and then indicate $\phi$ as a disjunction.

Example of such a latter calculus:

Our axiom set is {CpCqp, CCpqCCqrCpr, CCpCpqCpq} under the rules of detachment and substitution with no definitions.  Thus, the set {CpCqp, CCpqCCqrCpr, CCpCpqCpq, ApNp} is not complete, since neither $\vdash$ ApNp nor $\vdash$ NApNp.


Maximum number of 2 digit integers There are 90 cards with all two-digit numbers on them:
$$10,11,12,\dotsc,98,99$$
A player takes some of these cards. For each card taken she gets $1. However, if the player takes two cards that add up to 100 (say, 23 and 77), she loses all the money. How much could she get?

She can make at most $50.
Suppose there were 99 cards, numbered 1 to 99.  She would be able to take number 50, as well as one of 49 and 51, one of 48 and 52, ... and one of 1 and 99, for a total of 50 cards taken.  Because in this case she would be able to take cards the fifty cards numbered 50-99, she obviously can do so in our case as well (where cards 1-9 have been removed).  She can't make more money, because any additional card taken will pair with one of the cards she already took to total 100

If she has nine cards already, and all have partners on the table, then she has nine chances to lose 9 dollars and 72 chances to win 1, so she might as well stop.  The calculation changes depending on the number of unpaired cards she has.
If she has $m$ cards with partners on the table, and $n$ without partners, she has $m$ chances of losing $m+n$, and $90-2m-n$ chances of continuing, with $+1$ on that card.  The average is $\frac1{90}(-m^2-mn+90-2m-n)$, which is positive if $m+n\lt(90-m)/(m+1)$


In Russian roulette, is it best to go first? Assume that we are playing a game of Russian roulette (6 chambers) and that there is no shuffling after the shot is fired.

I was wondering if you have an advantage in going first?

If so, how big of an advantage?

I was just debating this with friends, and I wouldn't know what probability to use to prove it. I'm thinking binomial distribution or something like that.

If $n=2$, then there's no advantage. Just $50/50$ if the person survives or dies.

If $n=3$, then maybe the other guy has an advantage. The person who goes second should have an advantage.

Or maybe I'm wrong.

For a $2$ Player Game, it's obvious that player one will play, and $\frac16$ chance of losing.  Player $2$, has a $\frac16$ chance of winning on turn one, so there is a $\frac56$ chance he will have to take his turn. (I've intentionally left fractions without reducing them as it's clearer where the numbers came from)

Player 1 - $\frac66$ (Chance Turn $1$ happening) $\times \ \frac16$ (chance of dying) = $\frac16$

Player 2 - $\frac56$ (Chance Turn $2$ happening) $\times \ \frac15$ (chance of dying) = $\frac16$

Player 1 - $\frac46$ (Chance Turn $3$ happening) $\times \ \frac14$ (chance of dying) = $\frac16$

Player 2 - $\frac36$ (Chance Turn $4$ happening) $\times \ \frac13$ (chance of dying) = $\frac16$

Player 1 - $\frac26$ (Chance Turn $5$ happening) $\times \ \frac12$ (chance of dying) = $\frac16$

Player 2 - $\frac16$ (Chance Turn $6$ happening) $\times \ \frac11$ (chance of dying) = $\frac16$

So the two player game is fair without shuffling.
Similarly, the $3$ and $6$ player versions are fair.

It's the $4$ and $5$ player versions where you want to go last, in hopes that the bullets will run out before your second turn.

For a for $4$ player game, it's:
P1 - $\frac26$,
P2 - $\frac26$,
P3 - $\frac16$,
P4 - $\frac16$

Now, the idea in a $2$ player game is that it is best to be player $2$, because in the event you end up on turn six, you KNOW you have a chambered round, and can use it to shoot player $1$ (or your captor), thus winning, changing your total odds of losing to P1 - $\frac36$, P2 - $\frac26$, Captor -  $\frac16$

For 2 players and n chambers player 1 loses if the round is in the odd-numbered chamber. Hence if $n$ is even, chances are $50/50$ and if n is odd, chances of losing are $(\frac{n+1}{2n},\frac{n-1}{2n}$) - it's a disadvantage to go first.


Why is $(A^\top A \mathbf{x}, \mathbf{x}) = (A \mathbf{x}, A \mathbf{x})$? Let $(\mathbf{x}, \mathbf{y})$ denote the inner product between $\mathbf{x}$ and $\mathbf{y}$, and let $A$ be a real matrix.  Why is $(A^\top A \mathbf{x}, \mathbf{x}) = (A \mathbf{x}, A \mathbf{x})$?

Using the scalar product it's easy to see that
$$
\begin{align}
(A^\top A \mathbf{x})^\top \mathbf{x} &= \mathbf{x}^\top A^\top A \mathbf{x} \\
&= (A \mathbf{x})^\top A \mathbf{x}
\end{align}
$$
but using only the properties of the inner product I fail to see how one gets the result.

Edit:

To be clear, I'm asking how one comes to the conclusion that $(A^\top A \mathbf{x}, \mathbf{x}) = (A \mathbf{x}, A \mathbf{x})$, independently of how the inner product is defined (ie. only using the axioms of the inner product), so if your answer relies on a particular definition of the inner product such as $(\mathbf{x}, \mathbf{y}) := \mathbf{x}^\top\mathbf{y}$, it's not good.

Edit 2:

The equality I'm asking about can be found in the book Linear Algebra Done Wrong (pdf) by Sergei Treil (p. 172), although here I'm interested only in the case of real matrices in the book it covers the complex case as well.

Well, I don't think what you're asking can be proven, because I don't think it's true for all inner products on $\mathbf{R}^n$. Therefore, you have to specifically use the defined inner product.

We have that $\langle x,y\rangle$ is an inner product on $\mathbf R^n$ if and only if
$$\langle x,y\rangle = x^\intercal My$$
for a symmetric, positive definite matrix $M$, see here. Hence, we have that
$$\langle A^\intercal A x,x\rangle = (A^\intercal A x)^\intercal M x = x^\intercal A^\intercal A Mx = (Ax)^\intercal A Mx $$
whereas
$$\langle Ax, Ax\rangle = (Ax)^\intercal M Ax.$$

So they are only equal in general if $M$ and $A$ commute, which happens e.g. when $M = I_n$ is the idendity matrix which gives the standard inner product on $\mathbf R^n$.

In your work, we have $(Ax)^T Ax$. The inner product is $(x, y) = x^T y$. Now set $x=Ax, y=Ax$, and it becomes clear. You transpose the first argument and multiply by the second.


How to find the shaded area of a triangle formed by the tangent of a circle? Hey guys,
The last question of the chapter has stumped me once again! I need to find the area of part of this triangle. I’m not sure how to attack the question considering that I’m not given the radius of the circle and I’m not sure how the measurement of 4cm will help me?
Thanks,
Scratch Cat

One way is to find the difference in areas of a right-angled triangle and a circular sector.
Finding both areas requires finding the radius of the circle, which may be found through the triangle
$$\begin{align*}
\frac{r}{r+4} &= \sin 35^\circ\\
r &= (r+4)\sin 35^\circ\\
(1-\sin 35^\circ)r &= 4\sin 35^\circ\\
r &= \frac{4\sin 35^\circ}{1-\sin 35^\circ}
\end{align*}$$

Step 1
Step 1: Using our knowledge that $∠BAC = ∠DAE = 35^o$ and that $∠ADE = 90^o$ (tangentiality), we can calculate section $AD = AE/cos(35) \approx 4.88$
Step 2
Step 2: Using Thales' Theorem, we know that $∠ADF=90^o$.
Thus, we can simply calculate $AF=AD/cos(35) \approx 5.96$
Then, using simple trigonometry again we obtain $AB = AF/cos(35)\approx7.28 $
Step 3
Step 3: Invoking Thales' theorem again, we obtain that $∠ABC = 90^o$.
We then calculate $AC=AB/cos(35)\approx8.88$
Similarly, $FB=AF*tan(35)\approx4.17$
Step 4: The area of ABC is then = $0.5*FB*AC = 18.54$
The radius of the large circle = $8.88-4 \approx 4.88$
The area of the large circle = $\pi*r^2 \approx 74.91$
Step 5: However, we only need to know the part within the triangle. As $∠ABC = 90^o$ and $∠BAC=35^o$, $∠BCA = 55^o$.
Thus, the relevant area $ \approx 55/360*74.91 = 11.44$
Answer: The answer is then $18.54-11.44 = 7.10cm^2$


Self-Contained Proof that $\sum\limits_{n=1}^{\infty} \frac1{n^p}$ Converges for $p > 1$ To prove the convergence of the p-series

$$\sum_{n=1}^{\infty} \frac1{n^p}$$

for $p > 1$, one typically appeals to either the Integral Test or the Cauchy Condensation Test.

I am wondering if there is a self-contained proof that this series converges which does not rely on either test.

I suspect that any proof would have to use the ideas behind one of these two tests.

We can bound the partial sums by multiples of themselves:

$$\begin{eqnarray}
S_{2k+1}
&=&
\sum_{n=1}^{2k+1}\frac{1}{n^p}\\
&=&
1+\sum_{i=1}^k\left(\frac{1}{(2i)^p}+\frac{1}{(2i+1)^p}\right)\\
&<&1+\sum_{i=1}^k\frac{2}{(2i)^p}\\
&=&1+2^{1-p}S_k\\
&<&1+2^{1-p}S_{2k+1}\;.
\end{eqnarray}$$

Then solving for $S_{2k+1}$ yields

$$S_{2k+1}<\frac{1}{1-2^{1-p}}\;,$$

and since the sequence of partial sums is monotonically increasing and bounded from above, it converges.

(See also: Teresa Cohen & William J. Knight, Convergence and Divergence of $\sum_{n=1}^{\infty} 1/n^p$, Mathematics Magazine, 52(3), 1979, p.178. https://doi.org/10.1080/0025570X.1979.11976778)


3$\times$3 and 2$\times$2 matrices that satisfy $A^2 = -I$, $I$ being the identity matrix. I am looking for all square matrices $A$ of order 2 and 3 that satisfy $A^2 = - I$, $I$ being the identity matrix of the corresponding order.

There is no $3\times 3$ matrix $A$ with real entries that satisfies the equation $A^2=-I$.  Suppose there were.  We take determinants, to get $$\left\lvert A\right\rvert^2=\left\lvert A^2\right\rvert=\left\lvert -I\right\rvert=-1$$
But $\left\lvert A\right\rvert$ is real, and no real number squares to $-1$.

In the $2\times 2$ case, by a similar argument we must have $\left\lvert A\right\rvert=\pm1$.  There are solutions, e.g. $A=\left(\begin{smallmatrix}0&-1\\1&0\end{smallmatrix}\right)$.  To find them all use @Mitch Hughes' approach (which has been, unfortunately, deleted).  To summarize, calculate a system of equations from $$\left(\begin{smallmatrix}a&b\\c&d\end{smallmatrix}\right)^2=-I$$

There are essentially infinite solutions that satisfy A^2 = -I for 2x2 and 3x3 matrices.

They can be solved in similar manners of creating systems of equations and then solving those to produce general solutions. I will solve for 2x2 matrices.
For standard square matrices we have a matrix A such that

A =\begin{bmatrix}a&b\\c&d\end{bmatrix}

by multiplying the matrix by itself, we get the following matrix

A^2 = \begin{bmatrix}a^2+bc&ab+bd\\ac+cd&bc+d^2\end{bmatrix}

so we have the following set of conditions that we must statisfy:

a^2+bc = -1
b(a+d) = 0
c(a+d) = 0
d^2 + bc = -1

By solving this system we will be able to see that conditions that a,b,c and d must satisfy

We will first begin by subtracting the first and last equations to get:

a^2-d^2 = 0

so therefore, we can see that a = +/- d

We will first explore the first case when a = d

We immediately see through the 2nd and 3 equations that b and c must equal 0 since a+d will never be 0, unless a = d = 0 (discussed later)

However, knowing that b=c=0, we see that that the 1st and 4th equations will never hold true since a^2 and d^2 will be positive, but it must equal -1.

Now we explore the case when a = -d

Say that a = an arbitrary constant t (then d = -t)

Through the 2nd and 3rd equation we see that c and d could essentially be anything too, since a+d = 0 and the equation would always be true.
Thus the only condition it needs to satisfy are:

bc + t^2 = -1 (since a^2 = d^2 = t^2)

so bc = -(1+t^2)

Now we pick another arbitrary constant s for c

so b = -(1+t^2)/s

As a result we have the following conditions for a,b,c, and d that satisfy A^2 = -I

a= t
b= -(1+t^2)/s
c= s
d= -t

which can also be written as:

A =\begin{bmatrix}t&-(1+t^2)/s\\s&-t\end{bmatrix}

for any values t and s (notice that case of a=d=0 is included here)

As a result, there are infinite a,b,c,d that satisfy A^2 = -I that all lie on a plane in space (since geometric multiplicity of 2)


Minimisation of Concave Function $f\left(\mathbf{x}\right):\mathbb{R}_+^n\rightarrow\mathbb{R}_+$ is a concave monotonically increasing function to be minimised over the feasible region
$\sum_{i=1}^n x_i=1$ and $x_i\geq 0\quad\forall1\leq i\leq n$.

Given that the feasible region is a convex polytope, is it possible to say anything about the optimal $\mathbf{x}^*$? I have a hunch that at $\mathbf{x}^*$, at least one of the inequality constraints will be satisfied with equality. In other words $\mathbf{x}^*$ will be right on the edge of the feasible region. But can't prove it. Am I right there?

If it helps to consider a special case, the function is

$f\left(\mathbf{x}\right)=\log|\mathbf{I}_k+\sum_{i=1}^{n}x_i\mathbf{A}_i|$

where $\mathbf{A}_i\in\mathbb{C}^{k\times k}$ are all positive semidefinite Hermitian matrices.

um,this is obviously a economics problem.You may resort to Lagransian.If you want to know more ,confer to Microeconomic Theory by Mas-colell，Whiston and Green.


Understanding continuity of probabilities I am reading a book called All of Statistics by Larry Wasserman that includes a theorem called "continuity of probabilities."

It says that if $A_{n} \rightarrow A$ then $P(A_{n}) \rightarrow P(A)$ as $n \rightarrow \infty$

I am trying to understand what that statement means in English and to understand its import. How can one probability imply another?

I understand that $A_{n}$ can be broken into lots of disjoint $B_{n}$ which can be added up to form $A_{n}$ -- but I don't see the big picture or how that relates to sample spaces.

If we have $A_1 \subseteq A_2 \subseteq A_3 \subseteq \dots$ and $A=\cup_{n=1}^\infty A_n,$ we write $A_n \nearrow A.$   Given this assumption, a probability measure satisfies $\lim_{n\to \infty} P(A_n) = P(A).$  The word "continuity in probability" is analogous to continuity at point for a function (e.g.  As $x$ approaches $c$,  we have $f(x)$ approaches $f(c).$ )

I was trying to find the use of this theorem in a practical sense and came up with a following example.

Suppose you have some sort of distribution from $[0,2]$ on a real number line.
Now you want to find the probability of a number lying between $[0,1]$
Now,
Consider an event defined by the following equations and the occurrence of these events are governed by the above probability distribution in the range of $[0,2]$:
$A_n = [0, 2^{-\frac1n} )$. Therefore
$$A_1 = [0,.5)$$
$$A_2 = [0,.71)$$
$$A_3 = [0,.8)$$
 and so on and as $n$ tends to infinity
$A_n = [0,1)$

Here $P(A_n)$ would be some function giving a value. This value is the probability of the occurrence of An. This would also give the probability of the number lying in the interval $[0,1]$.

This is my understanding of this theorem as I am also reading the same book.


The determinant of adjugate matrix Why does $\det(\text{adj}(A)) = 0$  if  $\det(A) = 0$? (without using the formula $\det(\text{adj}(A)) = \det(A)^{n-1}.)$

Use :
$$\mathrm{adj}(A)\times A=A \times \mathrm{adj}(A)=\mathrm{det}(A)\mathrm{I}_{n}.$$

Let's assume that $\mathrm{det}(A)=0$. If $A=0$, then $\mathrm{adj}(A)=0$ as well and, obviously, $\mathrm{det}\big( \mathrm{adj}(A) \big)=0$. Otherwise, there exist $x \in \mathbb{R}^{n}$, $x \neq 0$ such that $Ax \neq 0$. It follows that $\mathrm{adj}(A) Ax=0$. As a consequence, $\mathrm{Span}(Ax) \subset \mathrm{ker} \; \mathrm{adj}(A)$ (which means that $\mathrm{ker} \; \mathrm{adj}(A) \neq \lbrace 0 \rbrace$). Therefore, $\mathrm{adj}(A)$ is not invertible and $\mathrm{det}\big( \mathrm{adj}(A) \big)=0$.

In that case, shall we use the following? $[adj(A)]^{-1} = adj(A^{-1})$ ?

Assume that $adj(A)$ is non-singular. Then inverse of adj(A) exists.

But we know $[adj(A)]^{-1}=adj(A^{-1})$ which is contradiction because it is given that $A$ is singular so $A^{-1}$ does not exist and hence adjoint of $A^{-1}$ also does not exist.

hence the proof.


Can the Taylor series be used to calculate this integral? The Wiki article here

says that non-elementary (definite) integrals can often be calculated using the Taylor series.  Can the Taylor series be used to calculate the integral of $$\ln\left[\left(\sin(x) + \frac{\sin(2x)}{2} +\frac {\sin(3x)}{3}\right)^2+0.1\right]$$ from $0$ to $2\pi$?  If so, how?

Sometimes a power series can be integrated term by term, i.e. sometimes $$\int_a^b\sum_{k=0}^\infty c_k(x-a)^k\ dx=\sum_{k=0}^\infty\int_a^b c_k(x-a)^k\ dx.$$ When is "sometimes"? When the series converges uniformly on $[a,b]$, for one.

Thus your first question reduces to whether the power series of the function converges uniformly on $[0,2\pi]$.


Probability of picking a certain fruit from a tree A tree has 20 fruits. 15 of which have no seeds and the rest do have seeds. A bird eats 5 of these fruits picked at random.

a) If i pick one fruit from whats left on the tree whats the probability it has seeds

b) Given that this one fruit i pick has seeds whats the probability that the bird had consumed at least one with seeds.

What i have so far:

So before the bird eats anything the probability of a fruit with no seeds is 15/20=0.75 and the probability of fruit with seeds is 5/20=0.25.

For part a is the probability that the fruit i pick(i.e. 6th fruit picked) has seeds is given by (15/20)* (14/19)* (13/18)* (12/17) *(11/16) *(5/15)=0.065

Am i on the right track with part a? As for part b im not sure where to begin.

a)

The probability that you calculated is the probability that the bird only picked fruits without seeds and you picked one fruit with seeds. That is not the correct answer to the question.

The correct answer is $\frac5{20}$. Think of it like this: all fruits are placed randomly in a row. Now what is the probability that on spot number $6$ (corresponding to your pick after the $5$ picks of the bird) a fruit is placed that has seeds? It will not differ from the probability of any other spot to achieve a fruit that has seeds.

b)

Let $E$ denote the event that the bird consumed none with seeds and let $F$ denote the probability that you picked a fruit with seeds. Then: $$P(E\mid F)P(F)=P(E\cap F)$$where $P(F)=\frac5{20}$ (the probability calculated at a) and $P(E\cap F)=\frac{15}{20}\frac{14}{19}\frac{13}{18}\frac{12}{17}\frac{11}{16}\frac{5}{15}$ (the probability that you calculated in your answer).

This allows you to find $P(E\mid F)$ and also $P(E^{\complement}\mid F)=1-P(E\mid F)$

Note that $P(E^{\complement}\mid F)$ is the probability that you are looking for.

For a more elegant route see the comment of Henry.

Since there bird eats only five fruits, it's relatively easy to consider all possible cases. Let's write

$P_0 = P(\text{bird eats zero fruits with seeds}) = \frac{15}{20} \frac{14}{19} \frac{13}{18} \frac{12}{17} \frac{11}{16} = \frac{1001}{15504}$
$P_1 = P(\text{bird eats exactly one fruit with seeds}) =$ $5\choose 1$$\frac{5}{20} \frac{14}{19} \frac{13}{18}\frac{12}{17} \frac{11}{16} = \frac{5005}{15504}$
$P_2 = P(\text{bird eats exactly two fruits with seeds}) =$ $5\choose 2$$\frac{5}{20} \frac{4}{19} \frac{13}{18}\frac{12}{17} \frac{11}{16} = \frac{715}{3876}$
$P_3 = P(\text{bird eats exactly three fruits with seeds}) =$ $5\choose 3$$\frac{5}{20} \frac{4}{19} \frac{3}{18}\frac{12}{17} \frac{11}{16} = \frac{55}{1292}$
$P_4 = P(\text{bird eats exactly four fruits with seeds}) =$ $5\choose 4$$\frac{5}{20} \frac{4}{19} \frac{3}{18}\frac{2}{17} \frac{11}{16} = \frac{55}{15504}$
$P_5 = P(\text{bird eats exactly five fruits with seeds}) =$ $5\choose 5$$\frac{5}{20} \frac{4}{19} \frac{3}{18}\frac{2}{17} \frac{1}{16} = \frac{1}{15504}$

So for part a, you can consider the five cases separately.

If the bird ate zero fruits with seeds, there are $5$ out of $15$ fruits with seeds left. So you have to multiply $\frac{1001}{15504}\frac{5}{15}$
If the bird ate one fruit with seeds, there are $4$ out of $15$ fruits left with seeds. So you have to multiply $\frac{5005}{15504} \frac{4}{15}$

etc ...


Evaluate $\frac{\partial^2}{\partial t^2} \left[ \prod_{j=1}^k (1+t+\dots+t^{d_j -1}) \right]$ at $t=1$ I need to find a "nice" formula for the evaluation of
$\frac{\partial^2}{\partial t^2} \left[ \prod_{j=1}^k (1+t+\dots+t^{d_j -1}) \right]$ at t=1, where $d_j \in \mathbb{N}$.

I have already proved that
\begin{equation}
\frac{\partial}{\partial t} \left[ \prod_{j=1}^k (1+t+\dots+t^{d_j -1}) \right]_{| t=1} =\frac{1}{2} \cdot \prod_{i=1}^k d_i \cdot \left[ \sum_{i=1}^k d_i -k \right]
\end{equation}

and now I'm looking for a similar concise formula for the second derivative at t=1.

Let $$f_i(x)=\sum_{n=0}^{d_i-1}x^n=1+x+\cdots+x^{d_i-1}$$ and let $$\Psi=\prod_{j=1}^kf_j(t)$$
We have that $$\Phi=\frac\partial{\partial t}\Psi=\Psi\left(\sum^k_{j=1}\frac{f_j'(t)}{f_j(t)}\right)$$ Therefore $$\frac{\partial^2}{\partial t^2}\Psi=\frac\partial{\partial t}\Phi=\Psi'\left(\sum^k_{j=1}\frac{f_j'(t)}{f_j(t)}\right)+\Psi\left(\sum^k_{j=1}\frac{f_j'(t)}{f_j(t)}\right)=\Psi\left(\left(\sum^k_{j=1}\frac{f_j'(t)}{f_j(t)}\right)^2+\left(\sum^k_{j=1}\frac{f_j'(t)}{f_j(t)}\right)'\right)$$
Finally after simplification
$$\Psi\left(\left(\sum^k_{j=1}\frac{f_j'(t)}{f_j(t)}\right)^2+\sum^k_{j=1}\frac{f_j(t)f_j''(t)-f_j'(t)^2}{f_j(t)^2}\right)$$
The rest is just computations of the derivative at of $f_j$ at $t=1$ which should simplify nicely in terms of $d_j$.

Further Calculations
$$f_j(1)=d_j$$
$$f_j'(1)=\frac{d_j(d_j-1)}{2}$$
$$f_j''(1)=\frac{d_j(d_j-1)(d_j-2)}{3}$$
Thus you may finally plug and chug.

After the plug and chug I get $$\frac{\partial^2}{\partial t^2}\Psi=\left(\prod^k_{j=1}d_j\right)\left(\left(\sum^k_{j=1}\frac{d_j-1}2\right)^2-\sum^k_{j=1}\frac{(d_j-1)(d_j-5)}{2}\right)$$

I've cheated, I admit:

$$\frac{\partial ^2\left(\prod_j^k \sum _{d=0}^j t^d\right)}{\partial t^2}$$
at $t=1$ seems to equal

$$\frac{1}{144} (k-1) k (k+1) (9 k+22) (k+1)!$$

I can show you the cheat, but cannot provide a proof.


Maximum possible value for the probability of an event. Consider two events $A$ and $B$ with the following given conditions:

$$P(A)=0.4\quad\textrm{and}\quad P(B)=0.7$$

The maximum and minimum value of $P(A\cap B)$ are?

Note: $P(X)$ denotes probability of occurrence of event $X$ and the other symbols have their usual meanings as in set theory.

Since $A \cap B \subset A$ and  $A \cap B \subset B$,

$$P(A \cap B) \leq \min(P(A),P(B)) = 0.4. $$

On the other hand

$$P(A^c \cup B^c) \leq P(A^c)+(B^c) = 0.6 +0.3 = 0.9 . $$

So

$$P(A \cap B) = 1- P((A \cap B)^c) = 1-P(A^c \cup B^c) \geq 1-0.9 = 0.1. $$

Now you can construct examples to show that both of these bounds can be attained (that is, in general, they are the best possible bounds).


How to isolate $x$ in $a^x + b^x = c$? (For use in medical statistics) I've broken a somewhat complex calculation down to the following mathematical equation to be solved:
$a^x + b^x = c$
How do I find $x$ when $a, b$ and $c$ are given as parameters?
I.e., if $a=3$, $b=4$ and $c=25$, then the solution is $x=2$.
(The question has nothing to do with geometry - this is just the simplest example I can come up with.)
Background
For specifying a person's overweight/underweight, the term Z-score is used and the Z-score is calculated based on the person's BMI, sex and age. For each sex and age interval, the constants $\lambda$, $\mu$ and $\sigma$ are given.
I have the formula $Z = \frac{(\frac{B}{ \mu})^\lambda -1}{\lambda\sigma}$, where $B$ is the person's BMI, for calculating the Z-score. Also, I have a set of $B$-values for a specific age and sex, for $Z \in \{-2, -1, 0, 1, 2\}$. For $Z=n$, let's call the $B$ data set $B_n$.
My task is, based on the above knowledge, to calculate $\mu$, $\sigma$ and $\lambda$ for a specific sex and age, so I can calculate the Z-score for a specific BMI, based on sex and age.
It is obvious that $\mu = B_0$. Moreover, using the Z-formula on $B_1$, I get $\sigma = \frac{(\frac{B_1}{B_0})^\lambda -1}{\lambda}$.
Using the Z-formula on $B_1$ and $B_{-1}$, I get $(\frac{B_{-1}}{B_0})^\lambda + (\frac{B_{1}}{B_0})^\lambda = 2$.
So, pretty simple, I just need to find how to isolate $x$ in the equation $a^x + b^x = c$...

There is almost surely no closed form expression, but a qualitative approach is possible.
If $b=a$, there is clearly a solution. Let us assume $\dfrac{b}{a}>1$, WLOG.
Taking the natural logarithm of both sides of the given equation:
$$\operatorname{ln}(a^x (1+\left(\tfrac{b}{a}\right)^x)=\operatorname{ln}(c)$$
$$\operatorname{ln}(a^x)+\operatorname{ln}(1+\left(\tfrac{b}{a}\right)^x)=\operatorname{ln}(c)$$
$$-x\operatorname{ln}(a)+\operatorname{ln}(c)=\operatorname{ln}(1+\left(\tfrac{b}{a}\right)^x)$$
which represents the equation verified by the abscissas of the possible intersection point(s) of the straight line (L) (magenta on the figure) and curve (C) (red on the figure) with resp. equations:
$$\begin{cases}y&=&-x\operatorname{ln}(a)+\operatorname{ln}(c)\\
y&=&\operatorname{ln}(1+\left(\tfrac{b}{a}\right)^x)\end{cases}$$
Please note that the second curve has a slant asymptote with equation $y=x \ln(b/a)$.

Therefore, 3 cases can occur with 0, 1 or 2 roots,

according to the resp. values of $a,b,c$. One cannot have more than two roots because curve (C) can be shown to be convex, and a straight line cannot intersect a convex curve in more than two points.

Fig. 1: The given case $a=3;b=4;c=25$ with a unique root at $x=2$.

Fig. 1: The case $a=0.5;b=2;c=2.5$ with two roots $x=-1$ and $x=1$.

We can do this assuming b is a function of a such that $b=a^{-1}$:
$$a^x+b^x=c=$$
$$a^x+a^{-x}=c$$
$$\frac{a^x+a^{-x}}{2}=\frac c2=$$
$$\frac{e^{x*ln(a)}+e^{-x*ln(a)}}{2}=\frac c2=$$
$$cosh(x*ln(a))=\frac{c}{2}$$
Now we can use the inverse hyperbolic cosine to solve for x:
$$x*ln(a)=x*ln(a)=2πin\pm cosh^{-1}\frac c2=$$
$$x=\frac{2πin\pm cosh^{-1}\frac c2}{ln(a)}=$$
$$x=\frac{2πin\pm ln(c+\sqrt{c-2}\sqrt{c+2})-ln(2)}{ln(a)}=$$
$$x=\frac{2πin}{ln(a)}\pm log_a(c+\sqrt{c-2}\sqrt{c+2})-log_a(2)$$
To finish off, this idea is from one of the comments and assuming that $n\in \Bbb Z$.
Another idea in the comments was to do $b=a^n$ so that a general solution can be found. Let us reset the variables so that no other ones are defined but our starting equation:
$$a^x+b^x=c=a^x+(a^x)^n=c$$
With $t=a^x$, this happens:
$$t+t^n=c=$$
$$t^n+t-c=0$$
It can be solved that:
$$t=\sqrt[n]{c-t}=$$
$$t= \sqrt[n]{c-\sqrt[n]{c-\sqrt[n]{...}}}$$
However, this path causes alternating pattern for $n\in\Bbb Z$. This problem makes us need to modify our solution:
$$-t= \sqrt[n]{c-\sqrt[n]{1+\sqrt[n]{c+...}}}$$ for odd n to avoid imaginary numbers, even though they may or may not bring about other roots of the polynomial, or  $$-t=\sqrt[n]{c+\sqrt[n]{c+\sqrt[n]{c+...}}}$$ for even n.
Because we defined  $t=a^x$, this advances to:
$$x=log_a\sqrt[n]{c\pm\sqrt[n]{c+\sqrt[n]{c+...}}}+\frac{2iπm}{ln(a)}; a,b,ln(a)\ne 0,b=\sqrt[n]a,n\in \Bbb Z$$
Use the positive branch for n=2k and the negative branch for n=2k+1, $k\in \Bbb Z$.
Here is proof of my answer for c=1: https://www.desmos.com/calculator/l8gr3pnb0u.
I apologize for whatever I did wrong that got a “down vote”. It would be significantly harder to figure out for $n\in \Bbb U$ or the universal set.
Please correct me and give me feedback!


Enumerable and not enumerable sets I am i bit stuck on this problem. Is it possible that sets A and B are not enumerable, but set $$A \cdot B  =\{x \cdot y \mid x \in A, y \in B\}$$ is enumerable? Defenition of emumerable set so, set is enumerable if there is an algorithm, that may enumerate it. Or, set is enumerable if there is exist semi-characteristic function which equals 1 if x in A, or not define if x not in A.
Thanks in advance

Yes, this is certainly possible.  Note that for any $S\subset\mathbb{N}$, if we take $A=\{1\}\cup S$ and $B=\{1\}\cup S^{c}$, then $A\cdot B=\mathbb{N}$, which is enumerable.  So we just need a non-enumerable set $S$ whose complement is also non-enumerable.  For instance, let $S$ be the set of (encodings of) Turing machines that accept finitely many inputs.  Then $S^{c}$ is the set of (encodings of) Turing machines that accept infinitely many inputs, plus those numbers that do not encode a Turing machine.  Neither $S$ not $S^{c}$ is enumerable, because the corresponding decision problem (does a particular Turing machine accept finitely many inputs?) is undecidable.

If $A$ and $B$ are subsets of the real numbers and $.$ means the product, the answer is no. Indeed, since $B$ is not enumerable, it contains a non-zero number $y_0$. Then $A.B$ contains the set $A.y_0 := \{x.y_0 | x \in A\}$ which is in bijection with $A$ since $y_0$ is invertible.


What are the odds of spinning matching items in a slot machine? Lets say we have a slot machine with $5$ reels. Each reel has $5$ different items on it.

What are the odds of spinning $2, 3, 4$ and $5$ matching items?

As I understand the probability of rolling a particular item in each reel is $\frac15$. I just don't know how to calculate the combined probability of these events. Is it simply $\frac15\cdot\frac15$ for two matching items or are things more complicated?

Yes but you need to multiply $1/25$ with $\binom 5r, r=\{2,3,4,5\}$ as you need to consider permutations of an arrangements on 5 distinct reels and hence consider choosing any two reels to be possible in multiple ways, i.e. $\binom 5r$

It's a bit more complicated.

Let's first answer the question what is the probability that there are $5$ matching items

The number of all possible spins is $5^5$.
The number of good spins is $5$.
Therefore, the probability of a good spin is $\frac{5}{5^5}=5^{-4} = 0.16\%$

Now, what about four matching items?

The set of all possible spins is now bigger. Let's say that each reel can spin $1,2,3,4$ or $5$. Now, how can we get $4$ same items?

Well, we can have four ones in the first four reels, and one non-one in the fifth reel. There are $4$ such results ($11112,11113,11114,11115$). Then, there are four results for four twos, and four for four threes and so on. So, there are $5\cdot 4=20$ possibilities for having $4$ equal items in the first four reels.

You then also have $20$ possibilities for $4$ equal items on reels $1,2,3,5$, and so on, with a total of $5\cdot 20=100$ good possibilities.

Therefore, the probability of a good spin is $\frac{100}{5^5} = 3.2\%$

OK, three matching items. Well:

there are ${5\choose 2}=10$ possible triplets of reels on which you
For each such triplet, you have $5$ possibilities for which item will appear on all three. You can then freely choose the other $2$ reels to be anything but the chosen item. Therefore, for each triplet, you have $5\cdot 4\cdot 4$ good spins.

So, $10$ possible triplets and $80$ good rolls for each triplet yields $800$ good spins. With $5^5$ total spins, the probability is
$$\frac{800}{5^5} = 25.6\%$$

What about the probability of two matching items? Well, that depends:

Do you want the probability that there is only one matching pair?
Do you allow for two pairs?
Do you allow one pair and one triple?


Looking for a book that discusses differential topology/geometry from a heavy algebra/ category theory point of view So I am trying to learn differential geometry, stuff like manifolds, Lie groups, Stokes' theorem and so on. I have read about many books which discuss these topics, such as Lee's Smooth Manifolds and Tu's Introduction to Manifolds. But I have not seen any that is heavy in category theory or algebra. For example, Lee briefly introduces some basic category theory but it's only in a small section and he never mentions it again. I want a book that really uses category theory. The same with algebra. Most of these books try to minimize the algebra as much as possible. For example, Lee develops tensor products but I am pretty sure it is not in a very general/developed form. So is there a book which introduces differential geometry/topology from an algebraic / category theoretic point of view.

The book you're looking for is our friend @Wedhorn's Manifolds, Sheaves and Cohomology.

Wedhorn's background is algebraic geometry, a subject in which he has already written (with his colleague Görtz ) a quite popular book, and his background shows in the book I recommend.
In Chapter 4 manifolds are presented as suitable ringed spaces i.e. topological spaces endowed with a  sheaf of rings (sheaves having been explained in chapter 3) and then the author introduces tangent spaces, Lie groups, bundles, torsors and cohomology.
This approach has been advocated since at least 50 years ago but Wedhorn's is one of the very rare books that consistently adopts the ringed space approach to manifolds.
The prerequisites in topology, categories, homological algebra and differential calculus are presented in appendices, so that the book is quite self-contained.

Apart from its elegance and efficiency the ringed space approach to manifolds allows for a smoother (!) introduction to the more dificult theory of schemes (or analytic spaces) and is thus also an excellent investment for  ulterior study of more advanced material.

There is an extensive treatise with applications to many sciences (not just physics) titled ¨Applied Differential Geoometry, A modern Introduction¨ from Ivancevic.


Nyquist Stability Criterion The rational function $b(s) = (s+3)(s-4)^{-1}$ is the frequency response function (FRF) of a system $B$. Is $B$ stable?

I understand that a system is unstable if there are poles in the closed right half plane. Would we say this system is not stable because of the pole at s = 4? Or does the zero at s = -3 interfere with the notion of stability? If someone could explain what is correct, with a bit more conceptual explanation it would be appreciated.

The system is unstable if any bounded input gives unbounded outout.

Physically, an unstable system whose natural response grows without bound can cause damage to system or property or human life.
Poles in the left half plane yield either exponential decay or sinusoidal responses, thus lhp poles indicate stability.

The system is stable if it doesn't have any right half plane poles. We say that the system is unstable because for values of s close to 4, the system tends to infinity and such large values make systems unstable.
The zero at s=-3 has nothing to do with the notion of stability. You can read control theory - a complete course on this.
Being more specific - Read Nyquist Stability criterion and Root locus plots for better visualisation.

Plot the Root locus diagram for the system and then analyze the stability.


Proving that (4-2/1)(4-2/2)...(4-2/n) in an integer. We have to prove that $(4-2/1)(4-2/2)...(4-2/n)$ is an integer for $n\in\mathbb{N}$. Can we do this by induction?

We prove for $n = 1$, which is trivial as $(4-2/1) = 2$ which is clearly an integer.

next we assume that the statement is true for some integer $n$
i.e. that $(4-2/1)(4-2/2)...(4-2/n) = p$ where p is an integer, and we prove for $n+1$.

So we need to show that $(4-2/1)(4-2/2)...(4-2/n)(4-2/(n+1))$ is also an integer. Which is the same as saying $p\times(4-\frac{2}{n+1})$ is an integer.

We can rewrite $p$ as $\dfrac{2^n(1\times3\times5\times...\times(2n-1))}{n!}$ so essentially all we need to prove is that $n+1$ divides the numerator of $p$. So assume that $n$ is even, which implies that $n+1$ is odd and certainly less than $2n-1$ so therefore it will divide one of the odd numbers in the numerator of $p$. If $n$ is odd, then $n+1$ is even and will not have more factors of $2$ than $2^n$. So if the prime factorization of $n+1$ contains $k$ factors of 2, $k<n$ or $k=n$,  implying that $2^k$ divides $2^n$. Also the prime factorisation of $n+1$ will not have have an odd factor that is greater than $2n-1$ as it already has a factor of 2. Therefore $n+1$ divides the numerator of p.

Can we do this? It seems like it is not entirely correct! I may have made a big mistake in my reasoning.

Any help is appreciated.

We can write $p$ on this form

$$p=\frac{2^n(1\times 3\times 5\times \cdots\times(2n-1))}{n!}=\frac{(2n)!}{n!n!}={2n\choose n}\leftarrow \text{integer}$$

The product is
$$ 2 {{2n-1}\choose {n-1}} $$
which is an integer for integer $n$.

This is also equal to
$$ 2n \choose n$$
as pointed out by another member earlier.


Solve $2\ddot{y}y - 3(\dot{y})^2 + 8x^2 = 0$ Solve differential equation

$$2\ddot{y}y - 3(\dot{y})^2  + 8x^2 = 0$$

I know that we have to use some smart substitution here, so that the equation becomes linear.
The only thing I came up with is a smart guessed particular solution: $y = x^2$. If we plug this function in, we get:
$$2\cdot2\cdot x^2 - 3(2x)^2 +8x^2 = 4x^2 - 12x^2 + 8x^2= 0$$

I made a mistake. The coefficients where different in the exam:
$$
\begin{cases}
    3\ddot{y}y + 3(\dot{y})^2 - 2x^2 = 0, \\
    y(0) = 1, \\
    \dot{y}(0) = 0.
\end{cases}
$$

Does it make the solution easier?

Let $z=e^{ky}$. Then $z'=ky'e^{ky}$ and $$z''=ky''e^{ky}+k^2(y')^2 e^{ky}\text.$$

If we put $k=-\frac{3}{2}$ then $z''=-\frac{3}{2}y''e^{ky}+\frac{3}{2}^2(y')^2 e^{ky}$, or $\frac{2}{3}z''=-y''e^{ky}+\frac{3}{2}(y')^2 e^{ky}$, or $$-\frac{4}{3}z''=2y''e^{ky}-3(y')^2 e^{ky}\text{, or even}$$ $$-\frac{4z''}{3z}=2y''-3(y')^2\text.$$

Your original equation is now $$\frac{4z''}{3z}=8x^2\text,$$ which I hope will get you closer.

Indeed...

Given $z''=6x^2z$, expanding $z$ as a power series in $x$ yields $a_n=\frac{a_{n-2}}{(n+1)(n+2)}$ times a constant, which makes one think that $z$ must be something to do with $e^{x^2}$.

Put $$z=e^{ax^2+bx+c}$$ and differentiate twice, and you get something like $$z''=2az+4a^2x^2z+2abxz+b^2z\text,$$ which is enticing but has an embarrassing $xz$ term. Fortunately that is the only term in which $b$ isn't squared. So if you take $$z=e^{ax^2+bx+c}+e^{ax^2-bx+c}$$ and differentiate it twice, you will get rid of the unwanted $xz$ term and you will have constraints on $a$ and $b$ which make $z$ satisfy the differential equation.

General note: All this has been done on the back of one and a half envelopes, so do check it.


Computational complexity of sizes and number of orbits of a group acting on a set Given a group action of a group $G$ on a set $X$, is there any way to relate the number of orbits, i.e. $|X/G|=|\{\{g\cdot x:g\in G\}:x\in X\}|$, to the sizes of the orbits, i.e. $|\{g\cdot x:g\in G\}|$ for $x\in X$?

I'm in particular interested in the relation between the computational complexity of these two questions. To be more precise, for a fixed group action, if I know that it's $\#P$-Complete to find the size of an orbit (given an element in $X$), can I then make any statement of the computational complexity of counting the number of orbits?

Note: I've now asked the same question on mathoverflow, since after two weeks there was only an related answer here.

Consider the following question: Given two elements of X, are they in the same orbit under the action of G? This problem is Graph Isomorphism complete.


What is the sum-capacity for a non-symmetric interference channel for information theorists? This question is dedicated for people who are experts in information theory.

An interesting result for a two user interference channel in information theory, is the sum-capacity to within one bit. It however was derived for the symmteric case.

I was wondering if anyone knows what the sum-capacity bound for the two user channel is, say for example user #1 wants to communicate with its own receiver at rate $R_1$ and user two communicates with receiver 2 at a rates $R_2$.

And let us say I have the following channel

$$Y_1= h_{11}X_1 + h_{12}X_2+Z_1$$
$$Y_2= h_{22}X_1 + h_{21}X_2+Z_2$$
where $h_{ij}$ for $i,j\in[1,2]^2$ are channel gains from receiver $j$ to transmitter $i$ and $Z_{i}$ for $i\in[1,2]$ are additive white gaussian noise mean zero variance one.

Can anyone help me write down what the sum-capacity for this general 'non' symmetric gaussian interference channel is?

See ''Sum Capacity of the Vector Gaussian Broadcast Channel and Uplink–Downlink Duality'' by  Viswanath and David, in IEEE Transactions on Information Theory v49, No. 8, Aug 2003.

These people found that for a MIMO broadcast with $N$ antennas and $K$ users, with additive iid CCS Gaussian noise $\eta$, with a flat-fading $N\times K$ channel (gain) matrix $H$,

(All this makes the reception $\vec{y}=H^{\dagger}\vec{x}+\vec{\eta}$)

the sum capacity is:

$$\underset{D\in \mathcal{A_1}}{\sup} \log \det ( I + HDH^{\dagger})$$

with $\mathcal{A_1}$ being the set of nonnegative $N \times N$ diagonal matrices, whose trace is less than or equal to your power constraint at the broadcaster.

The 'interference' in your case is cooked into $H$.

For, two user non-symmetric Gaussian interference channel capacity (not just sum-capacity) is know to withing 1 bit. This result was show by Etkin, Tse, and Wang.

If you have Network Information Theory book by El Gamal look at chapter 6 they have outer bound that is tight within in 1 bit.


Find $\,n\bmod 42\,$ given $\,7\mid 3^n+4n+1$ I have been working on a problem relating to remainders: "The number $3^n+4n+1$ is divisible by $7$. Find the remainder of n when it is divided by $42$." I have been unable to find a strategic way to solve this problem and so far, I have only found the remainders, $4$, $8$, $13$, $21$, $23$, and $24$. Can anyone confirm whether this is correct?

Key Idea $ $ We eliminate nasty exponential dependence on $\,n\,$ in $\,3^{\large n}\,$ by noting - by Fermat - that it's periodic $\!\bmod 7\!:\ 3^{\large 6}\equiv 1\,\Rightarrow\, 3^{\large n}\! =3^{\Large i\large +6k}\equiv \color{90f}{3^{\Large i}}\,$ by mod exponent reduction. Thus we get a simpler linear congruence by replacing the function $\ g(n) = \color{#90f}{3^{\large n}}\,$ by constants $\,\color{90f}{3^{\Large i}},\ i = 0,1,\ldots,5$.

Doing so $\ n = \color{}i\!+\!6\,\color{#c00}k\,\Rightarrow$ $\! \bmod\color{#c00} 7\!:\,\ \overbrace{0\equiv f(i\!+\!6k)\equiv \color{90f}{3^{\Large\color{}i}}\!+\! 4(i\!+\!6k)\!+\!1}^{\Large{\ \ \ \ 7\ \, \mid\, \  f(n)\ \ =\ \  \color{#90f}{3^{\LARGE n}}\ \ +\ \  4\,n\ \ +\ \ 1}} \equiv f(i)\!+\!3k\!$ $\iff\!$
 $ 3k\equiv -f(i)\!\equiv\!6f(i)\!\!\iff\!\!$ $\color{#c00}{k \equiv 2f(i)}^{\phantom{|^I}}\!\!\!,\, $ so $\, n^{\phantom{|^I}}\!\!\! = i\!+\!6\color{#c00}k = i\!+\! 6(\color{#c00}{2f(i)\!+\!7}j)\equiv \bbox[5px,border:1px solid #0a0]{i\!+\!12f(i)}^{\phantom |}\!\pmod{\!42},\, $ e.g.
 $\,i\!=\!0^{\phantom{|^{|^|}}}\!\!\!\Rightarrow n\equiv 0\!+\!12f(0)\equiv\bbox[5px,border:1px solid #0a0]{24},\,$
$\ i\!=\!1\Rightarrow n^{\phantom{|^I}}\!\!\!\!\equiv 1\!+\!12f(1)\equiv\bbox[5px,border:1px solid #0a0]{13},\,$ $\, i\!=\! 2,3,4,5\Rightarrow \bbox[5px,border:1px solid #0a0]{n\equiv \ldots\phantom{i}\!\!\!}$

Remark $ $ The same idea works generally. If $\,g(n)\,$ has period $\,\ell,\,$  so $\ g(i+ \ell k) = g(i),\,$ then we can solve $\,f(n,g(n)) \equiv a\pmod{\!m}\,$ by using this periodicity to eliminate the function $\,g(n)\,$ as above.

Doing so $\ n = i+\ell\, k\,\Rightarrow\bmod m\!:\ a\equiv f(n,g(n))\equiv f(n,g(i))\iff \ldots\, n\equiv r_i$

Combining $\,\ \begin{align}  &n\equiv\, i\pmod{\!\ell}\\ &n\equiv r_i\!\!\!\pmod{\!m}\\\end{align}\!\! \overset{\large \rm CRT\!\!}\iff n\equiv s_i \pmod{\!\ell m},\ $ with $\,s_i\,$ computable by CRT,

assuming $\,\ell,\,m\,$ are coprime. In our solution above we did the CRT step in parallel by substituting $\,n = i + \ell k\,$ not only into $\,g(n)\,$ but everywhere for $\,n\,$ in $\,f,\,$ then solve that for $\,k \equiv k_i\pmod{\!m},\,$ so $\, n = i +\ell (k_i + mj)\equiv i + \ell k_i \pmod{\!\ell m}$.


Finding a linear map $T:\mathbb R^3 \to \mathbb R^3$ that satisfy conditions about the kernel and image Finding a linear map $T:\mathbb R^3 \to \mathbb R^3$ that satisfy conditions about the kernel and image

  1)find linear map $T:\mathbb R^3 \to \mathbb R^3$ so that $\operatorname{Im} (T)=\{(x,y,z)|2x-y+z=0\}$

  2)find linear map $T:\mathbb R^3 \to \mathbb R^3$ so that $\operatorname{Ker} (T)=\{(x,y,z)|2x-y+z=0\}$

I choose for (1) : $T(x,y,z)=(x+y+z,3x+3x+3z,x+y+z)$

and for (2):$T(x,y,z)=(2x-y+z,4x-2y-2z,6x-3y+3z)$

so Does this answer correct and Is there any formal way to find $T$?

thanks

(1) Your answer is wrong. The image of your map $T$ is$$\left\{(x,y,z)\in\mathbb{R}^3\,\middle|\,2x-y+z=0\wedge x=z\right\}.$$You can solve the problem by finding a basis $\{e,f\}$ of $\left\{(x,y,z)\in\mathbb{R}^3\,\middle|\,2x-y+z=0\right\}$ (for instance $e=(1,2,0)$ and $f=(0,1,1)$) and define $T$ such that $T(e_1)=T(e_2)=e$ and $T(e_3)=f$, where $\{e_1,e_2,e_3\}$ is the standard basis of $\mathbb{R}^3$.

(2) You were almost right. The second $-2$ shouldn't have a minus sign in it. It should have been$$T(x,y,z)=(2x-y+z,4x-2y+2z,6x-3y+3z).$$


Characteristic value or eigenvalues and determinant I am having semester in linear algebra. And have recently got acquainted to eigenvalues.

What is the relation between eigenvalues and determinant? Going through answers of some questions I found suggestion of using eigenvalues to determine determinant of matrix. I just know $\det(eI-A)=0$ where e is eigenvalue of A. How to use this fact to calculate determinant of A?

You know that the eigenvalues are the roots (zeros) of the characteristic polynomial, i.e. of the polynomial $(x) = \det(xI - A)$. Watch what happens of you evaluate that polynomial at $0$, i.e. compute $p(0)$. You get $p(0) = \det(0I-A) = \det(-A)$. Now, $\det(-A) = (-1)^n\det A$ (assuming that A is a $n\times n$ matrix).

So how do you find $p(0)$, if you only have the eigenvalues of $A$. Well, since the eigenvalues of $A$ are the zeros of $p(x)$, you can write $p(x)$ as a product $$
  p(x) = \alpha(x - \lambda_1)(x - \lambda_2)\ldots(x - \lambda_n)
$$
where $\lambda_1,\ldots,\lambda_n$ are the zeros of $p(x)$, i.e. the eigenvalues of $A$. Note that some particular $\lambda$ will appear more than once in this list, if $p(x)$ has a zero with an order higher than 1 at $\lambda$. For characteristic polynomials, you also know that $\alpha = 1$, because the coefficient of $x^n$ in the characteristic polynomial of a $n\times n$ matrix is always $1$. Now watch what happens if you evaluate $p$ at $0$. You get $$
  p(0) = (0 - \lambda_1)(0 - \lambda_2)\ldots(0 - \lambda_n) = (-1)^n\lambda_1\ldots\lambda_n \text{.}
$$

Combining these results, you have $$
  p(0) = (-1)^n\det A = (-1)^n\lambda_1\ldots\lambda_n
  \implies
  \det A = \lambda_1\ldots\lambda_n \text{.}
$$

Write the characteristic polynomial $\chi_A$ as $\chi_A(X)=\prod_{\lambda\text{ eigenvalue}}(X-\lambda)$; what is $\chi_A(0)$? You can express it as the constant coefficient (easy to compute, function of the eigenvalues only), or with your expression above (taking your "$e$" to be 0))

(in the product, the $\lambda$'s are not necessarily distinct)


prove ABCD is also a square What we know is that:
1. ABCD is a quadrilateral.
2. The red area is a square.
3. AH=BE=CF=DG

The question is prove that ABCD is also a square.
I have realised that the four triangles here AHG, DGF, EFC and HBE have the same length hypotenuse and also AH = DG = CF = BE, so if I can prove ∠ A, B, C, D are 90°, then four triangles are congruent. Then I will know that four sides, AB,BC,CD,DA are the same length then I can prove it.
The problem is that I dont know how to prove angle A,B,C,D are 90 degree.
Thanks!

If one among the angles at $A$, $B$, $C$, $D$ is a right angle, then it is easy to prove they are all right angles and $ABCD$ is a square. Suppose then none of them is a right angle: at least one of them ($\angle GAH$, for instance) must then be obtuse. I'll show that this leads to a contradiction.

Let $N$ be the foot of the perpendicular from $G$ to line $AH$. As $\angle GAH>90°$ then $AH<NH$.

Let $M$ be the foot of the perpendicular from $E$ to line $BH$: we have then $EM\le EB$. But triangles $EMH$ and $HNG$ are congruent (because they have $\angle NGH=\angle MHE=\pi/2-\alpha$, $\angle NHG=\angle MEH=\alpha$ and $GH=HE$), thus:
$$
AH < NH = EM \le EB
$$
which is in contradiction with the hypothesis $AH=EB$.

Hint: extend side $AD$ and $EF$. Let's say they intersect at point $X$. Then you can see that $\angle{XFG}=90°$ and $\angle{DXF}\cong\angle{AGH}$. Can you take it from here?


Probabilities - Traffic Light problem Exercise :

Let it be that on a road trip, the probability of a traffic light having the same color as the previous one, is $p$. If the first traffic light is red with probability $a$ and green with $1-a$, calculate the probability of the third traffic light being green.

Attempt :

Let $R$ be the event of the traffic light being red and $R^c$  the event of the traffic light being green.

Then, for the first traffic light, it is :

$$P(R) = a$$

$$P(R^c) = 1-a$$

Also, let $S$ be the event of the next traffic light having the same color as the previous one. So :

$$P(S) = p$$

Now, to me it seems like it could be something involving Bayes Theorem, especially because we have $3$ traffic lights to check, which goes perfect with the denominator of Bayes, but I really cannot see how to continue on solving this one, basically I do not know how to start and choose between each case (green/red). I would really appreciate a thorough explanation/solution.

This actually has nothing to do with Bayes' theorem.  Let the probability that the $n$th traffic light is red be $a_n$.  Then we have $a_1 = a$.  To obtain a recursive relation for $a_n$, we have two cases: either the previous light is red (probability $a_{n-1}$, and then probability $p$ that the $n$th light is also red), or the previous light is green (probability $1-a_{n-1}$, and then probability $1-p$ that the $n$th light changes to red).  This yields $$a_n = pa_{n-1} + (1-p)(1-a_{n-1}).$$

We now have a recursive relation which we can use to solve for $a_3$ by working up from $a_1$.

Note: your problem is an example of what's called a Markov Chain.  You can read more about it here.


Distance point on ellipse to centre I'm trying to calculate the distance of a certain point of an ellipse to the centre of that ellipse:

The blue things are known: The lengths of the horizontal major radius and vertical minor radius and the angle of the red line and the x-axis. The red distance is the desired result. It is not given where on the ellipse the point is. It can be anywhere on the ellipse. Is this problem possible? If so, in which can this be solved? Thanks in advance!

After reading Kaj Hansen's comment and trying a bit this is what I did, it still won't work though.

In a triangle, $tan(\theta)=\frac{\text{opposite side}}{\text{adjecent side}}$. The slope of a line is $\frac{\Delta y}{\Delta x}$. Therefor the slope of the red line is $\tan(\theta)$; the formula of the line is $y=\tan(\theta)\cdot x$.

The formula of the ellipse is $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$. When I put the two formulas together I get
$$\frac{x^{2}}{a^{2}}+\frac{(\tan(\theta)\cdot x)^{2}}{b^{2}}=1$$

After a bit of rearranging:

$$x=\pm \sqrt{\frac{a^{2}\cdot b^{2}}{a^{2}\cdot (\tan(\theta))^{2}+b^{2}}}$$

$$$$

$$y=\tan(\theta)\cdot x$$

$$y=\pm \tan(\theta) \cdot \sqrt{\frac{a^{2}\cdot b^{2}}{a^{2}\cdot (\tan(\theta))^{2}+b^{2}}} $$

Now with the help of Pythagoras' theorem $c=\sqrt{a^{2}+b^{2}}$ the red line should be

$$\sqrt{\left ( \sqrt{\frac{a^{2}\cdot b^{2}}{a^{2}\cdot (\tan(\theta))^{2}+b^{2}}}\right ) ^{2}+\left ( \tan(\theta) \cdot \sqrt{\frac{a^{2}\cdot b^{2}}{a^{2}\cdot (\tan(\theta))^{2}+b^{2}}} \right )^{2}}$$

which can be simplified:

$$\sqrt{\frac{a^{2}\cdot b^{2} \cdot (\tan(\theta))^{2}+a^{2}\cdot b^{2}}{a^{2}\cdot (\tan(\theta))^{2}+b^{2}}}$$

This, however, does not give the right answer. Let's try something:

$a=2$; $b=1$; $\theta=\frac{1}{2}\cdot \pi$ (The point is the point where the ellipse intersects with the minor radius)

$$\sqrt{\frac{2^{2}\cdot 1^{2} \cdot (\tan(\frac{1}{2}\cdot \pi))^{2}+2^{2}\cdot 1^{2}}{2^{2}\cdot (\tan(\frac{1}{2}\cdot \pi))^{2}+1^{2}}}$$

$$\sqrt{\frac{4 \cdot (\tan(\frac{1}{2}\cdot \pi))^{2}+4}{4\cdot (\tan(\frac{1}{2} \pi))^{2}+1}}$$

But wait, $\tan(\frac{1}{2}\cdot \pi)$ is undefined. The formula cannot be filled in completely, which is a requirement. I need a formula that can be filled in for every value of $\theta$ on the domain $[0,\frac{1}{2}\cdot \pi]$

If $a$ is the length of the horizontal axis and $b$ the vertical axis, then the equation for the ellipse is $(x/a)^2 + (y/b)^2 = 1$.

From here, we can also get an equation for the red line since its slope is $\tan(\theta)$ and $y$-intercept is $0$.

To find the point where the line intersects the ellipse, simply take the equation of the line, $y = mx$, and plug it into the equation for the ellipse, $(x/a)^2 + (y/b)^2 = 1$, and solve for $x$ (note that this will give two possible values).  Then, you can plug the appropriate $x$-value back into the ellipse formula to get the corresponding $y$-coordinate.

Now that you have the point of intersection, simply use the Pythagorean theorem to obtain the length of the red segment.

Convert the equation of ellipse from rectangular to polar coordinates.

$ (x/a)^2 + (y/b)^2 = 1 $ , $ x = r cos(\theta), y = r sin(\theta)$.

$ (cos(\theta)/a )^2+ (sin(\theta)/b)^2 =1/r^2 $.

$(a,b,\theta )$ are given. Find out 1/$r^2$ and $r$.


Parabola describing projectile motion. I am trying to create a function that will generate a parabola that describes projectile motion.

Here are my inputs:

The starting x-y coordinate of the throw
The initial x-y velocity vector.

I need an output for an equation of a parabola that fits this motion.

By Newton's second law $F = ma = m(0,-g)$, which es equivalent to the set of equations:
\begin{align}
\ddot x(t) &= 0\\
\ddot y(t) &= -g.
\end{align}
After integrating these, we get that:
\begin{align}
x(t) &= x_0+v_{x}t\\
y(t) &= y_0+v_yt-\frac{gt^2}{2},
\end{align}
where $(x_0,y_0)$ is the initial position of the object and $(v_x,v_y)$ its initial velocity. These equations define a parabola.

If you want to get an equation of the form $y = f(x)$, use that $t= \frac{x-x_0}{v_x}$, then:
$$
y = y_0+v_y\frac{x-x_0}{v_x} - \frac{g(x-x_0)^2}{2v_x^2},
$$
where $x\in[x_0,x_1]$ and $x_1$ is where the projectile reaches the ground again.

EDIT1:

Write down vertical acceleration equilibrium and integrate two times,remembering to include vertical component of projectile velocity $V$ shot at $ \alpha $ to x-axis as follows.

The accelerations in x- and y-(vertical) directions are:

\begin{align}
\ddot x(t) &= 0\\
\ddot y(t) &= -g
\end{align}
First integration for velocity with components of initial $V$ components as  :
\begin{align}
\dot x(t) &= V \cos \alpha \\
\dot y(t) &= V \sin \alpha -g  \, t \\
\end{align}
Second integration is for position starting at origin with boundary condition ( x=0, y=0 ):
\begin{align}
\dot x(t) &= V \cos \alpha \,t \\
\dot y(t) &= V \sin \alpha \,t -g t^2/2 \,  \\
\end{align}

Now eliminate $t$ to find trajectory as a parabola reaching a maximum point before hitting ground.

$$ y = x \tan \alpha - g x^2 / 2 /(V \cos \alpha)^2 $$

It is possible to find maximum range and height using using above relations. Maximum range occurs at $ \alpha = 45^0 $ and of course maximum height $ V^2 / 2g $ when shot vertically up.

It is an interesting fact I found that as $\alpha $ is varied, the envelope of all parabolas is another parabola.


Solving $2^x - 3^x + 6^x =0$. Are there any known methods to solve
$$2^x - 3^x + 6^x = 0,$$ where $x$ is either in closed form, perhaps in terms of special functions, or to give inequalities on the answers, where $x\in\mathbb{C}$ ?

My thoughts: $x\in\mathbb{C}$ would presumably lead to many solutions. My first attempt was the following:

$$e^{x\log 2}-e^{x\log 3}+e^{x\log 6}=0.$$ Letting $x=a+ib$ then gives the two equations

$$2^a\cos(b\log 2)-3^a\cos(b\log 3)+6^a\cos(b\log 6)=0$$
and
$$2^a\sin(b\log 2)-3^a\sin(b\log 3)+6^a\sin(b\log 6)=0,$$
corresponding to the real and imaginary parts respectively. From then on it gets a bit messy. My feeling is there are no known methods except numerical ones?

Equations like this have infinitely many complex solutions.

Let $a = \log 2-\log 3$ and $b = \log 6 - \log 3 = \log 2$. Then let $f(z) = \exp(az) + \exp(bz)$. We want to solve $f(z)=1$

Since the two coefficients are real and have opposite signs, for $\Re(z) \to \infty$, $f(z) \sim \exp(bz)$ and for $\Re(z) \to - \infty$, $f(z) \sim \exp(az)$.

In fact, if you note $r_+$ the positive real solution to $\exp(bx) = \exp(ax) +1$ and $r_-$ the negative real solution to $\exp(ax) = \exp(bx) + 1$ (in fact, it is $-1$), all the solutions to $f(z) = 1$ must be in the region $r_- \le \Re(z) \le r_+$

Then the number of solutions with $|\Im(z)| < B$ is the number of turns around $1$ made by $f(z)$ as you go along the rectangle whose vertices are $r_\pm \pm iB$
(more precisely, it's $\frac 1{2i\pi} \int f'/(f-1) dz$)

On the right side of the rectangle, $f$ is homotopic to $\exp(bz)$ without going through the disk, so the integral is $2ibB + O(1)$

On the left side, same thing except we go downwards, so we get $-2iaB + O(1)$.

Summing up everything, assuming the top and bottom integrals are also a $O(1)$, the number of $1$s with $|\Im(z)| < B$ is $\frac{(b-a)B}\pi + O(1) = \frac {B\log 3} \pi + O(1)$ .

For a more detailed analysis,

let $H(x) = a$ for $x<0$ and $H(x) = b$ for $x>0$, and define

$I(y_1,y_2) = \int_{\Bbb R} \left(\frac {a\exp(a(z-iy_1)) + b\exp(b(z-iy_2))}  {\exp(a(z-iy_1)) + \exp(b(z-iy_2)) - 1} - H(z)\right) dz$.

Since the integrand converges to $0$ exponentially at $\pm \infty$, this integral is well-defined whenever the integrand doesn't have a pole,
which means whenever we don't find a solution to the shifted equation on the real line. Away form those solutions, $I$ is even holomorphic asa function of two complex variables.

Moreover, if $dt > 0$ is real, then a change of variables gives
 $I(y_1+idt,y_2+idt) = \int_{\Bbb R} \left(\frac {a\exp(a(z-iy_1+dt)) + b\exp(b(z-iy_2+dt))}  {\exp(a(z-iy_1+dt)) + \exp(b(z-iy_2+dt)) - 1} - H(z) \right) dz
 \\ = \int_{\Bbb R} \left(\frac {a\exp(a(z-iy_1)) + b\exp(b(z-iy_2))} {\exp(a(z-iy_1)) + \exp(b(z-iy_2)) - 1} - H(z-dt)\right) dz
 \\ = I(y_1,y_2) + \int_{\Bbb R} (H(z) - H(z-dt)) dz
 \\ = I(y_1,y_2) + \int_0^{dt} (b-a)dz
 \\ = I(y_1,y_2) + (b-a)dt$

Since this is true for all positive real $dt$ and $I$ is holomorphic,
it is also valid for any complex $dt$ (as long as we don't go over a solution),
replacing $dt$ with $-idt$ gives us $I(y_1+dt,y_2+dt) = I(y_1,y_2) + i(a-b)dt$.

Moreover, as a function of two real variables, $I$ is $2\pi/a$-periodic in the first and $2\pi/b$-periodic in the second,
so we only need a nice picture of $I$ on a fundamental rectangle.
And to do this we need the picture of where the integral jumps (by $2i\pi$) because of a solution to the shifted equation.

Let $x \in \Bbb R$. A solution at $z=x$ happens when $1 = \exp(a(x-iy_1)) + \exp(b(x-iy_2))$.
With the triangle inequality and the law of cosines, this is only possible when $x \in [r_- ; r_+]$
with $2\exp(ax)\cos(ay_1) = 1 + \exp(2ax) - \exp(2bx), 2\exp(ax)\cos(by_2) = 1 + \exp(2bx) - \exp(2ax), \sin(ay_1)\sin(by_2) \le 0$.

Plotting those points give this graphics :

Finding a solution at $x+iy$ to the original problem is equivalent to reaching a discontinuity (so reaching one of the black curves) at $I(y,y)$. For example this shows that there are two solutions with imaginary part in $[0; 2\pi/b]$ because the diagonal crossed two curves so far.

Since $a/b$ is irrationnal, when we follow that diagonal using $I(y_1+dt,y_2+dy) = I(y_1,y_2) + i(a-b)dt$ and adding back $2i\pi$ whenever you cross the curve, the trajectory is dense in the corresponding torus, and therefore by continuity, $I$ is determined by this up to an additive constant.

In fact, now we can check that $I(y_1,y_2) = I(0,0) + iay_1 - iby_2 \mod (2i\pi)$. So $I$ was secretly a piecewise affine function ! (though as you input complex arguments, the black curve will change shape and you might have more complicated things happening, for example it might cross itself ... )

We also see that if we "approximate" the curves with straight lines, then the number of solutions we find with imaginary part between $0$ and $B$ would be exactly $\lfloor (b-a)B/\pi + k \rfloor$ for some constant $k$. For the actual equation, this is still $(b-a)B/\pi + O(1)$ and you can find the best lower and upper bound for the $O(1)$ by measuring precisely how far the curve gets away from the approximation line, and how far the origin is from the first line.

To find the distribution of the real part of the solutions, (so the expected number of roots with real part in some given interval in $[r_-,r_+]$), thanks to Weyl's equidistribution theorem you "only" need to highlight the points of the curves corresponding to that interval and see how much "shadow" they generate when you project them along the diagonal compared to the rest of the curve. This is a bunch of ugly calculus so I'll skip it.