Probability result in well-known paper appears to use std deviation correction of sqrt(n/(n-2.5)) The below problem is cited in the 1972 paper "Subjective Probability: A Judgment of Representativeness" by Daniel Kahneman and Amos Tversky.  The paper provides an answer to the problem for the purpose of comparing that answer with the impressions of test subjects to whom the problem was posed.

The average heights of adult males and females in the US are, respectively, 5 ft 10 in. and 5 ft 4 in. Both distributions are approximately normal with a standard deviation of about 2.5 in.
An investigator has selected one population by chance and has drawn from it a random sample.
What do you think are the odds that he has selected the male population if
(i) the sample consists of a single person whose height is 5 ft 10 in.?
(ii) the sample consists of 6 persons whose average height is 5 ft 8 in.?
........ The correct odds are 16 in case (i) and 29 in case (ii).

Bayes Theorem and the Gaussian PDF give odds of 17.8 and 317 for cases (i) and (ii), calculations below.  Since the problem provides a population standard deviation, using a sample standard deviation adjustor does not seem in order.  However, in case(ii) I checked a few adjustors $\sigma_{adj} = \sigma \cdot \sqrt{\frac{n}{n-k}}$ for $k = 1$ (Bessel's correction) and $k=1.5$ (alternative correction for normal distribution).  Neither returned the authors' given answer, although each moved the calculated odds in that direction.

The provided answer for case(ii) IS obtained using an adjustor based on k=2.5!

The question: what method could the authors be using that would provide their stated answer?  Why would k=2.5 be used?

Solution using Bayes Theorem and the Gaussian PDF

Case (i)

Using Bayes Theorem $$P(A|B)=\frac{P(B|A)\cdot P(A)}{P(B)}$$
$$$$

$$P(male | 5'10)=\frac{P(5'10 | male)\cdot P(male)}{P(5'10)}$$
$$$$

$$P(female | 5'10)=\frac{P(5'10 | female)\cdot P(female)}{P(5'10)}$$
$$$$
Assuming $P(male)=P(female)$ which is not explicitly stated in the problem ...
$$$$
$$odds=\frac{P(male | 5'10)}{P(female | 5'10)}=\frac{P(5'10 | male)}{P(5'10 | female)}$$
$$$$

Using the Gaussian PDF $$f(x)={\frac 1{\sqrt{2\pi}\sigma}e^\frac {-(x-\mu)^2}{2\sigma^2}}$$
yields ...
$$\frac{P(5'10 | male)}{P(5'10 | female)}=\frac{{\frac 1{\sqrt{2\pi}\cdot 2.5}e^\frac {-(70-70)^2}{2\cdot 2.5^2}}}{{\frac 1{\sqrt{2\pi}\cdot 2.5}e^\frac {-(70-64)^2}{2\cdot 2.5^2}}}=\frac{{e^0}}{{e^{-2.88}}}=17.8$$
$$$$
Case (ii)

Applying the same methodology to case (ii) with an adjusted $\sigma=\frac{2.5}{\sqrt6}=1.0206$ gives $$odds=\frac{P(5'8\,avg | male)}{P(5'8\,avg | female)}=\frac{{\frac 1{\sqrt{2\pi}\cdot 1.0206}e^\frac {-(68-70)^2}{2\cdot 1.0206^2}}}{{\frac 1{\sqrt{2\pi}\cdot 1.0206}e^\frac {-(68-64)^2}{2\cdot 1.0206^2}}}=\frac{{e^{-1.92}}}{{e^{-7.68}}}=317.4$$

$$$$
So there is a small difference (17.8 vs. 16) from the given answer in case (i) and a very large difference (317 vs. 29)in case (ii).

What method could be arriving at the stated answers of 16 and 29?

Thanks for posting this. I've been bugged by my inability to come up with the answer to case ii. The whole population SD switch to sample SD seems so odd (along with the -2.5 correction). Also odd to me is using the height of normal dist (PDF) instead of the area under a range of scores (CDF) as the basis for the probability estimates given that the probability that a person is exactly 5'10" is, strictly speaking, zero; i.e., P(5'10"|male) = 0. Same for female of course.

The answers aren't any better when using CDF values either. So it seems clear that T&K used a PDF in their calculation.


Direct image of a set for a function definded on $\mathbb{R}^2$ to $\mathbb{R}$ defined by $f((x,y))=x-9$

i want to find $f(A)$ where $$A=\{(x,y)\in\mathbb{R}^2, (x-1)^2+(y+1)^2<1\}$$

How to do ?

Can i say that $$f(A)=]1-\sqrt{1-(y+1)^2}, 1+\sqrt{1-(y+1)^2}[$$ with $-2<y<0$


What is "sampling from a distribution"? Exercise 4.11.3 of Grimmett and Stirzaker's Probability and Random Processes reads "Use the rejection method to sample from the gamma density $\Gamma(\lambda,t)$ where $t (\geq 1)$ may not be assumed integral."

What exactly is this exercise asking for? More generally, what is meant by "sampling from a distribution" (or "sampling from a density")? Section 4.11 of the given textbook refers to "sampling" over and over, but doesn't define it.

It seems, based on the examples, that it means to simply find a random variable with the given distribution. If that's correct, why is it called "sampling"?

Am I supposed to use a computer for this exercise?

In view of the context, no computer is involved. Rather, one asks for a method to generate a random variable $X$ with a given distribution (in your case, Gamma) from a collection, possibly infinite, of independent random variables, which are either uniformly distributed over $(0,1)$, or some simple transformations of these.

Assume for example that one asks for a standard gaussian random variable. A popular approach, called the Box-Muller method, is to use $U_1$ and $U_2$ i.i.d. uniform on  $(0,1)$ and to set $X=\sqrt{-2\log U_1}\cos(2\pi U_2)$. See here for more explanations and examples.

In the Box-Muller method, one uses $U_1$ and $U_2$ to get one random variable $X$. Acception-rejection methods are different because they use a random number of random variables to get $X$. Their general principle is to use a given number of i.i.d. random variables, say two uniform random variables $U_1$ and $U_2$, and to test if these satisfy a well chosen property, say $\mathcal Q(U_1,U_2)$. If  $\mathcal Q(U_1,U_2)$ holds, one accepts $(U_1,U_2)$ in the sense that one sets $X=\Phi(U_1,U_2)$ for a suitable function $\Phi$ and the game is over. Otherwise, that is, if $\mathcal Q(U_1,U_2)$ does not hold, one rejects $U_1$ and $U_2$, and one performs the same test with two other uniform random variables $U_3$ and $U_4$, say. If  $\mathcal Q(U_3,U_4)$ holds, one accepts $(U_3,U_4)$ in the sense that one sets $X=\Phi(U_3,U_4)$ and the game is over. Otherwise one rejects $U_3$ and $U_4$ and one turns to $U_5$ and $U_6$, and so on. One can also keep some information from the rejected random values, for example their number, to make the value $X$, but the procedure above is always the rough idea.

Hence, in rejection methods, the total number of random variables $U_n$ used to produce one value of $X$ is random. In your case, you are given a possibly infinite sequence $(U_n)_{n\geqslant1}$ of i.i.d. uniform random variables and you are asked to devise an algorithm to produce one random variable $X$ with the desired Gamma distribution, using almost surely a finite number of random variables $U_n$.

A good and relatively short introduction is the first chapter (The acceptance rejection method with applications; generating a standard normal random variable) of the lecture notes of this course. Another source, written for high energy astrophysicists and remarkably down-to-earth, is here.

The second link provides the code of an algorithm to generate Gamma distributions from what they call a Lorentzian distribution, which is also known as a standard Cauchy distribution. This is quite convenient since the latter is simply the distribution of $\tan(\pi U)$ with $U$ uniform on $(-1,1)$. I would rather leave you the pleasure of putting these bits and pieces together to concoct your algorithm, unless you get stuck at some point.

I believe that you would need to use something like MATLAB for that. You need to input the proper values and you can generate a set of random values that you can work from.


If $A,B\in M(2,\mathbb{F})$ and $AB=I$, then $BA=I$ This is Exercise 7, page 21, from Hoffman and Kunze's book.

  Let $A$ and $B$ be $2\times 2$ matrices such that $AB=I$. Prove that
  $BA=I.$

I wrote $BA=C$ and I tried to prove that $C=I$, but I got stuck on that. I am supposed to use only elementary matrices to solve this question.

I know that there is this question, but in those answers they use more than I am allowed to use here.

I would appreciate your help.

I will give a sketch of a proof. Let $A=
 \left( \begin{array}{cc}
a & b  \\
c & d  \end{array} \right)
$ and  $B=
 \left( \begin{array}{cc}
x & x  \\
z & w  \end{array} \right)
$ such that $AB=I.$ Then we get $\left\{\begin{array}{c}
ax + bz = 1 \\
cx + dz = 0 \\
\end{array}\right.$ and $\left\{\begin{array}{c}
ay + bw = 1 \\
cy + dw = 0 \\
\end{array}\right.$

I will assume that $a\neq 0$ (since there is no $B$ such that BO=I.) Then we have $x=\frac{1}{a}-\frac{bz}{a}$ and we get $(ad-bc)z=-c$. Let suppose that $ad=bc$. If $b=0$ or $c=0$ then $d=0$  and  we would have $A=
 \left( \begin{array}{cc}
a & b  \\
0 & 0  \end{array} \right)$, or $A=
 \left( \begin{array}{cc}
a & 0  \\
c & 0  \end{array} \right)$, or $A=
 \left( \begin{array}{cc}
a & 0  \\
0 & 0  \end{array} \right)$ but in any case there is no $B$ such that $BA=I$ (It is easy to prove that).  So we have $(a,b,c,d)\neq (0,0,0,0)$. Then we have $a=\frac{bc}{d}$, but in this case the systems above do not have solution. Then $ad-bc\neq 0$ and we get $z=\frac{-c}{ad-bc}$. In the end we will find that $B=
 \frac{1}{ad-bc}\left( \begin{array}{cc}
d & -b  \\
-c & a  \end{array} \right).$ It is easy to check that $BA=I.$ Now if $a=0$ then we have $b\neq0$ and $\dots$

I don't know how to solve the exercise in a different way. This is my best effort.

$AB = I$ implies that $ABAB = I$ and $AABB = A(I)B = AB = I$, hence $ABAB = AABB$. Since $\det(AB) = \det(A)\det(B) = 1$, the determinant of $A$ and the determinant of $B$ are units, so that $A$ and $B$ have inverses (using the adjoint matrix thing), hence $AABB = ABAB$ implies  $BA=AB = I$.

Hope that helps,

Note : This proof works when $\mathbb F$ is an arbitrary commutative ring with unity. You didn't specify what $\mathbb F$ was so that I'm stating in which generality this proof holds.


Geometry of 3 points getting closer to each other This is originally a physics problem but I am not interested in solving it, and my question has more to do with maths hence me being here and not on physics.stackexchange.

  Three points are located at the vertices of an equilateral triangle whose side equals $a$. They all start moving simultaneously with velocity $v$ constant in modulus, with the first point heading continually for the second, the second for the third, and the third for the first. How soon will the points converge?

I made a diagram for the movement after a period of time $dt$

$(1)=(2)=(3)$ since the velocites are the same.

One method of solving this problem consists of saying that $a$ will shrink after $dt$ by a value of $vdt+v\cos{(60)}dt$ where $v\cos{(60)}$ is the velocity of one point towards the point that it's not heading towards.

How can we deduce this mathematically? It seems intuitive that if two bodies are heading towards each other with a given constant velocity then the decrement would be as stated, but, and especially trying to use the diagram, I can't seem to deduce it "with rigour".

Thank you for your time!

The complicated version is to write coordinates for each point, so point $1$ would be $(x_1,y_1)$.  You can write differential equations for each point, so
$$x_1'=v\frac {x_2-x_1}{\sqrt{ (x_2-x_1)^2+(y_2-y_1)^2}}$$
and similarly for the others, then solve the coupled differential equations.  This throws away the knowledge that the triangle stays equilateral, which I think is the "non-rigorous" part.  Of course it will come out to be that in the final solution.

Due to polar symmetry each of the figures joining them is an equilateral triangle.

That is, they are shrinking at a constant rate, rotating at a constant rate and the ratio of these rates ( velocity components in circumferential and radial directions) is constant.

$$ \frac{v_{\theta}}{v_{r}}$$

is a constant. If both velocity components are not constant, then we have the instantaneous triangle distorted, not being equilateral. I.e.,

$$ \frac{r\cdot d\theta/dt}{-dr/dt} = \tan \alpha, $$

where $\alpha$ is constant ratio.

$$ \frac{r\cdot d\theta}{-dr} = \tan \alpha $$

Integrating

$$ r= r_{o}e^{- \cot \alpha.\theta} $$

The angle $\alpha $ is half of equilateral triangle angle $60^{\circ}$ due to symmetry.  One of three curves spiralling inwards  plotted with $r_{o}=1$ is

$$ r= r_{o}e^{- \sqrt{3}\theta} $$

This is a logarithmic spiral.

It can go to the center $(r=0)$ only when $ \theta\rightarrow \infty $.

We can twice polar rotate the curve about origin by $ 2\pi/3$ to see the other two loci. That is, the three points approach center arbitrarily close, but can never meet at a finite time... never to converge.

EDIT1:

If the velocity is directed between two vertices/along sides  of an equilateral triangle without rotation then the variable triangle becomes smaller and progressively becomes minimized at the sides midpoints (in-circle tangent points) and becomes bigger again without coming near each other. So this option can be ruled out.


Finding integer solutions of $m^2-n^5 = m - n$ How to list all integer solutions of

$m^2-n^5 = m - n$

Here $m$ and $n$ are some positive integers.

Also, I want to know the name of this type equations (if name exist).

Regards
Rosy

Write your equation as $m^2 - m = n^5 - n$.
You want $m = (1 + \sqrt{4 n^5 - 4 n + 1})/2$ to be an integer.  Trying the first $10^6$ values of $n$, we find that $4n^5 - 4n + 1$ is a square for
$n = 1, 2, 3, 30$, corresponding to $m = 1, 6, 16, 4930$.

The curve $x^2 - x - y^5 + y$ has genus $2$ (according to Maple), so by Faltings's theorem there are only finitely many rational points, and in particular only finitely many integer solutions.  I suspect that the solutions I listed are all the positive integer solutions, but  I don't know if it's possible to prove that.

Let start by eliminating the trivial solutions of the equation: $$m^2-n^5 = m - n$$ which is obviously equivalent to $$m^2-m=n^5-n$$
For instance, $m=n$ or $mn=0$
For I am sure you are looking the non-trivial ones. So we will suppose that $\gcd(m,n)=1$

First observation: According to Fermat's Little Theorem, we have:
$$m^2 \equiv m \pmod 2$$
$$n^5 \equiv n \pmod 5$$
In other words, there exist 2 non-zero integers $s,t \in \mathbb{Z}$ such that:
$$m^2=m+2s$$
$$n^5=n+5t$$
Therefore,
$$m^2-n^5=m-n+2s-5t=m-n$$
As a result,$$2s=5t$$ From this, it is clear that $2|t$ and $5|s$
Therefore, $$2s=5t\equiv 0 \pmod {10}$$
Then, we can write $$ m^2-m=10k$$
$$n^5-n=10k$$ with $k \in\mathbb{Z} \neq 0$
The first equation can be easily solved by using the quadratic formula:
$$(m-\frac{1}{2})^2=10k+\frac{1}{4}=\frac{40k-1}{4}$$
We must have $40k+1=u^2$ ($u \in\mathbb{Z} \neq 0$),in other words:$$40k=u^2-1=(u-1)(u+1)$$
Clearly, $u$ is odd. Hence, $\gcd(u-1),u+1)=2$. We deduct that one of those prime factors is a multiple of $20k$. Then we write: $$u \pm1=20k'$$ or $$u=20k'\pm 1$$
We then obtain the following values of $m$
$$m= \pm{\frac{20k'+1}{2}}+\frac{1}{2}=\pm 10k'+1$$
$$m= \pm{\frac{20k'-1}{2}}+\frac{1}{2}=\pm 10k'$$

Second Observation: From the system of equations$$ m^2-m=10k$$
$$n^5-n=10k$$ by equating $10k$ we obtain after factorization:
$$m(m-1)=n(n^4-1)$$. According to Gauss lemma, since $\gcd(m,n)=1$ then, $n| m-1$ So there exists an non-zero integer $h$ such that $$m-1=nh$$
From that, we can deduct the values of $n$ knowing the above values of $m$. NOT SURE WHERE TO GO FROM HERE.


Solve $x-1 \ge \sin(x)$ I got as far as $-1 = \sin(x)-x$.

I don't know what to do next. Pretty sure I am forgetting some simplification rule.

There is no analytical solution to the problem.

Numerical solution suggests $x\ge 1.9345632$

Others have suggested there is no analytical solutions, however;

Let $x=\frac{5\pi }{2}$, then $\frac{5\pi }{2}-1\geq \sin(\frac{5\pi }{2})=1$. Just an example, you can go from here.

It seems there are infinite solutions.


Gaussian integral via power series? It is well known that the Gaussian integral is $\sqrt \pi$, i.e.,
$$ \int_0^\infty e^{-x^2} dx = \frac{\sqrt \pi}2.$$
This is something we typically learn in calculus with polar coordinates.  We also learn how to integrate $e^{-x^2}$ via power series, e.g.,
$$ \int_0^t e^{-x^2} dx = \sum_{n=0}^\infty (-1)^n \frac{t^{2n+1}}{(2n+1)n!}.$$
Hence one has
$$ \frac{\sqrt \pi}2 = \lim_{t \to \infty} \left(\sum_{n=0}^\infty (-1)^n \frac{t^{2n+1}}{(2n+1)n!}\right).$$

Question: Is there a nicer way to rewrite the expression on the right, e.g. with "only one infinity" involved, to get an interesting expression/algorithm for computing $\pi$?

Of course the goal would be "pass the limit through the sum" but this is evidently non-trivial as the tail of the series explodes with $t$.  My limited experience with analysis suggests one could try to split up the sum in various ranges ($n \ll t$, $n \approx t$, $n \gg t$) but that middle ranges could get hairy.

In addition to changing the "upper bound" $t$ of integral,
you can make the integrand depends on $t$. This allows you
to get rid of the tail of the integrand for finite $t$.

For any $n \in \mathbb{Z}_{+}$, let $f_n : [0,\infty) \to \mathbb{R}$ be the function

$$f_n(x) = \begin{cases} (1 - \frac{x^2}{n})^n, & x \le \sqrt{n}\\0, & x \ge \sqrt{n}\end{cases}$$

For any fixed $x \in [0,\infty)$, it is easy to see $f_n(x) \le f_{n+1}(x)$ whenever $x \ge \sqrt{n}$.
When $x < \sqrt{n}$, we can apply AM $\ge$ GM to $n$ copies of $1 - \frac{x^2}{n}$ and one copy of $1$ and get

$$1 - \frac{x^2}{n+1} = \frac{n}{n+1}\left(1 - \frac{x^2}{n}\right) + \frac{1}{n+1} \ge \left( 1 - \frac{x^2}{n}\right)^{n/n+1} \implies f_{n+1}(x) \ge f_n(x)$$

This means $f_1(x), f_2(x), \ldots$ is a sequence of pointwise non-decreasing,
non-negative functions. Since its pointwise limit equals to $e^{-x^2}$, we can use Monotone convergence theorm to convert the integral on $e^{-x^2}$ to a limit of single variable:

$$\begin{align}\frac{\sqrt{\pi}}{2}
= \int_0^\infty e^{-x^2} dx
&= \int_0^\infty \lim_{n\to\infty} f_n(x) dx
\stackrel{\rm MCT}{=}
\lim_{n\to\infty} \int_0^\infty f_n(x) dx\\
&= \lim_{n\to\infty} \int_0^{\sqrt{n}}\left(1 - \frac{x^2}{n}\right)^n dx
= \lim_{n\to\infty} \sqrt{n} \int_0^1 \left(1 - t^2\right)^n dt\\
&= \lim_{n\to\infty} \sqrt{n}\sum_{k=0}^n \frac{(-1)^k}{2k+1}\binom{n}{k}
\end{align}$$

The Expression Can Be Written As Zeta Function and Zeta Function At infinity is 1 hence using proper methods it may be evaluated


Help solve recurrence : f(x) = 2f(x-1) + x While trying to analyze the time complexity of the heapify algorithm ,
I came up with the following recurrence for it's time complexity in terms of heap height:

T(h) = 2T(h-1) + h

Could you help me approach this recurrence?

To solve the recurrence $$a_h=2a_{h-1}+h \tag{1}$$ write $$b_h=a_h+h$$ Adding $h$ to both sides of $(1)$ gives $$b_h=2b_{h-1}+2$$  Can you take it from here?


Give an example of a continuous function $f : [0, ∞) \mapsto [0, ∞)$ such that $\int_{0}^{\infty}f(x)dx$ exists but $f$ is unbounded. Give an example of a continuous function $f : [0,  ∞) \mapsto [0, ∞)$ such that  $\int_{0}^{\infty}f(x)dx$ exists but  $f$ is unbounded.

I have been thinking about this. And I have come to the conclusion that I will need to construct a function, $f$, such that $f$ is a sequence of triangles of increasing height, but decreasing base.
I obviously need $f$ such that both the height of the triangles, and sum of the bases tends to infinity.
But I also need that the $\sum (\text{height} \times \text{base}) \leq \infty $

Building off your idea, but allowing most of the triangles to have zero height (otherwise it will not converge): Choose some divergent series. I'll assume we're dealing with the harmonic series $\sum \frac{1}{n}$. The nth triangle will have base $\frac{1}{n}$.

Now in general, start with $f \equiv 0$, i.e. suppose the general triangle we have is degenerate. But let the $2^n$th slot have a triangle of height $2^{n/2}$.

Or more generically: have some shape that has area $1$ from $0$ to $1$, then some shape that has area $\frac{1}{2}$ from $1$ to $2$, $\frac{1}{4}$ from $2$ to $3$, smoothing as necessary. You can make your function $0$ as much as you want, but you need to make your shapes get taller. You can think of fitting a triangle of height $k$ in the $k$th slot, making the base the necessary width to give area $\frac{1}{2^k}$ in that slot.

You can do many more along these lines.

Let $f= \cases{ x^{-1/2} , & $[0,1)$ \\ x^{-2} , & $[1,\infty )$ }$, then $\int_0^\infty f(x)dx = 3$. Is that what you are looking for?

EDIT: In the spirit of mixedmath: Let $Tr(x,h,w)=\max (1-|2x/w -1| , 0 )h$. Then let $h_i=i^{1/4}$ and $w_i=1/i^{2+1/4}$. Let $f=\sum_{i=1}^\infty Tr(x-i,h_i,w_i)$ such that
\begin{equation}
\int_0^\infty f(x) dx = \int_0^\infty\sum_{i=1}^\infty Tr(x-i,h_i,w_i) dx= 1/2\sum_{i=1}^\infty i^{-2}=\pi^2/12
\end{equation}
with questionable convergence.


Inverse Laplace transformation of $\frac{s^2}{(s^2+1)^2}$ I am a bit lost how to compute the inverse Laplace transformation of $$ \frac {s^2}{(s^2+1)^2}$$ I think it will be some combination of sine and cosine (oscillation-like), however I ran into some weird expressions when I tried to expand the expression as $$ \frac {s}{(s^2+1)}  \frac {s}{(s^2+1)}$$
Could someone please give some suggestions? Thanks in advance!

The original problem was for $\frac{s^2}{(s^2 + 1)^2}$
Consider the Laplace transforms of $\sin(a t)$ and $t \, \cos(a t)$ which are
\begin{align}
\sin(a t) &\doteqdot \frac{a}{s^2 + a^2} = \frac{a \, s^2 + a^3}{(s^2 + a^2)^{2}} \\
t \, \cos(a t) &\doteqdot \frac{s^2 - a^2}{(s^2 + a^2)^{2}}.
\end{align}
Now one can notice that
$$ \frac{1}{2 a} \, \left( \sin(a t) + a \, t \, \cos(a t) \right) \doteqdot \frac{s^2}{(s^2 + a^2)^{2}}. $$
A second method is to use the convolution theorem,
$$ \bar{f}(s) \, \bar{g}(s) \doteqdot \int_{0}^{t} f(t-u) \, g(u) \, du $$
where $\bar{f}(s)$ is the transform of $f(t)$. Since
$$ \cos(a t) \doteqdot \frac{s}{s^2 + a^2}$$
then
\begin{align}
\frac{s^2}{(s^2 + a^2)^2} &\doteqdot \int_{0}^{t} \cos(a u) \, \cos(at - au) \, du \\
&\doteqdot \left[\frac{2 a t \, \cos(a t) - \sin(a (t - 2 u))}{4 a} \right]_{0}^{t} \\
&\doteqdot \frac{1}{2 a} \, \left( \sin(a t) + a \, t \, \cos(a t) \right).
\end{align}
Modified form $\frac{s^3}{(s^2+1)^2}$
By using :
\begin{align}
\cos(a t) - a t \, \sin(a t) &\doteqdot \frac{s \, (s^2 - a^2)}{(s^2 + a^2)^2} \\
\cos(a t) + a t \, \sin(a t) &\doteqdot \frac{s \, (s^2 + 3 a^2)}{(s^2 + a^2)^2}
\end{align}
then it can be shown that
$$ \frac{1}{2} \, ( 2 \, \cos(a t) - a \, t \, \sin(a t) ) \doteqdot \frac{s^3}{(s^2 + a^2)^2}. $$


How is the group $O(p,q)$ defined wrt quadratic forms? Given a symmetric matrix $Q$ defining a quadratic form with signature $(p,q)$, the group of matrices $X$ which preserve $Q$, i.e.
$$
Q(Xu, Xv) = Q(u,v),
$$
is denoted $O(p,q)$.

So, I understand that, for instance, for a non-degenerate indefinite binary quadratic form $Q$, the group of matrices preserving the form is $O(1,1)$, and this is a group of dimension $1$.

However, I am confused because surely the group of matrices preserving a quadratic form depends on the quadratic form? For instance the matrices preserving
$$
A = \begin{pmatrix}
  1 & 1 \\
  1 & -1 \\
\end{pmatrix} \quad \quad \text{and} \quad \quad B = \begin{pmatrix}
  1 & 1 \\
  1 & -2 \\
\end{pmatrix}
$$
are going to be different - in this case
$\begin{pmatrix}
  1 & 2 \\
  2 & 5 \\
\end{pmatrix}$
preserves $A$ but not $B$.

With this in mind, how is $O(p,q)$ defined? I can find no explanation suggesting $O(p,q)$ depends on the choice of $Q$, but if it is the same group regardless of the choice of $Q$, then what differentiates, say, $O(1,1)$ from $GL_2(\mathbb{N})$?

If $A \in \operatorname{GL}_n(\mathbb{R})$ is a matrix representing a bilinear form, then the group of transformations preserving this form is $\mathscr{O}(A) = \{ X \in \operatorname{GL}_n(\mathbb{R}): X^t A X = A\}$.

Now let $A, B \in \operatorname{GL}_n(\mathbb{R})$ be two matrices representing a quadratic form with signature $(p,q)$ (where $p+q = n$). Then Sylvester's law of inertia tells you that $A = P^tBP$ for some $P \in \operatorname{GL}_n(\mathbb{R})$.

I leave it an exercise for you to show that $\mathscr{O}(A) = P^{-1} \mathscr{O}(B) P$.

That is, the groups $O(p,q)$ you get with different forms are usually not the same, but they will be conjugate in $\operatorname{GL}_n(\mathbb{R})$.

$O(p,q)$ depends of the choice of the signature of the quadratic form $Q(x)=\sum_{i=1}^{i=p}x_i^2-\sum_{i=p+1}^{i=n}x_i^2$ with $p+q=n$. Here the signature is $(p,q)$. You can always find a basis where the quadratic form has that expression. Two quadratic forms with the same signature yields to the same automorphisms group.


Showing that $c_n=\cos\frac\pi{2^n}$ satisfies the recurrence $c_{n+1} = \sqrt{\frac{1+c_n}2}$ From the identity $\cos(\alpha) = \sqrt{\frac{1+\cos(2\alpha)}{2}}$ with $0 \leq \alpha \leq \frac{\pi}{2}$, show that the sequence $c_n = \cos(\frac{\pi}{2^n})$ satisfies the recurrence relation $$c_{n+1} = \sqrt{\frac{1+c_n}{2}}$$

What I did was:
Say that $\alpha_n = \frac{\pi}{2^n}$. We first must prove $0 \leq \alpha_n \leq \frac{\pi}{2}$. By induction:
Base case is trivial: $0 \leq \frac{\pi}{2^1} \leq \frac{\pi}{2}$.
For $p(n) \Rightarrow p(n+1)$:
$0 \leq \frac{\pi}{2^n} \leq \frac{\pi}{2} \iff 0 \leq \frac{\pi}{2^n}\frac{1}{2} \leq \frac{\pi}{2}\frac{1}{2} \iff 0 \leq \frac{\pi}{2^{n+1}} \leq \frac{\pi}{4} < \frac{\pi}{2}$
So it is true that $\{a_n\}^\infty_{n=1} \in [0, \frac{\pi}{2}]$ and we can write $c_n = \cos(a_n)$.
Therefore:
$$c_1 = \cos(\alpha_1) = \cos(\frac{\pi}{2}) = 0$$
$$c_n = \cos(\alpha) \iff 1+c_n = 1+\cos(\alpha)\iff \frac{1+c_n}{2} = \frac{1+\cos(\alpha)}{2}$$
$$\iff \sqrt{\frac{1+c_n}{2}} = \sqrt{\frac{1+\cos(\alpha)}{2}}$$
But $\sqrt{\frac{1+c_n}{2}}$ would be the following term for $c_n$, meaning $c_{n+1} = \sqrt{\frac{1+c_n}{2}}$.
This was my atempt to prove it but I can't tell if it actually proves the recurrence. How could I prove recurrence relations in sequences like this?

You're doing much more work than you have to. As @Surb notes, you just need to verify $\cos\tfrac{\pi}{2^{n+1}}=\sqrt{\tfrac{1+\cos\tfrac{\pi}{2^n}}{2}}$, which is the case $\alpha=\tfrac{\pi}{2^{n+1}}$.

Let $\alpha = \frac{\pi}{2^{n+1}}$
$c_{n+1} = \cos\frac{\pi}{2^{n+1}} = \cos\alpha$ and $c_n = \cos\frac{\pi}{2^n} =  \cos\frac{2\pi}{2^{n+1}} = \cos(2\alpha)$
Now, $\cos(\alpha) = \sqrt{\dfrac{1+\cos(2\alpha)}{2}} \Rightarrow c_{n+1} = \sqrt{\dfrac{1+c_n}{2}}$


Proving limit of $\sqrt{n} (\sqrt[n]{n} -1)$ is $0$ as $n \to \infty$. I managed to prove that limit to $\sqrt[n]{n} -1$ tends to $0$, but couldn't go anywhere from there. Could somebody help?

We have
\begin{align}
\ln(\sqrt{n}(\sqrt[n]{n}-1)) &= \ln(\sqrt{n}) + \ln(\exp{\frac{\ln(n)}{n} -1})\\
&= \ln(\sqrt{n}) + \ln(\frac{\ln(n)}{n} + o(\frac{\ln(n)}{n}))\\
&= \ln(\frac{\ln(n)}{\sqrt{n}} + o(\frac{\ln(n)}{\sqrt{n}}) \rightarrow - \infty
\end{align}
Thus
$$
\sqrt{n}(\sqrt[n]{n}-1) \rightarrow 0
$$
as $n \rightarrow + \infty$.

Using l'Hôpital's rule, the derivative of $f(n)=\sqrt{n} (\sqrt[n]{n} -1)$ is:

$$
f'(n)=n^{1/n + 1/2} \frac{1-\log(n)}{n^2} + \frac{\sqrt[n]{n} -1}{2 \sqrt{n}}.
$$

Now taking $\displaystyle  \lim_{n\to\infty}$ of the above expression, we get $\displaystyle 0 + \lim_{n\to\infty}  \frac{\sqrt[n]{n} -1}{2 \sqrt{n}}$, and we know that $\displaystyle \lim_{n\to\infty} \sqrt[n]{n} = 1$, which yields the desired result.


Integers between $1$ to $1000$ satisfying a condition that $1/2 * {2n \choose n}$ is even For how many integers $n,$ for $1 \leq n \leq 1000,$ is the number $\frac{1}{2}\left(\begin{array}{c}2 n \\ n\end{array}\right)$ even?
I'm thinking about expanding the top into $(2n \cdot (2n-1) \cdot ... \cdot (n+1))$ and the bottom is simply $n!$ and that expression would be $0 \mod 4$, but I'm not sure how to approach it after that point.

Here is my solution with $p$-adic valuation.
I claim $ \frac{1}{2} {2n\choose n}$ is odd if and only if $n = 2^{k}$ for some $k$.
This is true since $v_{2}(2^{k}!) = 2^{k} - 1$, which implies $n = 2^{k}$ as $v_{2}\left({2n\choose n}\right) = v_{2}(2n!) - 2v_{2}(n!)$. Now considering the binary representation of $n$, we see $v_{2}(n!)$ can be written as the sum of terms of the form $v_{2}(2^{i}!)$ from which the result follows.
By complementary counting, the answer is therefore $1000 - \lceil\log_2(1000)\rceil= 990$.

It is simpler to find $n$s where the term is not even.  A quick substitution of $n = 2^i$ for $i = 0, 1, 2, \ldots , 9$ shows this is the case.
If $n = 2^i$ we have $\frac{1}{2} {2 \cdot 2^i \choose 2^i} = \frac{2^i (2^{i+1}-1)(2^{i+1}-2) \cdots (2^{i+1}-2^i)}{2^{i-1}!}$.  The number of factors of $2$ in the numerator are the same as the number of factors of $2$ in the denominator, so the term is not even.
We need not consider any other numbers where there are no factors of $2$, as these simply cannot be even.
Thus the total number is $1000 - 10 = 990$.


How to know when an author is using natural log? I was always taught that "log" is log base ten and "ln" is log base $e$, and there are plenty of sources that use this. But a lot of other sources use "log" as the natural log, and some books I have don't seem to specify which system they use and I can't quite figure it out sometimes. Is there a good way to know when somebody is using one or the other, or are there particular fields which tend to use one system or the other?

Just look for instances that give away which log they are referring to.

If you are using a precalc or a calc textbook, the chances that log refer to base 10 is slim to none.


If something is neither true nor false, what is it? I saw the other day something similar to the following:

  one of the following is true.
  the above is false.
  $ 1 + 1 = 5 $

You can probably see the problem with this. I can clearly state that $3$ is false, but what would I call $1$ and $2$?

To clarify, I really meant if there were some state between true and false that could make these consistent.

These statements taken together are called inconsistent.  That means that they cannot all be simultaneously true.  But the first and second are neither true nor false without broader context.  Using the language of first order logic, they might be said to be formulas with "free variables."  Here's an example of a formula with free variables
$$ 4x+3y=9$$
This is neither true nor false because I haven't told you what $x$ or $y$ are.  If I use quantifiers to get rid of all the free variables, then I have a sentence which may be true or false:
$$\forall x\forall y (4x+3y=9)$$
$$\forall x\exists y (4x+3y=9)$$
$$\exists x\exists y (4x+3y=9)$$
The first statement is false, while the second two are true.

The answer you are looking for is "undecidable".
What is neither true nor false is called an undecidable proposition. This is about the liar paradox and Gödel's incompleteness theorems.

Every attempt to establish the truth of the first proposition leads to a contradiction in the second. And you would never want inconsistency, ever. Not in mathematics and logic.

So, you decide it as "undecidable" to avoid contradiction.


Solving a trigonometric equation: $2 \sin(3a)=\sqrt{2}$ I have the following equation :

$2 \sin(3a)=\sqrt{2}$

Not sure how to solve it (Because it's a transformed sin function, meaning 6 solution with 3 cycles in $2\pi$) after a moment I finally found that each solution can be expressed as a multiple of $\frac{\pi}{12}$ (I aided myself with a graph of the function.)

For example :

$\frac{\pi}{12} \frac{3\pi}{12} \frac{9\pi}{12}$  etc.

My question would be : Is this the standard way of solving transformed functions of this sort ? Is there a sort of symbolic way without the need to use the graph ?

Is my way complicated for nothing ? Thank you !

Hints:

$$2\sin3a=\sqrt2\iff \sin 3a=\frac{\sqrt2}2=\frac1{\sqrt2}\iff 3a=\begin{cases}\frac\pi4\\{}\\\frac{3\pi}4\end{cases}+2k\pi\;,\;k\in\Bbb Z\;\ldots$$

We have $2sin(3A)=\sqrt 2$.

Multiplying both sides by $\sqrt 2$ , we have

$2 \sqrt 2sin(3A)=2$

$\implies sin(3A)=\frac {1}{\sqrt2}$. Let $3A=\alpha$.Then

$sin(\alpha)=\frac {1}{\sqrt2}$.Clearly such a value for $\alpha$ which satisfies the equation is $\alpha=30$.But $\alpha = 3A$ $\implies 3A =30$.$\implies A=10$.Hope it helps.Any doubts are welcomed.


Basic math, Confusion working out chance of failure. While I am quite technical, I am not a mathematician.  So please be forgiving if this question seems overly simple. My probability abilities left me long ago.

I have a problem which I was arrogant enough to think was simple until I done a few calculations that got me confused.

Lets say we have 2 systems (System A (SA), and System B(SB)). Upon which our overall system relies on.

A typical real world example would be deciding if you want your user to be able to download a library from google server - which is more likely to be cached on the users computer, or download the same library from your own server hosted in the cloud by amazon (on which all your other files reside anyway). This library is obviously something your own system depends on. Now you want to work out the risk of failure considering you have your libraries split across 2 places, instead of on one. However real world is not relevant, I just use that for context so you know where I came across the problem.

Assumption 1: The overall risk of failure if:

SA = 5% and SB = 3% = overall = 15%?

I am assuming this is correct, if not please let me know as this simple math is the basis for my question. Essentially, it is a multiplier, not a divisor or addition?

With assumption 1 assumed to be correct for now, my question is:

As we get closer to either number being 1, the risk of failure grows smaller but is still bigger than either one as long as both are above 1:

SA = 1% x SB = 3% = Overall = 3%

This seems simple, but it's based on assumption 1 being correct.

But once we go below 1 on any or both systems, the risk of failure actually goes down by multiplication, not up:

SA = 0.5% x SB = 0.5% = overall = 0.25%

This doesn't seem right to me, I would have thought it would be 2.5%. I believed the risk of failure should be greater than either of the systems failing combined but this math shows otherwise.

Is there some simple rule for working this kind of thing out consistently for numbers both greater and less than 1 which always give the larger result?

If we change to a divisor, the inverse happens:

SA = 5% / SB = 3% = Overall = 1.67% (smaller than both)
SA = 0.5% / SB = 0.5% = Overall = 1% (bigger than both)

This also introduces the potential of division by zero. As there is a possibility (however remote) that a system will never fail, and that system could in theory be SB. This (along with the number inversion) is the reasons that this approach doesn't feel right.

Or should I be using addition?

SA = 5% + SB = 3% = Overall = 7%
SA = 0.5% + SB = 0.5% = Overall = 1%

While addition consistently gives me larger numbers, I am unsure if this is correct for my circumstances as the last answer doesn't come close to the 2.5% expected, hence my uncertainty on this approach too.

Subtraction would always give a smaller result so I didn't even consider that.

Another reason I am asking is because google-fu comes up with all kinds of crazy formulas for assessing risk of failure on engineering systems, but even my probability at college didn't use formulas like that, so google seems to be failing me.

I'd appreciate any help.

For independent events $A, B$,

$$ P(A\text{ and }B) = P(A) P(B). $$

For any two events,

$$ P(A\text{ or }B) = P(A) + P(B) - P(A\text{ and }B). $$

This can be derived either by considering a Euler-Venn diagram or as mentioned in a comment by @calculus.

Then in your example

$$ P(A\text{ or }B) = P(A) + P(B) - P(A) P(B) = 0.05 + 0.03 - 0.05 \cdot 0.03 = 0.0785 = 7.85\%. $$

If the overall failure occurs if SA AND SB (both) fail, then you should multiply probabilities, but if the overall failure occurs if SA OR SB (at least one) fails, then you should add the probabilities up.


If permutation matrices are conjugate in $\operatorname{GL}(n,\mathbb{F})$ are the corresponding permutations conjugate in the symmetric group? There is a standard embedding of the symmetric group $S_n$ into $\operatorname{GL}(n,\mathbb{F})$ (for any field $\mathbb{F}$) that sends each permutation in $S_n$ to the corresponding permutation matrix.  As this is a group homomorphism, certainly, conjugate permutations map to conjugate matrices.  What about the converse?  Suppose that $A$ and $B$ are similar permutation matrices in $\operatorname{GL}(n,\mathbb{F})$ coming from permutations $\alpha$ and $\beta$ in $S_n$.  Does it follow that $\alpha$ and $\beta$ are already conjugate in $S_n$?  (I thought that I had a proof, but I noticed that the argument seems to fail in positive characteristic.)

I think the following argument works independently of the field. The dimension of the fixed point subspaces of $A$ and $B$ are equal to the total number of cycles (including cycles of length $1$) of $\alpha$ and $\beta$ so, if $A$ and $B$ are similar matrices, then $\alpha$ and $\beta$ must have the same total numbers of cycles.

The same applies to $\alpha^p$ and $\beta^p$ for any prime $p$, and the number of their cycles are equal to $a_1+a_2p$ and $b_1 + b_2p$, where $a_1$ and $a_2$ are the numbers of cycles of $\alpha$ of length not divisible by $p$ and divisible by $p$ respectively, and similarly for $b_1,b_2$ with $\beta$. So $a_1+a_2p=b_1+b_2p$, and since we already know that $a_1+a_2=b_1+b_2$, we get $a_1=b_1$, $a_2=b_2$. In other words, for any prime $p$, $\alpha$ and $\beta$ have the same numbers of cycles of lengths divisible by $p$.

Now, by an inductive argument on the number of prime factors of $k$, we can show that for any integer $k>0$, $\alpha$ and $\beta$ have the same numbers of cycles of lengths divisible by $k$, and that implies that $\alpha$ and $\beta$ have the same cycle types, so are conjugate in $S_n$.

For example for $k=12$ let $a_i$ be the number of cycles of $\alpha$ of length $j$, where $\gcd(j,12)=i$, for $i=1,2,3,4,6,12$. By induction, we know $a_1+a_2+a_3+a_4+a_6+a_{12}$, $a_2+a_4+a_6+a_{12}$, $a_3+a_6+a_{12}$, $a_4+a_{12}$ and $a_6+a_{12}$. The total number of cycles of $\alpha^{12}$ is $a_1+2a_2+3a_3+4a_4+6a_6+12a_{12}$ so that number, together with the already known sums, is enough to determine $a_{12}$ and hence to prove that $a_{12}=b_{12}$.

Yes, they both permute the standard basis and have the same cycle structure.

It's easier to see if you view them as linear transformations: Each cycle corresponds to a stable subspace of the total space and the cycle decomposition corresponds to the stable decomposition.


Show that $121^{n}-25^{n}+1900^{n}-(-4)^{n}$ is divisible by 2000. My question comes from the British Mathematical Olympiad (Round 1) paper from 2000:
Show that $121^{n}-25^{n}+1900^{n}-(-4)^{n}$ is divisible by 2000 for any natural $n$.
My immediate idea was to write $121^{n}-25^{n}=(11^{n}
+5^{n})(11^{n}-5^{n})$ but this doesn't seem to help.
My next idea was to use induction; the base case is easy but I can't see how the $k^{th}$ case would imply the $(k+1)^{th}$. Any help would be appreciated.

Do it by induction. Does it work for $n = 1$? Assume it works for some $n = k$ and show it holds for $n = k + 1$.


$k$ such that $n,2n,\dots,kn$ have odd sum of digits is the following statement true or not?

for any $k\in\mathbb{N}$ there exists $n\in\mathbb{N}$ such that all numbers $n,2n,\dots,kn$ have odd sum of digits?

I have no idea... it may turn out hard (for even sum of digits it would be very easy)

Hint : if we consider the number $n=9\cdots9$ with $2k-1$ of $9$'s. so for every element $0\leq i\leq 10^k$ we have the sum of digits of $ik$ is odd (is equal to 9(2k-1)) which proves your statement. I will provide a proof if this is necessary.

Hint: Having an odd sum of digits means is equivalent to the condition that the number mod 9 is odd. This means that it's sufficient to consider $n \in \{0,\ldots,8\}$.


Find limit of an integral with variable upper bounds. $$\lim_{x\to0+}\frac1x\int\limits_{-x}^x\sqrt[t^2]{\cos t}\,\mathrm dt$$

As I understand  $\displaystyle\int_{-x}^{x} \sqrt[t^2]{\cos(t)}\,\mathrm dt = 2\sqrt[x^2]{\cos(x)}$. But I am not sure if it is right and what to do next.

The integrand is defined over $[-\pi/2,\pi/2]\setminus\{0\}$. However the function has a removable singularity at $0$, because
$$
\lim_{x\to0}(\cos x)^{1/x^2}=e^{-1/2}
$$
which can be proved by considering
$$
\lim_{x\to0}\frac{\log\cos x}{x^2}=\lim_{x\to0}\frac{1}{2}\frac{\log(1-\sin^2x)}{\sin^2x}\frac{\sin^2x}{x^2}=-\frac{1}{2}
$$
If we extend the function, we get a continuous function, let's call it $f$, and we have
$$
\int_{-x}^x (\cos t)^{1/t^2}\,dt=\int_{-x}^x f(t)\,dt
$$

Now you have
$$
\lim_{x\to0}\frac{1}{x}\int_{-x}^x f(t)\,dt
$$
with $f$ continuous in a neighborhood of $0$. Now apply the fundamental theorem of calculus with the advantage that, in this case,
$$
\int_{-x}^x f(t)\,dt=2\int_0^x f(t)\,dt
$$

Using L'Hospital we get

  $$\lim_{x\to0+}\frac1x\int\limits_{-x}^x\sqrt[t^2]{\cos t}\,\mathrm dt =
\lim_{x\to0+}\frac2x\int\limits_{0}^x\sqrt[t^2]{\cos t}\,\mathrm dt =
\lim_{x\to0+}2\sqrt[x^2]{\cos x}\,\mathrm  = 2e^{-\frac12} $$
  because
  $$\lim_{x\to0+}\frac{ln(cosx)}{x^2} =
\lim_{x\to0+}\frac{-sinx}{2xcosx} = -\frac12 $$


Square wave transform One can define an analogous transform to the Fourier transform that uses square waves as the basis instead of sinusoids. Everything seems to work out in parallel and I imagine one can even come up with a FFT analogue.

We can define the square wave functions in a number of ways. The most direct seems to be

$$sq(t) = { 1 \mbox{ if } \frac{\{t\}}{2\pi} < 0.5, \mbox{ else} -1 }$$
$$cq(t) = sq(t + \pi/2)$$

here $sq(t)$ and $cq(t)$ are analogous to $sin(t)$ and $cos(t)$ resp.

The spectrograms are almost identical(well, there is some artifacting in the square wave case) and I was thinking they would be drastically different(the square wave case being more condensed).

Anyone think they can come up with an FFT version or have some fast hardware they could do some computations of various test cases and send me the spectrums if they are different? Or anyone can prove that they should be identical(or close)?

I am using

$$\int f(\tau) w(t - \tau) (cq(\tau) + isq(\tau)) d\tau$$

as the transform.

The problem with simply turning sine waves into square waves with "sign" is linear independence. Specifically, this set of square waves is not linearly independent. Pender and Covey (1992) came up with a square wave transform (SWT) composed of sets of phase-shifted waves that are linearly independent though not orthogonal. The transform is interesting because the forward transform is order N, like the Haar Transform, while the inverse transform is order N log N like the FFT and others. The Pender/Covey SWT is also very fast because it only uses sums and bit shifts.


Wedge product and cross product - any difference? I'm taking a course in differential geometry, and have here been introduced to the wedge product of two vectors defined (in Differential Geometry of Curves and Surfaces by Manfredo Perdigão do Carmo) by:

Let $\mathbf{u}$, $\mathbf{v}$ be in $\mathbb{R}^3$. $\mathbf{u}\wedge\mathbf{v}$ in $\mathbb{R}^3$ is the unique vector that satisfies:

$(\mathbf{u}\wedge\mathbf{v})\cdot\mathbf{w} = \det\;(\mathbf{u}\;\mathbf{v}\;\mathbf{w})$ for all $\mathbf{w}$ in $\mathbb{R}^3$

And to clarify, $(\mathbf{u}\;\mathbf{v}\;\mathbf{w})$ is the 3×3 matrix with $\mathbf{u}$, $\mathbf{v}$ and $\mathbf{w}$ as its columns, in that order.

My question: is there any difference between this and the regular cross product or vector product of two vectors, as long as we stay in $\mathbb{R}^3$? And if there is no difference, then why introduce the wedge?

Cheers!

I would have left this as a comment, but apparently I can't comment yet.

To answer your final question, why introduce the wedge, the point is that the wedge product (as explained in the Wikipedia article) is a notion that generalizes to $\Bbb R^n$ and indeed any vector space---in general the output is what is called a "bivector". Now, it just so happens that in $\Bbb R^3$ there is a natural identification between bivectors and vectors, and the vector that corresponds to the (bivector) wedge product is indeed the cross product, but this does not work in other vector spaces.

So, to sum up, the notion of cross product is specific to $\Bbb R^3$, whereas wedge product makes sense in any vector space.

There is a difference. Both products take two vectors in $\mathbb{R}^3$. The cross product gives a vector in the same $\mathbb{R}^3$ and the wedge product gives a vector in a different $\mathbb{R}^3$. The two output vector spaces are indeed isomorphic and if you choose an isomorphism you can identify the two products. However this isomorphism is a choice, or to put it another way depends on fixing a convention.

In higher dimensions the wedge product gives a vector in a vector space of higher dimension and so no identification is possible.


Existence of minimum in bounded but open set According to the Extreme Value Theorem, a continuous function achieves at least one minimum and one maximum whenever the set is bounded and closed (i.e. compact).

In my case, I have a bounded and open set, thus, although an infimum exists, a minimum is not assured.

But consider the extra information:

my continuous function is like $z=f(x,y)$, with $f: A \rightarrow \Bbb R$, and $A \subset \Bbb R^2$
from the plot I see that at the boundary the function is far from the minimum, and the minimum is at the interior. In other words, the infimum is not at the boundary.

The question is:

How can I formalise this graphical insight in order to apply the Extreme Value Theorem to prove that my function necessarily has a minimum? I am thinking on the following logic:

continuous function +
Bounded and open set +
Infimum not at the boundary +
Extreme Value Theorem

=> Minimum necessarily exist.

I would say yes.

From what you said, it seems that you can continuously extend your function $f$ on $A$ to a function $\tilde{f}$ on $\bar{A}$ (where $\bar{A}$ is the minimum closed set containing $A$) and apply the Extreme Value Theorem to $\tilde{f}$.

Similarly if you know that the infimum is not at the boundary than you know that the infimum is in $B \subset A$ with $B$ closed (and thus compact) and apply the Extreme Value Theorem to $f$ restricted to $B$.


Limit of exponential series Find the following limit:
$$\lim_{n \rightarrow \infty} \sqrt[n]{ \left( 3^n-2^n+1/2^n \right ) x^n }$$
How to calculate this limit? I am lookin for the solution that would involve use of squeeze theorem.

First note that
$$ 3x \sqrt[n]{1 - \left(\frac 23\right)^n + \left(\frac 16\right)^n} = \sqrt[n]{(3^n - 2^n + 1/2^n)x^n}. $$
Now
$$ 1 - \left(\frac 23\right)^n + \left(\frac 16\right)^n \le 1 $$
and, as the left hand side tends to one, finally (that is for all large enough $n$):
$$ \frac 1n \le 1 - \left(\frac 23\right)^n + \left(\frac 16\right)^n $$
That gives, by the squeeze theorem, that
$$ 1 = \lim_n \frac 1{\sqrt[n]n} \le \lim_n \sqrt[n]{1 - \left(\frac 23\right)^n + \left(\frac 16\right)^n} \le \lim_n 1 = 1 $$
Hence
$$ \lim_n\sqrt[n]{(3^n - 2^n + 1/2^n)x^n} = 3x $$


Sections of box tensor product of sheaves on the diagonal Let $A,B$ be two (say coherent) sheaves on some scheme/variety $X$ and consider their product $A\boxtimes B$ on the diagonal $X\times X$. Is it always

$$H^0(A\boxtimes B)\simeq H^0(A)\otimes H^0(B) $$

(I guess this is a general fact but I am really interested in the case $A$,$B$ locally free and $X$ smooth)

If $X = \text{Spec}(R)$ then $X \times X = \text{Spec}(R \otimes R)$, $A,B$ are just $R$-modules and $A \boxtimes B $ is just $A \otimes_R B$. If $X$ is not affine, there is an affine cover of $X$ and one can apply the previous argument again.


examples of random variables that are not measurable Let $(\Omega,\mathcal{F},\mathbb{P})$ a probability space and $(\Omega',\mathcal{F}')$ a measurable space. Furthermore let
\begin{equation}
X:(\Omega,\mathcal{F},\mathbb{P}) \rightarrow (\Omega',\mathcal{F}')
\end{equation}
be any function. Can someone give examples where $X$ is not measurable?

Small note: we only call such a function a random variable when $X$ is measurable.

A very simple example would be $\Omega = \Omega' = \{0, 1\}$, with $\mathcal F = \{\emptyset, \Omega\}$ and $\mathcal F' = \mathcal P(\Omega)$, and $X = \mathrm{id}_\Omega$. Then $\{1\}$ is measurable in $(\Omega', \mathcal F')$, but $X^{-1}(\{1\}) = \{1\}$ is not measurable in $(\Omega, \mathcal F)$.

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space for a fair coin flip $(\{H,T\}, 2^{\Omega}, \mathbb{P}(\omega)=\frac12)$, and let $(\Omega',\mathcal{F}')$ be a (probable?) measurable space for a die roll $(\{1,2,3,4,5,6\}, 2^{\Omega'})$. Let $X(\omega) = 1_{H}(\omega)$, a random variable indicating payoff of flipping heads in the fair coin.T hen $X$ is $\mathcal{F}$-measurable but not $\mathcal{F'}$ measurable.


What is the number of arrangements of 5 different things taken 3 at a time in which 2 particular thing always occur? I know , there is formula to apply, s!(r-s+1)( n-s P k-s).

I tried with example,
n=5 ; r=3 and s=2.
I'm getting answer as 2!*(3-2+1)*3 = 2*2*3 = 12.

But, if I solve manually with example,
consider n = {1,2,3,4,5}
then possible arrangements are as follows:

145
154
245
254
345
354
415
425
435
451
452
453
514
524
534
541
542
543.
Total count is 18.
Where am I going wrong

The formula is designed in a way such that it only accounts for permutations, where the chosen fixed elements appear together i.e. as a block. For example, in your case $451$ is allowed, whereas $415$ is not because $4$ and $5$ are the chosen elements. If you count only those, there are indeed $12$ of them.

The formula that deals with what you want is $r!{n-s \choose r-s}$. In this case, the number would be $3!{3 \choose 1}=18$, exactly what you got.

2 necessary items may be placed in 3P2 ways
And remaining 1item out of 3 items in 3! Ways
Hence answer =3P2*3P1=3*3!=18


Linear Algebra Dimensions of $\mathrm{Nul}(A)$ Suppose that $T \colon \mathbb{R}^3 → V$ is a linear transformation from $\mathbb{R}^3$ to some vector space $V$. Let $A$ represent the matrix of $T$. What are the possible dimensions of $\mathrm{Nul}(A)$? What would those dimensions mean geometrically?

The nullity of T depends upon the Codomain as well but the fact is it can't be greater than 3 due to sylvesters law of nullity

$NullA$ is exactly as solving $Ax = 0$. Which is exactly $KerT$. Thus, Its dim is smaller or equal to 3.


Are equations correct? counterexamples Check if these equations are correct:
(function is measurable)

)  $ f((A)\bigcup(B) ) = f(A) \bigcup f(B)$
) $ f((A)\bigcap(B)) = f(A) \bigcap f(B)$

I took counterexamples. For the second case i took  $f(x)= x^2$ and $ A=[-1;2], B=[-2;3]$ then $(A\bigcap B)= [-1;2] , f(A\bigcap B) = [1;4] , f(A)=[1;4], f(B)=[4;9] $ so $f(A\bigcap B) \neq f(A) \bigcap  f(B) = [4]$
For the first case i took the same function and i get $A\bigcup B = [-2;3]$    then $ f(A\bigcup B) = [4;9]$ and $ f(A)=[1;4], f(B)=[4;9]  $  but
$f(A)\bigcup f(B)= [1;9]$ . So again this equation is not correct.  Am i right?

In general,

$$
f(A \cup B) = f(A) \cup f(B)
$$

and

$$
f(A \cap B) \subseteq f(A) \cap f(B)
$$

For the first one: if $y \in f(A \cup B)$, either $y$ is the image of some element in $A$, or the image of some element in $B$, so respectively $y\in f(A)$ or $y\in f(B)$, which means that $f(A \cup B) \subseteq f(A) \cup f(B)$. Reciprocally , if $y \in f(A) \cup f(B)$, either $y\in f(A)$ or $y\in f(B)$, and let's assume without loss of generality that $y\in f(A)$. Since $A \subseteq A \cup B$, $y$ is the image of an element of $A \cup B$, and therefore we obtain $f(A) \cup f(B) \subseteq f(A \cup B)$. This completes the proof.

As for the second one, if $y \in f(A\cap B)$, $y = f(x)$ with $x\in A\cap B$. We can immediately tell that $x$ is both in $A$ and in $B$. So $y$ is in $f(A)$ and $f(B)$ respectively and therefore in $f(A) \cap f(B)$, which  proves the claim.

Your second example proves that the equality in the second case does not always hold, and I'll leave you to examine why the first one is not an actual counterexample.

You need to know that your function is isomorphism injection,byjection,continuous)otherwise you can't speak about equalities


proper way to approach learning higher (university level) mathematics I have been struggling to self-learn some somewhat higher mathematics- mostly university level mathematics. However, I've looked up other questions and they didn't mostly line up with what I am personally struggling with.

The problem I am facing simply put is that most of the math books at this level rely on proofs and examples more than tedious repetitive- mostly computational- exercises at the end of each section/chapter like in earlier math subjects. Problem is, I feel frustrated that I can't verify my solutions OR WORSE reproduce the proofs I analyze and study. Sure I can understand the proofs just fine (most of the time), but revising them the next day feels harder let alone reproducing them without taking a peek. I feel hesitant on whether I should take the material as is and move on to the next section/chapter or fret over the material until it becomes a second nature (probably takes weeks and highly inefficient; one page per week at worst).

How do you think I should approach this without feeling like I am skipping or just spending my time inefficiently? please advise.

Note: English isn't my first language so when I write proofs I tend to be redundant and less compliant to the common writing formats.

Advanced mathematics tends not to have "tedious repetitive - mostly computational- exercises at the end of each section/chapter". Rereading abstract proofs is probably not a good way to use your time. Spend as much as you can on the examples. Take each theorem and see what it says about all the examples you know to which it applies. See how the proof works in each particular case. Get several books and mine them for more examples. The ones that occur in all the books will be the important ones. The rarer examples will be instructive too.

Study counterexamples too. See how theorems fail when one or more of the hypotheses is false.

You need not master section $n$ before reading ahead into section $n+1$. You can and should circle back from time to time. The earlier material will probably become clearer that way.

(Your English here is just fine.)

University level...I think it depends on your current nation and your field of study as well.

For example, I learned in my engineering department in the university in Republic of Korea these subjects;

ODE: ordinary differential equation

PDE: partial differential equation

One of books widely used in the engineering course is Advanced Engineering Mathematics written by Erwin Kreyszig. This book could show you what subjects are taught in the university around the world.


I get a contradiction in the theory of free abelian groups. What am I doing wrong? Hi: The definition I'll use is this: Let $F$ be an abelian group and $X$ a subset of $F$. Then $F$ is a free abelian group on $X$ if for every abelian group $G$ and every function $f$ from $X$ to $G$ there is a homomorphism $\phi$ from $F$ to $G$ that extends $f$.

Let $G$ be a finite group and $X$ a subset of $G$. Let $F$ be the free abelian group on $X$. Then $F=\langle X\rangle$ and so $F\subseteq G$. That is, every finite group has an infinite subgroup. What am I doing wrong?

EDIT: It will be easier to make myself clear working with free groups. I'll quote from Derek Robinson, A Course in the Theory of Groups, 2nd ed.

From this a free group is not only always free on a subset but additionally that subset generates it. If $G$ is a group and $X$ is a subset, however, indeed there will exist a free group on $X$ but I am unable to show it will be generated by $X$ based in the above quote. Which is very natural, of course. Thanks for the posts. Honestly none of the feedback, up to now,  throws light in the paradox (paradox for me, of course).

You are using $\langle X \rangle$ to mean two different things, and conflating them:

You are using it to mean the free abelian group on $X$.
You are using it to mean the subgroup of $G$ generated by $X$.

These are not the same thing, but you assume that they are.

Consider $G=\mathbb{Z}/5\mathbb{Z}$ and $[1]\in G$, let $X=\{[1]\}$ $F=<X>=G$, let $f:X\rightarrow\mathbb{Z}$ defined by $f([1])=1$, $f$ cannot be extended to $G$ in $g$, since we must have $g(5.[1])=5.f([1])=5$ and $g(5.[1])=g([0])=0$ since $g$ is a morphism of groups.


$1$ is not congruent because of Fermat's Last Theorem? I would like someone to explain something I did not understand. I was reading a page called "nuking the mosquito" where they give very complex proofs for very simple results.

The proof I want to talk about deals with the irrationality of $\sqrt{2}$.

If $\sqrt{2}$ is rational, then there is a right angeled equilateral triangle with sides $(\sqrt{2},\sqrt{2},2)$. The area of this triangle is $1$, and if we assume $\sqrt{2}$ is rational then $1$ is congruent.

Then the writer says "Hence $1$ is congruent. This contradicts Fermat's Last Theorem with exponent $4$"

I don't doubt that $1$ is not congruent, I just don't see the connection to FLT. Would anyone shed some light to what I'm missing?

If $\sqrt{2}$ is rational, then there exists an isosceles triangle has side lengths $(\sqrt{2},\sqrt{2},2)$. This triangle has area $1$; hence $1$ would be a congruent number. This contradicts Fermat with $n=4$; see Theorem $2.1$ in http://www.math.uconn.edu/~kconrad/ross2007/congnumber.pdf for the relation to Fermat.


Countability of local maxima on continuous real-valued functions I am working through a bank of previous exams and couldn't figure a problem out to my satisfaction.

  Let $f(x) : \mathbb{R} \to \mathbb{R}\,$ be a continuous function.

  Show that $f$ can have at most countably many strict local maxima.
  Assume that $f$ is not monotone on any interval. Then show that the local
  maxima of $f$ are dense in
  $\mathbb{R}$.

For each $\delta>0$, the set of all $x\in\mathbb{R}$ such that $f(y)<f(x)$ for all $y$ with $0<|x-y|<\delta$ is countable.  This can be seen by noting that the set contains at most one element of the interval $[k\frac{\delta}{2},(k+1)\frac{\delta}{2}]$ for each integer $k$, and these intervals cover $\mathbb{R}$.  The set of strict local maxima is a countable union of such sets, for example taking $\delta=\frac{1}{n}$ as $n$ ranges over the positive integers.
Suppose that $f$ is a continuous function that is not monotone on any interval.  We want to show that $f$ has a local maximum in every interval.  The argument is the same if the interval is $(0,2)$, so to slightly reduce notation let's work there.  Note that not being monotone in any interval means that each interval contains pairs $x_1<x_2$ and $y_1<y_2$ such that $f(x_1)<f(x_2)$ and $f(y_1)>f(y_2)$.  So $(0,1)$ contains a pair $x_1<x_2$ such that $f(x_1)<f(x_2)$, and $(1,2)$ contains a pair $y_1<y_2$ such that $f(y_1)>f(y_2)$.  Because $f$ is continuous, there is a point $x_0\in [x_1,y_2]$ where $f$ has its maximum value.  Because $f(x_2)>f(x_1)$ and $f(y_1)>f(y_2)$, $x_0$ is not an endpoint of $[x_1,y_2]$.  Therefore $f$ has a local maximum at $x_0$.

For #1, prove that you can find disjoint intervals around each local maximum. Then pick a rational number in each interval.

Edit: This answer is wrong but the comments below are instructive, so I'll leave it here.


How to solve for theta? $$-12\sinθ-5\cosθ=0$$

Is it possible? or do I need to generate a graph? I heard the Newton Raphson Method can be used...not sure how to proceed please advise.

I'm trying to calculate the minimum and maximum values of: $y=12\cosθ-5\sinθ$

The equation above is a derivitive of $y$.

Hint: Check if $\theta=\pi/2,3\pi/2$ (I'm considering the interval $[0,2\pi]$ ) are solutions, and then divide by $\cos$ and solve for $\tan$.

y = 12cosx -5sinx

y'= -12sinx -5cosx

put y'=0

-12sinx -5cosx = 0

12sinx + 5cosx =0  (multiplied by -1 both sides )

12sinx = -5cosx

sinx/cosx=-5/12

tanx=-5/12

X = tan inverse (-5/12)

Now can you proceed?


No Adjacency Combinatorics Problem via Generating Function I would like to find the generating function solution for the following combinatorics/probability problem. I have a combinatorial solution and the generating function deduced thereof. But I can not fathom the direct interpretation of it. I would like to reverse the current solving process and directly deduce the following or some other generating function, then get the relevant coefficient as the solution.

In a stack of $W$ white cards, $G$ green cards and $R$ red cards, what is the number of arrangement for no green card is next to a red card?

One solution is to view the $W$ white cards as dividers and leaving $W+1$ slots in between them together with both ends. For each $r\in \{1,2,\dots,R\}$ and $g\in\{1,2,\dots, G\}$, designate non-overlappingly $r$ slots for inserting the red cards and $g$ slots for the green cards. There are ${W+1\choose r,g}$ combinations. For each of these combinations, we insert $R$ red cards into the previously designated $r$ red slots, and there are ${R-1\choose r-1}$ ways of insertion. Do the same for the $g$ green cards and there are ${G-1\choose g-1}$ ways of insertion. Multiplying them and summing over $r$ and $g$, we arrive at the desired total number of arrangements:

$$\sum_{r=1}^R\sum_{g=1}^G{W+1\choose r,g}{R-1\choose R-r}{G-1\choose G-g}.$$

But this is the coefficient of $x^Ry^G$ in $(1+x+y)^{W+1}(1+x)^{R-1}(1+y)^{G-1}$.

Now how do I deduce this generating function directly from the problem?

(The notations are slightly changed here)

We may use a directed graph for such pattern avoiding problems.

Write the matrix as

\begin{align*}
A = \left(\begin{array}{r|rrr}
  & W & G & R \\
\hline
W & w & g & r \\
G & w & g & 0 \\
R & w & 0 & r \\
\end{array}\right)
\end{align*}

$W,G,R$ are the states to indicate the three colors and $w,g,r$ are the formal variables to indicate the colors.

To obtain the recurrence formula, find the characteristic polynomial,

$$x^{3} - \left(g + r + w\right) x^{2} + g r x + g r w = 0$$

If we indicate $F(a,b,c)$ as the number of ways to arrange the cards, where a,b,c are the number of white, green and red cards respectively, then
$$F(a,b,c) = F(a-1,b,c)+F(a,b-1,c)+F(a,b,c-1)-F(a,b-1,c-1)-F(a-1,b-1,c-1)$$
and set the necessary boundary conditions

To obtain the generating function, find
$$\left(I-x\, A\right)^{-1}$$
Each entry will have a g.f. for that entry, and we require the sum of the first row, which is

$$G(x) =\frac{1-g\, r\, x^{2}}{1\, -\, {\left(g\, +\, r\, +\, w\right)+\, g\, r\, x^{2}+g\, r\, w\, x^{3}}}$$

If there are $n=a+b+c$ cards in total, find $[x^n]$ and find the required $[w^a\, g^b\, r^c]$

Richard Stanley's "Enumerative combinatorics" discusses such methods (transfer matrices)

This is not an answer to the question but some of this information may help.  Lets call a permutation of n (white green or red) cards where no red card is adjacent to a green card a valid arrangement.  The bivariate generating function (b.g.f.) for the number of length n valid arrangements that have exactly k white cards is : c(x,y) = $-((1 + x)/(-1 + x + x y + x^2 y))$.  The coefficient of y^k*x^n is the number of valid arrangements having a total of n cards where exactly k of the cards are white.  c(x,y) does not tell us anything about the number of red and greeen cards in the arrangements (other than that the number of red or green cards is n - k).

I solved this system of equations: a(x, y) == y x a(x, y) + y x b(x, y)
and b(x, y) == 1 + x + x b(x, y) + 2 x a(x, y). I derived these equations directly (i.e. via the symbolic method) from the properties of such arrangements.  a(x,y) is the b.g.f. for valid words that start with a white card.  b(x,y) counts the valid words that are empty or start with a red card or start with a green card.The generating function c(x,y) = a(x,y) + b(x,y).

The first few rows of the triangle are:

1;

2, 1;

2, 4,  1;

2, 8,  6,  1;

2, 12, 18, 8,  1;

2, 16, 38, 32, 10, 1;

2, 20, 66, 88, 50, 12, 1;

  This is Sloane's OEIS A113413.

Also interesting is if you let W = R = G then you get the sequence that starts:
2, 12, 92, 780, 7002, 65226, 623576, 6077196,...  which is A103882


Understanding why $x^{\log_b(y)} = y^{\log_b(x)}$ According to wikipedia, we have that

$$
x^{\log_b(y)} = y^{\log_b(x)}
$$

because

$$
x^{\log_b(y)} = b^{\log_b(x) \log_b(y)} = b^{\log_b(y) \log_b(x)} = y^{\log_b(x)}
$$

But what justifies that first leap?

$$
x^{\log_b(y)} = b^{\log_b(x) \log_b(y)}
$$

It starts from the fact that $x$ is equal to $b^{\log_b(x)}$. To elaborate, the log to the base $b$ of $x$ is defined as the number which, when $b$ is raised to this number, gives back $x$. Now, since $x = b^{\log_b(x)}$, we can always replace any occurrence of $x$ in any equation whatsoever with $b^{\log_b(x)}$. So, breaking that first leap into two steps, we first have:

$x^{log_b(y)} = (b^{\log_b(x)})^{log_b(y)}$

From here, we can apply a known rule of exponents which is that $(s^r)^t = s^{rt}$. Applying that rule we get:

$(b^{\log_b(x)})^{log_b(y)} = b^{\log_b(x)log_b(y)}$

Which is the first step in the above.

Method-1:
Notice, we know that $\large \color{red}{\frac{1}{\log_m(n)}=\log_n(m)}$ & $\color{red}{m^{\log_m(n)}=n}$

Now, we have
$$\large x^{\log_b(y)}=y^{\log_b(x)}$$ $$\large \iff x^{\frac{1}{\log_b(x)}}=y^{\frac{1}{\log_b(y)}}$$
$$\large \iff x^{\log_x(b)}=y^{\log_y(b)}$$
$$\large \iff b=b$$
 Method-2: we have $$\large x^{\log_b(y)}=y^{\log_b(x)}$$ taking log with base $x$ on both the sides, we get
$$\large \log_x(x^{\log_b(y)})=\log_x(y^{\log_b(x)})$$ $$\log_b(y)\log_x(x)=\log_b(x)\log_x(y)$$
$$\frac{\log_b(y)}{\log_b(x)}=\log_x(y)$$
$$\log_b(y)\cdot \log_x(b)=\log_x(y)$$
$$\log_x(y)=\log_x(y)$$


Do Zero degree angles exist? If an angle is the measure of distance between to points (Edit: Ok, admittedly bad phrasing. A measure of rotation between two intersecting lines, or points, etc.), is there such a thing as a zero degree angle? I asked a math nerd friend about this years ago and he still hates me for it but couldn't prove it either way at the time. Is this merely pedantic or delusional thoughts regarding definition of "angle" (Edit: based on answer below regarding Euclidian geometry, this all now seems very likely a colossal misunderstanding) or am I missing something? I do not think zero degree angles exist. Is there proof of this either way?

Edit: A better question would be why Euclidian zero degree angles were thought to not exist, or why he avoided them?

I suspect that behind this question lies the notion that an angle should look like some sort of wedge shape, created by two distinct lines meeting at a point, whereas a 'zero angle' just looks like part of a line.

In fact, that was very much the view taken by Euclid in Euclid's Elements, his seminal work of classical geometry, in which he states

  "...a plane angle is the inclination of the lines to one another, when two lines in a plane meet one another, and are not lying in a straight-line."

This definition appears to rule out a zero angle, and indeed a 180-degree angle — in fact, the description of an angle by any number of degrees assumes a concept of angle measurement that Euclid avoided: instead he compared their size by mapping them onto each other, or added them by mapping them adjacently to one another, and was thus able to make remarkable progress.

The idea of measuring angles has been of practical interest for thousands of years for purposes such as astronomy, navigation and telling the time, but the use of numerical measurements as a tool for studying theoretical geometry didn't really take off until the work of Descartes and Fermat in the 17th century. Their work led to the study of geometry through algebraic functions and number, and the development of a more numerical understanding of angle was a natural part of that process.

In short, if you think of angles according to numerically-measured size, it is convenient to have a definition that extends along the number line, but if you're thinking of them from a purely spatial perspective, then such an extended definition is perhaps unnecessary and potentially confusing.

Wouldn't a zero angle be a point. Once you leave that point then you have created an acute, right, obtuse, straight, or reflex angle.


Prove: $\frac{(2n)!}{n!(n+1)!}\in \mathbb{N}, \forall n\in\mathbb{N}$ Prove: $\frac{(2n)!}{n!(n+1)!}\in \mathbb{N}, \forall n\in\mathbb{N}$

I used induction, for $n=k$ assume that

$\frac{(2k)!}{k!(k+1)!}\in \mathbb{N}$

For $n=k+1$

$\frac{(2k+2)!}{(k+1)!(k+2)!}\in \mathbb{N}$

$x=\frac{(2k+2)!}{(k+1)!(k+2)!}\cdot \frac{k!(k+1)!}{(2k)!}\notin \mathbb{N}$

If $x \notin\mathbb{N}$ then how to prove $\frac{(2n)!}{n!(n+1)!}\in \mathbb{N}, \forall n\in\mathbb{N}$

The Catalan numbers:
$$ C_n = \frac{1}{n+1}\binom{2n}{n} $$
satisfy the recurrence relation:
$$ C_{n+1} = \sum_{i=0}^{n} C_i C_{n-i} $$
hence they are integers, since $C_0=C_1=1$.

$$\frac{(2n)!}{n!(n+1)!} = \frac{(n+2)\cdots (n+n)}{n!}$$

since for every number $k\le n$ there exists a number $m$ such that  $ n+2 \le mk\le2n$, for any factor in the denominator there is that factor (and then some) in the numerator. Hence it's an integer

Here is why I don't think induction is a good idea:

$$\frac{(2k+2)!}{(k+1)!(k+2)!} = \frac{(2k+2)(2k+1)(2k)!}{(k+1)(k+2)(k+1)!k!} = \frac{(2k+2)(2k+1)}{(k+1)(k+2)} \cdot \frac{(2k)!}{(k+1)!k!} = $$$$=
\frac{2(2k+1)}{k+2} \cdot \frac{(2k)!}{(k+1)!k!}$$

Now if $\displaystyle \frac{2(2k+1)}{k+2}$ was an integer, induction would work perfectly. But it's not an integer! So you need to show somehow that $k+2$ divides the factor on the right.. But then the induction hypothesis is useless and you are better off just proving the result directly! ;-)


ABCD is a rectangle and M is the midpoint of CD.The inradii of triangles ADM and ABM are 3 and 4 respectively.Then find the area of the rectangle. ABCD is a rectangle and M is the midpoint of CD.The inradii of triangles ADM and ABM are 3 and 4 respectively.Then find the area of the rectangle.
Let $AD=a,DM=b$ so $AM=\sqrt{a^2+b^2}$
In a triangle ,inradius$=r=\frac{abc}{4Rs}$
where $a,b,c$ are the side lengths and R,r,s are circumradius,inradius and semiperimeter respectively.
Using this relation in the triangle $ADM$,$r=\frac{ab\sqrt{a^2+b^2}}{4(\frac{a+b+\sqrt{a^2+b^2}}{2})\frac{\sqrt{a^2+b^2}}{2}}$
So,$3a+3b+3\sqrt{a^2+b^2}=ab$
Squaring both sides and after simplification we get
$6(a+b)=18+ab$
And the area of the rectangle is $2ab$.I need to find $2ab,$I am not able to find it.Please help me.

continuing from your method: $AB=2b$ and $AM=MB=\sqrt{a^2+b^2}$

Now $$Ar(\Delta ABM)=ab=r_{ABM}\times s_{ABM}=4\frac{(2b+2\sqrt{a^2+b^2})}{2}$$ So

$$ab=4(b+\sqrt{a^2+b^2}) \tag{1}$$ and from your equation we have

$$3a+3(b+\sqrt{a^2+b^2})=ab \tag{2}$$ $\implies$

$$3a+\frac{3ab}{4}=ab \implies 3+\frac{3b}{4}=b \implies b=12$$

In the rectangle, if $R$ is the point on $AB$ opposite to $M_y$, then $AR=x$.

If P be the point of contact of AM with bigger circle, then $AP=x$, and $Q$ is point of contact of $AD$ with smaller then $AQ=x$, as pair of tangents from A
So in triangle BMC, $PM=x+3$, $BR=x$ and $BM=2x-3$.

From the Pythagorean equation,
$X=9$, $AB=18$, and $AD=12$.


Law of Cosine: Trigonometry Word Problem The longest side of a triangular lot is 21 meters. If another side of this lot is 17 meters and if the angle formed by thus side and third side is 95 degrees, how long is the third side?

Hint:  Draw a picture.  A 95 degree angle is a little more than right, and the longest side will be opposite it.  Label what you know.  Write down the law of cosines and identify the sides with the variables.  Solve.  What step is the problem?


Dimensions of $Pol(\mathbb{Z}_3)$ (polynomial vector space) How many dimensions does $Pol (\mathbb{Z}_3)$ have, where $Pol (\mathbb{Z}_3)$ is a vector space of polynomial functions with one variable ($f: x \mapsto \sum^n_{k=0} \lambda_kx^k$).

My "guess" would be: infinite. Since $$\lambda_{n+1}x^{n+1} = \sum^n_{k=0} \lambda_kx^k$$ only holds for $\lambda_1 = \lambda_2 = \dots = \lambda_n = \lambda_{n+1} = 0$. So, the search for the biggest linear independent generator set never ends. So $dim(Pol (\mathbb{Z}_3)) = \infty$.

Let me denote by $\mathbb{Z}_3[X]$ the vector space of polynomials over $\mathbb{Z}_3$ in one indeterminate and by $\mathcal{Fun}(\mathbb{Z}_3)$ the vector space of functions from $\mathbb{Z}_3$ to itself. Any polynomial $p(X) = \sum_{i=0}^n p_iX^i$ in $\mathbb{Z}_3[X]$ induces an element of $\mathcal{Fun}(\mathbb{Z}_3)$ by evaluation as follows
$$\mathbb{Z}_3 \to \mathbb{Z}_3, \qquad z \mapsto \sum_{i=0}^{n}p_i\cdot z^{i}.$$
This defines a surjective morphism of vector spaces
$$\mathbb{Z}_3[X] \to \mathrm{Pol}(\mathbb{Z}_3) \tag{$\star$}$$
(in fact, this is the definition of $\mathrm{Pol}(\mathbb{Z}_3)$) which however is not injective. For example, the polynomial $X^3$ is sent to the function
$$[0]_3 \mapsto [0]_3, \qquad [1]_3\mapsto [1]_3, \qquad [2]_3\mapsto [8]_3 = [2]_3$$
which is the same as the image of the polynomial $X$ (in fact, by Fermat's Little Theorem).

By resorting to your argument, you proved that $\mathbb{Z}_3[X]$ is infinite dimensional. However, you may verify that any polynomial of the form $q(X)(X^3-X)$ is mapped to the zero function on $\mathbb{Z}_3$ by the assignment above. Conversely, if a polynomial $p(X)$ is mapped to the zero function then it admits $[0]_3,[1]_3,[2]_3$ as roots and hence, by the Factor Theorem,
$$p(X) = q(X)(X-[0]_3)(X-[1]_3)(X-[2]_3) = q(X)(X^3-X).$$
Summing up, the vector space of polynomial function on $\mathbb{Z}_3$ is $3$-dimensional.

As a bonus, you may check that $(\star)$ in fact exists for every field $K$ (and not only for $K=\mathbb{Z}_3$). However, it is bijective if and only if $K$ is infinite.

If $\mathbb{Z}_3$ means $\mathbb{Z}/3\mathbb{Z}$, then the dimension is not infinite. This is because you explicitly write "polynomial functions". A function is defined by its values on all inputs, and since $\mathbb{Z}/3\mathbb{Z}$ has only 3 elements, a function can only take 3 values. Therefore, there are exactly 9 functions on $\mathbb{Z}/3\mathbb{Z}$, so the dimension has to be finite. To find the dimension, try to look at the functions $1$, $x$, $x^2$ and so on.

If by $\mathbb{Z}_3$ you mean the 3-adic integers, then your argument works.


Two players placing coins on a round table with the goal of making the last move I came across this riddle during a job interview and thought it was worth sharing with the community as I thought it was clever:

  Suppose you are sitting at a perfectly round table with an adversary about to play a game. Next to each of you is an infinitely large bag of pennies. The goal of the game is to be the player who is able to put the last penny on the table. Pennies cannot be moved once placed and cannot be stacked on top of each other; also, players place 1 penny per turn. There is a strategy to win this game every time. Do you move first or second, and what is your strategy?

JMoravitz has provided the answer (hidden in spoilers) below in case you are frustrated!

Yes, I've seen this one before.  Assuming exactly one penny is allowed to be placed per turn:

   Go first and place a penny in the dead center of the table.  From then on, any move your opponent makes, place a penny in the mirror opposite location (i.e. rotated 180 degrees).  It stands to reason that if your opponent's move was valid, yours will be too.  Hence, you will always have an available move if your opponent does.  Since the table is only finitely large, there can only be finitely many turns, hence you will eventually win.

A more complete proof:

Suppose the table is described using polar coordinates with the center of the table as the origin ($r=0$).

   My first move is to place at $r=0$.  When my opponent makes a legal move at $(r,\theta)$ I attempt to place a coin at $(r,\theta+180^\circ)$.

Claim: I am always allowed to do so and such a move will always be valid.

  Proof: Suppose otherwise.  Then that implies that either the target location is not on the table (in which case my opponent's previous move will also have not been on the table and therefore was also invalid), or that target location would have a coin overlap with another previously placed coin.  As it could not have been the coin that my opponent has just placed on his last turn (as it is $180^\circ$ away), that implies that those coins must have been placed previously.  However... since my moves are always playing $180^\circ$ away from my opponent, that implies that there should be the same situation on the other side of the table and that my opponents coin also is overlapping the corresponding mirrored coins and therefore my opponents move was invalid.  Either way, we reach a contradiction implying that if my opponents move was valid that my move is also guaranteed to be valid too.

Update: OP changed rules at revision 3.

My attempt as spoiler below.

  I move first and cover the whole table with pennies.


Is there an easier way to solve this logarithmic equation? $$2\log _{ 8 }{ x } =\log _{ 2 }{ x-1 } $$

Steps I took:

$$\frac { \log _{ 2 }{ x^{ 2 } }  }{ \log _{ 2 }{ 8 }  } =\log _{ 2 }{ x-1 } $$

$$\frac { \log _{ 2 }{ x^{ 2 } }  }{ 3 } =\log _{ 2 }{ x-1 } $$

$$\log _{ 2 }{ x^{ 2 } } =3\log _{ 2 }{ x-1 } $$

$$2\log _{ 2 }{ x } =3\log _{ 2 }{ x-1 } $$

$$\log _{ 2 }{ x } =\frac { 3 }{ 2 } \log _{ 2 }{ x-1 } $$

$$\log _{ 2 }{ x } =\log _{ 2 }{ (x-1)^{ \frac { 3 }{ 2 }  } } $$

This method seems to be very inefficient and I don't know how I would go from here. Can someone please point me in the right direction. Hints only please. No actual solution.

Using a change of base: $$\log_b a = \frac{\log_n a}{\log_n b}$$

Change $\log_8 x^2$ base to 2: $$\log_8 x^2 = \frac{\log_2 x^2}{\log_2 8} = \frac{2\log_2 x}{3} $$

Let $\log_2 x = a$

$$ \frac{2a}{3}= a-1 \ \implies \ a=3$$

$$\log_2 x = 3\ \implies \ x=2^3 = 8$$


Parking cars and vans into car-, van, and car/van- parking spots The problem is simply:

You have C cars and V vans, and parking spots of which X allows cars, Y allow vans, and Z allow either. How many ways could you park these vehicles? (Cars and vans are indistinguishable amongst themselves, parking spots are numbered.)

Is there a directly computable solution to this question (i.e. not involving sums or recursion)?

Work so far:

I have started by finding the correct count.

$\sum_{i=0}^X {X \choose i} {Z \choose X-i} {Y+Z-X+i \choose V}$

This puts some (i) cars in the car spots and the remainder in shared spots, for varying amounts of i and adds up for each choice of i. When i is specified, this is a directly computable combinatorics problem.

Expand the polynomial $(1+s)^X(1+t)^Y(1+s+t)^Z.$ Then look for the coefficient of $s^Ct^V$, which will give the number of ways to park $C$ cars and $V$ vans in the given devoted/either-or parking spaces. Admittedly this is not really a "closed form" solution, unless one knows how to extract the coefficient from the expansion.

I tried it for 2 each of car only, van only, and "either". For example there was $14$ as the coefficient of $st$ which would mean $14$ ways to park one car and one van in these six slots. To check manually, the car can go in a "car only" in two ways, after which there remain four positions for the van [so far 2*4=8], and on the other hand the car can go in one of the "either" spaces in two ways, this time leaving only three spots for the van, so this gives another 2*3=6. Altogether 8+6=14 which checks with the polynomial coefficient.

I'd be interested if the OP (or anyone) could work out some other cases to see if this polynomial expansion method is giving the right results, but it looks OK to me at this point. Again it would be of interest if a "closed form" could be obtained.

Assuming X intersection with Y is Z. X+Y-Z is the total number of potential spaces. If labellings of the parking spaces can be rearranged between types of vehicle allowed then (X+Y-Z)nCr(X-Z) is the potential car only spot numberings similarly for (Y-Z) but slightly less spaces to choose from so is (X+Y-Z-(X-Z))C(Y-Z) or (Y)nCr(Y-Z) and the remainder are forced to be in Z. So multiplying together our parking preset types have (X+Y-Z)nCr(X-Z) * (Y)nCr(Y-Z) choices. In X-Z each spot will develop 2 choices either a car or be empty so in total 2^(X-Z). Similarly for Y-Z spaces having 2^(Y-Z) potential van arrangements and in Z we have 3 options so is 3^Z. Since all intersections between the sets of spaces we have handled are empty sets the product of these is the number of potential car park states so together is,

(X+Y-Z)nCr(X-Z) * (Y)nCr(Y-Z) * 2^(X+Y-2*Z) * 3^Z

Unless Z was an independent category from X and Y in which case replace X-Z with X, Y-Z for Y, X+Y-Z for X+Y+Z and respective totals in the above.


How can I prove this statement by induction? I would like to prove that
$2^0 + 2^{-1} + 2^{-2} +...+ 2^{-n} \leq 2$
by induction.

I got the induction basis as $2^0 \leq 2$, but I couldn't find a induction step that could help me to prove that.

Any suggestions?
Thank you

This is one of those cases where in order for induction to work, you actually need to prove something stronger. Try replacing $\leq 2$ with $=2-2^{-n}$.


Working on Homotopy Equivalent Suppose I have spaces $X_{0}, X_{1},Y_{0},Y_{1}$ and $X_{0}$ is homotopy equivalent to $Y_{0}$ and $X_{1}$ is homotopy equivalent to $Y_{1}$.

I would like to show that $X_{0}$ x $X_{1} $ is homotopy equivalent to $Y_{0}$ x $Y_{1} $.

How do I approach this statement? I am fresh new to this topic so any input will be greatly appreciated.
Thanks!

Since $X_0 \simeq Y_0$ and $X_1 \simeq Y_1$, there are two pairs of functions
$$X_0 \overset{f_0}{\underset{g_0}{\rightleftarrows}} Y_0 \quad \text{and} \quad X_1 \overset{f_1}{\underset{g_1}{\rightleftarrows}} Y_1$$
such that the relevant composites are homotopic to the relevant identities.

From these functions, you can construct functions
$$X_0 \times X_1 \overset{f}{\underset{g}{\rightleftarrows}} Y_0 \times Y_1$$
where $f$ and $g$ are defined by
$$f(x_0,x_1) = (f_0(x_0),f_1(x_1)) \quad \text{and} \quad g(y_0,y_1)=(g_0(y_0),g_1(y_1))$$
for all $x_0,x_1,y_0,y_1$.

You need to check that $g \circ f \sim \mathrm{id}_{X_0 \times X_1}$ and $f \circ g \sim \mathrm{id}_{Y_0 \times Y_1}$.

In a similar way to how we constructed $f$ and $g$ from $f_0,f_1,g_0,g_1$, you can construct the two desired homotopies from the homotopies
$$g_0 \circ f_0 \sim \mathrm{id}_{X_0}, \quad g_1 \circ f_1 \sim \mathrm{id}_{X_1}, \quad f_0 \circ g_0 \sim \mathrm{id}_{Y_0} \quad \text{and} \quad f_1 \circ g_1 \sim \mathrm{id}_{Y_1}$$
I'll leave you to construct these homotopies.

acord your homotopy you to functions f0 and f1 and g0 g1 and g0(f0) in X0 and(g1(f1)) in X1 are nullhomotopys. I try to take the function f:=(f0,f1) and g:=(g0,g1) so g(f)=(g0(f0),g1(f1)) is nullhomotopy in X0*X1.


Arbitrarily discarding/cancelling Radians units when plugging angular speed into linear speed formula? Why is the radians implicitly cancelled?  Somehow, the feet just trumps the numerator unit.
For all other cases, you need to introduce the unit conversion fraction, and cancel explicitly.  Is it because radians and angles have no relevance to linear speed (v), so they are simply discarded?

Radians are dimensionless, i.e. there is nothing to cancel. This is because a radian is defined as a length over a length:
$$\frac{\text{radius}}{\text{circumference}}$$

So a radian is defined as

$1~\mathrm{radian} = \frac{\mathrm{radius}}{\mathrm{circumference}}$?

No, the radius over the circumference is a pure number equal to $1/(2\pi)$.

"One radian is the angle subtended at the center of a circle by an arc that is equal in length to the radius of the circle" is the more accurate definition. [1]

[1] http://en.wikipedia.org/wiki/Radian


Does the square or the circle have the greater perimeter? A surprisingly hard problem for high schoolers An exam for high school students had the following problem:

Let the point $E$ be the midpoint of the line segment $AD$ on the square $ABCD$. Then let a circle be determined by the points $E$, $B$ and $C$ as shown on the diagram. Which of the geometric figures has the greater perimeter, the square or the circle?

Of course, there are some ways to solve this problem. One method is as follows: assume the side lengths of the square is $1$, put everything somewhere on a Cartesian coordinate system, find the midpoint of the circle using the coordinates of $E$, $B$ and $C$, then find the radius of the circle, and finally use the radius to calculate the circle's circumference and compare it to the perimeter of the square.

The problem with that method is that ostensibly this problem is supposed to be very simple; it shouldn't require the student to know the formula for the midpoint of a circle given three coordinates. Therefore the question here is: does there exist a simple way to solve the problem without knowing any complicated geometric formulas?

Perhaps the examiner intended the students to notice the square is determined by a $(3, 4, 5)$ triangle, because $3 + 5 = 4 + 4$ (!):

Consequently, as several others have noted,
$$
\frac{\text{perimeter of the circle}}{\text{perimeter of the square}}
  = \frac{5 \cdot 2\pi}{4 \cdot 8} = \frac{\pi}{3.2}
 < 1.
$$

For an approach less dependent on inspiration, taking the origin of the coordinate system at the center of the circle seems easier than placing the origin at the center of the square. Without loss of generality, assume the circle has unit radius:

Equating the lengths of the horizontal and vertical sides of the square in this diagram, we read off
$$
x + 1 = 2y\quad\text{(or $x = 2y - 1$).}
$$
Invoking the Pythagorean theorem and substituting the preceding line,
\begin{align*}
0 &= x^{2} + y^{2} - 1 \\
  &= (2y - 1)^{2} + y^{2} - 1 \\
  &= 5y^{2} - 4y \\
  &= y(5y - 4).
\end{align*}
Clearly $y \neq 0$, so $y = 4/5$, $x = 3/5$, and we notice the Examiner's Favorite Triangle.

In this solution the green line is a perpendicular bisector of the blue line. Thus it intersects the y-axis at the center of the circle.

(Sorry for using absolute size of everything. I beg your indulgence in this regard. The idea should still be clear, if not clearer.)


Find the maximum value of the expression $ \frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}$ Find the maximum value of the expression $ \frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}$  where $x, y, z$ are real numbers satisfying the condition $x + y + z = 1 $.
My Attempt
$$
\begin{align}
\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2}&=\frac{1}{x+\frac{1}{x}}+\frac{1}{y+\frac{1}{y}}+\frac{1}{z+\frac{1}{z}}\\&\geq \frac{(1+1+1)^2}{x+\frac{1}{x}+y+\frac{1}{y} + z+\frac{1}{z}}\\&=\frac{9}{x+y+z +(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})}
\end{align}
$$
From AM-HM
$$(x+y+z)(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \geq 3^2 $$
$$\implies (\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \geq 9 $$
$$\implies (x+y+z)+ (\frac{1}{x}+\frac{1}{y}+\frac{1}{z}) \geq 10$$
From this point i am not being able o proceed as inequality signs are getting mixed. Can anyone help out? Thanks in advance.

You've already been given a hint to try Jensens inequality, which works fine in this case. Alternatively by AM-GM
$$9\frac{x}{9+(3x)^2}\leqslant 9\frac{x}{10(3x)^{2/10}}=\frac3{10}(3x)^{4/5}$$
Sum that up and observe by  power means:
$$\sum (3x)^{4/5} \leqslant 3\left(\frac{3x+3y+3z}3\right)^{4/5}=3$$
Equality is possible when $x=y=z=\frac13$, so these inequalities give you the maximum.

Ok rugi I will solve it using derivatives for your question consider x=y=z because for any condition like x+y+z=a we need to consider x=y=z to get maxima .So answer is 9÷10=0.9.This is for a <3.And when the degree in the numerator is less than the degree in denominator in which the value is to be found out


Unexpected yet accessible examples of the probabilistic method What are some examples of applications of the probabilistic method to areas where it might not be expected? I was just wondering how feasible it would be to present different ideas about the probabilistic method to those who don't necessarily have a lot of math background.

One really interesting example I found was hat games, and the applications of Hamming codes and probabilities to find really slick ways to think about different hat puzzles. I was just wondering if you had any other specific examples that don't take a lot of build up to get to some interesting results.

I plan to introduce the basics of probability and expected value, but I don't think I can get to topics like the Local Lemma and Martingales. Also, I don't think I'd like to go very deep into graphs or analysis either, given that a lot of people aren't very familiar with those concepts.

Honestly, I find almost every example of the probabilistic method unexpected. I think that my favourite cute example comes in the following easy-to-state puzzle:

"Prove that any 10 points in the plane can be covered with disjoint open unit discs."

(At first glance, this might seem super obvious, but to see that it isn't true for arbitrarily many points consider like a million points distributed in the unit square, which can't be covered)

The solution by the probabilistic method comes from taking an arbitrary collection of 10 points $x_1, \dotsc, x_{10}$ and dropping a hexagonal lattice of unit discs uniformly at random on the plane (this can be made precise with a little thought).

Letting $A_i$ be the event that $x_i$ is not covered by any of the discs in the lattice, we have that $\mathbf{P}(A_i)$ is the density of "holes" between the discs in the lattice. It is a straightforward geometric exercise to see that the area of each "hole" is $\sqrt{3} - \pi/2$ and so that their density is
$$
\frac{\sqrt{3} - \pi/2}{\sqrt{3}} \approx 0.0931.
$$

In particular, we have that
\begin{align*}
\mathbf{P}(\text{All points are covered})
&= \mathbf{P}\Big(\bigcap_{i=1}^{10} A_i^c\Big) \\
&= 1 - \mathbf{P}\Big(\bigcup_{i=1}^{10} A_i\Big) \\
&\geq 1 - 10 \cdot 0.0931 \\
&> 0.
\end{align*}

Since this probability is positive, the probabilstic method yields that there must be some orientation of the lattice, and so some collection of disjoint open unit discs that cover all 10 points.

This application seems to be beyond the scope of your question since it relies on Chebyshev's inequality, some understanding of the binomial distribution and a bit of algebraic manipulation, but I'd like to add it nonetheless since it is still relatively elementary and certainly unexpected - it may be of interest to other users ending up here through a search.

Weierstrass approximation theorem: Given a continuous function $f : [0, 1] \to \mathbb{R}$ and $\epsilon > 0$, we can construct a polynomial $p(x) = a_n x^n + a_{n-1} x^{n-1} + \dots + a_0$ such that for all $x\in[0,1]$ we have $|f(x) - p(x)| < \epsilon$.

A short proof (less than two pages) of this appears in Alon & Spencer's The Probabilistic Method, as one of the 'probabilistic lens'-intermezzos. The original proof is from Bernstein (1912) (written in French).


Using Matlab's curve fitting tool, I got the following fit:

with the parameters as below:
General model Exp2:
     f(x) = a*exp(b*x) + c*exp(d*x)
Coefficients (with 95% confidence bounds):
       a =       3.513  (3.237, 3.788)
       b =     -0.1935  (-0.2215, -0.1655)
       c =      0.9278  (0.7994, 1.056)
       d =   -0.004471  (-0.006014, -0.002928)

and the goodness of fit as below:
Goodness of fit:
  SSE: 0.2513
  R-square: 0.9903
  Adjusted R-square: 0.9893
  RMSE: 0.09473

which seems reasonable.


What are better approximations to $\pi$ as algebraic though irrational number? I know that $\pi \approx \sqrt{10}$, but that only gives one decimal place correct. I also found an algebraic number approximation that gives ten places but it's so cumbersome it's just much easier to just memorize those ten places.

What's a good approximation to $\pi$ as an irrational algebraic number (or algebraic integer if possible) that is easier to memorize than the number of places it gives correct?

EDIT: Algebraic number preferably of low degree, such as $2$ or $3$ (quadratic or cubic).

(expanding my comments)

Let's start with the fraction $\;\dfrac{355}{113}\,$ easy to remember with something like :
"doubling the odds to be near the pi" (whatever this may mean...).

It is easy to find starting with the continued fraction of $\pi$ and stopping just before the (relatively) large term $292$ :
\begin{align}
\pi&=[3; 7, 15, 1\color{#00ff00}{, 292, 1, 1, 1, 2, 1, 3, 1, 14,\cdots}]\\
\pi&\approx \frac{355}{113}\approx 3.141592\color{#808080}{035}\\
\end{align}

My next step will be to compute the continued fractions of the first powers of $\pi\,$ and stop the expansion before the first large term (as previously) to get :

$$\frac {355}{113},\; \sqrt{\frac{227}{23}},\; \sqrt[3]{31},\;\sqrt[4]{\frac{2143}{22}},\;\sqrt[5]{306},\cdots, \;\sqrt[11]{294204},\cdots$$

After the first power the most interesting term was the fourth :
\begin{align}
\pi^4&=[97; 2, 2, 3, 1\color{#00ff00}{, 16539, 1, 6, 7, 6,\cdots}]\\
\pi^4&\approx \frac{2143}{22}\\
\pi&\approx \sqrt[4]{\frac{2143}{22}}\approx 3.14159265\color{#808080}{258}\\
\end{align}

Mnemonic : think at "three ways to reverse two two" that I'll note $\pi\approx \sqrt{+ \negthickspace+/}$ :

Power way: two $\sqrt{}\;$ to reverse the double squaring $^2$ $^2$.
Incremental way : reverse two times two terms of the $2\times 2$ terms $\;\underbrace{12}\underbrace{34}$
Divide by $\,22$.

This solution is interesting because of the large (omitted) $16539$. Should we incorporate this term in the c.f. then the next numerator and denominator would have around $4$ additional digits (since $\log_{10}(16539)\approx 4.2\;$ and from the method to obtain the next fraction in the first link).
The precision will be better with this supplementary term (say $4.3$ digits more) but we needed $4+4$ more digits for this. Without this term we used $4+2=6$ digits for a result of $10$ digits (excellent), with this term we have $8+6=14$ digits for a result of $14$ digits (average for a c.f.).

Searching the largest terms at the beginning of a c.f. (excluding the first non-zero term) should thus be rather interesting! Unfortunately c.f. coefficients as large as $16539$ are rather uncommon.

This result was found by Ramanujan and is given too by Mathworld with many others.
$$-$$
Some additional results :

A palindrome for the fractional part of $\pi$ : $\frac 1{\large{\sqrt[5]{17571}}}\approx 0.1415926\color{#808080}{48}$ (with two more terms this becomes $\sqrt[5]{\dfrac{296}{5201015}}\approx 0.141592653589\color{#808080}{63}$). Another one : $\;\dfrac 1{\sqrt[8]{6189766}} \approx 0.141592653\color{#808080}{64}$.

We may too search continued fractions $\dfrac{\log\pi}{\log n}\,$ to obtain :
\begin{align}
7^{10/17}&\approx 3.141\color{#808080}{35}\\
6^{23/36}&\approx 3.1416\color{#808080}{09}\\
7^{58701/99785}&\approx  3.1415926535\color{#808080}{9651}\\
\end{align}

Other random solutions perhaps nearer to OP's question (with some usual c.f. for reference) :
\begin{align}
\frac{22}7 &\approx  3.14\color{#808080}{2857}\\
\frac{8.5^2}{23} &\approx  3.141\color{#808080}{30}\\
\sqrt[3]{31}&\approx 3.141\color{#808080}{38}\\
\sqrt{51}-4  &\approx  3.141\color{#808080}{428}\\
\sqrt{4508}-64   &\approx  3.141\color{#808080}{64}\\
4-\sqrt{\frac {14}{19}} &\approx 3.141\color{#505050}{60}\color{#808080}{49}\\
7-\left(\frac{55}{28}\right)^2 &\approx  3.1415\color{#808080}{816}\\
1+\left(\frac{60}{41}\right)^2 &\approx  3.1415\color{#808080}{82}\\
\sqrt{14434}-117 &\approx  3.1415\color{#808080}{83}\\
2+\sqrt[17]{9.5} &\approx  3.14159\color{#808080}{78}\\
5-\sqrt[5]{22+\frac{1}6} &\approx  3.14159\color{#808080}{62}\\
\sqrt{\frac{1961}2}-19 &\approx  3.1415\color{#808080}{898}\\
2+\sqrt[8]{\frac{75}{26}} &\approx  3.141592\color{#808080}{19}\\
\frac{355}{113} &\approx 3.141592\color{#808080}{92}\\
\sqrt[11]{294204} &\approx 3.1415926\color{#808080}{36}\\
\left(\sqrt{\frac {1731}{76}}-3\right)^2 &\approx 3.1415926\color{#808080}{65}\\
\sqrt{6}+\sqrt[3]{\frac {61}{184}}&\approx 3.1415926\color{#808080}{45}\\
\sqrt{35}-\sqrt[3]{\frac{6215}{291}} &\approx 3.14159265\color{#808080}{266}\\
\sqrt[4]{\frac{2143}{22}}&\approx 3.14159265\color{#808080}{258}\\
5-\sqrt[11]{913+\frac 16} &\approx  3.141592653\color{#808080}{37}\\
\sqrt{5}+\sqrt[4]{\frac{2323}{3455}} &\approx 3.141592653\color{#808080}{436}\\
\sqrt{4508-\frac 1{153}}-64 &\approx  3.1415926535\color{#808080}{28}\\
\sqrt[4]{\frac{788453}{95}}-\sqrt{41} &\approx  3.1415926535\color{#808080}{918} \\
\sqrt[4]{\sqrt{\frac{1087906}{63}}-34}&\approx 3.14159265358\color{#808080}{876}\\
\frac{5419351}{1725033}&\approx 3.141592653589\color{#808080}{815}\\
\sqrt{7}+\sqrt[8]{\frac{94680}{25912921}} &\approx 3.141592653589793\color{#808080}{309}\\
\sqrt{\sqrt{\frac{10521363651}{311209}}-174} &\approx 3.141592653589793238\color{#808080}{01}\\
\frac{21053343141}{6701487259}&\approx 3.141592653589793238462\color{#808080}{38}\\
\sqrt{\sqrt{\frac{20448668456155}{3958899937}}-62}  &\approx 3.14159265358979323846264338\color{#808080}{5}\\
\sqrt{12}-\sqrt[3]{\frac{626510899334}{18676834489131}} &\approx 3.1415926535897932384626433832\color{#505050}{80}\color{#808080}{4}
\end{align}

We could too use the integer relation algorithms as in Will Jagy's answer or this one but this seems more cumbersome for this problem.

It is not a low degree polynomial, but easy to remember for sure.
We know that $$\frac{1}{1+x}=1-x+x^2-x^3+\cdots$$
Substituting $x^2$, $$\frac1{1+x^2}=1-x^2+x^4-x^6+\cdots$$
But we also know that $\int\frac1{1+x^2}dx=\arctan x$.

So let us integrate both sides (from $x=0$ to $x=y$), $$\arctan y=y-\frac{y^3}3+\frac{y^5}{5}-\frac{y^7}{7}+\cdots$$
Substitute $y=1$ and we get $$\pi=4(1-\frac13+\frac15-\frac17+\cdots)$$.