Random diffusion coefficient in the Fourier equation I'm stuck on the following simple problem: It's given the Fourier equation:
$$\partial_t{u(x,t)}=\partial_x[k(t)\partial_xu(x,t)]$$where the diffusion coefficient $k(t)$ is a random variable with a gaussian probability density function: $N(0,\sigma)$. Because $k(t)$ is not depending on $x$ we can write the previous equation as:
$$\partial_t{u(x,t)}=k(t)\partial_{xx}u(x,t)$$
Given the boundary conditions:
$$u(0,t)=u_0,u(L,0)=u_L=u(x,0)$$
How can I solve (if it's possible) this stochastic PDE?
Thanks in advance.

Of course we use separation of variables:

Let $u(x,t)=X(x)T(t)$ ,

Then $X(x)T'(t)=k(t)X''(x)T(t)$

$\dfrac{T'(t)}{k(t)T(t)}=\dfrac{X''(x)}{X(x)}=-s^2$

$\begin{cases}\dfrac{T'(t)}{T(t)}=-s^2k(t)\\X''(x)+s^2X(x)=0\end{cases}$

$\begin{cases}T(t)=c_3(s)e^{-s^2\int_0^tk(t)~dt}\\X(x)=\begin{cases}c_1(s)\sin xs+c_2(s)\cos xs&\text{when}~s\neq0\\c_1x+c_2&\text{when}~s=0\end{cases}\end{cases}$

$\therefore u(x,t)=C_1x+C_2+\int_0^\infty C_3(s)e^{-s^2\int_0^tk(t)~dt}\sin xs~ds+\int_0^\infty C_4(s)e^{-s^2\int_0^tk(t)~dt}\cos xs~ds$

$u(0,t)=u_0$ :

$C_2+\int_0^\infty C_4(s)e^{-s^2\int_0^tk(t)~dt}~ds=u_0$

$\int_0^\infty C_4(s)e^{-s^2\int_0^tk(t)~dt}~ds=u_0-C_2$

$C_4(s)=(u_0-C_2)\delta(s)$

$\therefore u(x,t)=C_1x+C_2+\int_0^\infty C_3(s)e^{-s^2\int_0^tk(t)~dt}\sin xs~ds+\int_0^\infty(u_0-C_2)\delta(s)e^{-s^2\int_0^tk(t)~dt}\cos xs~ds=C_1x+C_2+\int_0^\infty C_3(s)e^{-s^2\int_0^tk(t)~dt}\sin xs~ds+u_0-C_2=C_1x+u_0+\int_0^\infty C_3(s)e^{-s^2\int_0^tk(t)~dt}\sin xs~ds$

$u(x,0)=u(L,0)=u_L$ :

$C_1x+u_0+\int_0^\infty C_3(s)\sin xs~ds=u_L$

$\int_0^\infty C_3(s)\sin xs~ds=-C_1x+u_L-u_0$

$C_3(s)=C_1\delta'(s)+\dfrac{2(u_L-u_0)}{\pi s}$

$\therefore u(x,t)=C_1x+u_0+\int_0^\infty\biggl(C_1\delta'(s)+\dfrac{2(u_L-u_0)}{\pi s}\biggr)e^{-s^2\int_0^tk(t)~dt}\sin xs~ds=C_1x+u_0+\int_0^\infty C_1\delta'(s)e^{-s^2\int_0^tk(t)~dt}\sin xs~ds+\int_0^\infty\dfrac{2(u_L-u_0)e^{-s^2\int_0^tk(t)~dt}\sin xs}{\pi s}ds=C_1x+u_0-\biggl.\dfrac{d}{ds}(C_1e^{-s^2\int_0^tk(t)~dt}\sin xs)\biggr|_{s=0}+\int_0^\infty\dfrac{2(u_L-u_0)e^{-s^2\int_0^tk(t)~dt}\sin xs}{\pi s}ds=C_1x+u_0-C_1x+\int_0^\infty\dfrac{2(u_L-u_0)e^{-s^2\int_0^tk(t)~dt}\sin xs}{\pi s}ds=u_0+\int_0^\infty\dfrac{2(u_L-u_0)e^{-s^2\int_0^tk(t)~dt}\sin xs}{\pi s}ds$

As far as I understand your question, you can just solve the equation as if $k$ was deterministic and then integrate with respect to the law of $k$.

By the way, you say that

  $k(t)$ is a random variable with a gaussian probability density function: $N(0,\sigma)$.

Does this mean that $k(t)$ is a (random) constant ?


Similarity of $3\times3$ matrices via Jordan canonical form Let matrices
$$A=\begin{bmatrix}
-3&3&-2\\
-7&6&-3\\
1&-1&2
\end{bmatrix},\quad
B=\begin{bmatrix}
0&1&-1\\
-4&4&-2\\
-2&1&1
\end{bmatrix},\quad
C=\begin{bmatrix}
0&-1&-1\\
-3&-1&-2\\
7&5&6
\end{bmatrix},\quad
D=\begin{bmatrix}
0&1&2\\
0&1&1\\
0&0&2
\end{bmatrix}
$$
My goal is to determine which of them are similar to each other. First, we observe that
$$\det(A)=\det(B)=\det(C)=4\neq\det(D)=0$$
Thus, $D$ cannot be similar to any of $A,B,C$. $\det(\lambda I-A)=0$ and similarly for $B,C$. Denote $\sigma(X)=\operatorname{spectrum of matrix }X$. With a slight abuse of notation, to denote that $\lambda_2=2$ has algebraic multiplicity of $2$, I found
$$\sigma(A)=\sigma(B)=\sigma(C)=\{\lambda_1=1,\lambda_2=2,\lambda_2=2\}$$
This means that all of them have the same Jordan Canonical Form which is (up to permutations of the diagonal Jordan blocks)
$$
J_A=J_B=J_C=\begin{bmatrix}
\lambda_1&0&0\\
0&\lambda_2&1\\
0&0&\lambda_2
\end{bmatrix}
=
\begin{bmatrix}
1&0&0\\
0&2&1\\
0&0&2
\end{bmatrix}
$$
Hence, $A,B,C$ are similar.
Is this justification correct? If yes, is there any faster way/trick to get to this result? Finding the eigenvalues by solving a 3-rd degree equation took me some time.

$B$ diagonalizes

$$
\frac{1}{2}
\left(
\begin{array}{rrr}
4 & -2 & 2 \\
-4 & 3 & -2 \\
-2 & 1 & 0 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & 1 & -1 \\
2 & 2 & 0 \\
1 & 0 & 2 \\
\end{array}
\right) = I
$$
$$
\frac{1}{2}
\left(
\begin{array}{rrr}
4 & -2 & 2 \\
-4 & 3 & -2 \\
-2 & 1 & 0 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
0 & 1 & -1 \\
-4 & 4 & -2 \\
-2 & 1 & 1 \\
\end{array}
\right)
\left(
\begin{array}{rrr}
1 & 1 & -1 \\
2 & 2 & 0 \\
1 & 0 & 2 \\
\end{array}
\right) =
\left(
\begin{array}{rrr}
1 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 2 \\
\end{array}
\right)
$$

Regardless, this was going to require a fair amount of work by hand. I am not so much a fan of doing things quickly as of doing things well and including alternate methods that may indicate errors. In this case, one may confirm that $B^2 - 3 B + 2 I = 0.$ For $A$ and $C$ the minimal polynomial agrees with the characteristic polynomial, which means that each eigenvalue occurs in a single Jordan block, and since $2$ has higher multiplicity this means they do not diagonalize. In contrast, for $B$ we see that $2$ must occur in more than one Jordan block. In fact, there are two, each dimension one, meaning there is a basis of two eigenvectors for eigenvalue $2.$ I have typed in one way  to diagonalize $B.$ The matrix on the right has columns that are all eigenvectors


The number of limit points of the set $\left\{\frac1p+\frac1q:p,q \in \Bbb N\right\}$ is which of the following: I am stuck on the following problem:

The number of limit points of the set $\left\{\frac1p+\frac1q:p,q \in \Bbb N\right\}$ is which of the following:

$1$

$2$

Infinitely many

Finitely many

If I take $p$ to be fixed (say=$k$) and let $q \to \infty$, then the limit point is given by $\frac{1}{k}$. Since $k$ is an arbitrary natural number, the number of limit points is infinite. The same case can be continued after taking $q$ to be fixed (say=$k_1$). I think option 3 is the right choice. Am I on the right track? Can someone give further explanation?

Your answer is correct. You have shown that there are infinitely many limit points, which is enough information to answer the multiple choice question.

You can prove that the set of limit points of this set is $\{0\}\cup\{1/k:k\in\mathbb N\}$. Since $\frac{1}{k} = \frac{1}{2k} + \frac{1}{2k}$, that means that $0$ is the only limit point that is not in your original set.

I think the answer is $\{0\}$ because limit points means nbd of open interval contain at least one points other than point. Let $k=1,2,3,...$ then $\frac{1}{k} = 1,1/2,1/3...$  the nbd of $\{0\}$contained in an open interval other than $\{0\}$. So, the limit point of $\frac{1}{k}$ is $\{0\}$.


In how many ways 11 items can be distributed among 3 peoples such that, sum of items received by any two is more than the third In how many ways $11$ items can be distributed among $3$ peoples such that, number of items received by any two person is more than the number of items received by the third person.

I have tried with case by case. But it will be cumbersome when the number of people and no. of items are large. Is there any general strategy to solve this problem, which can be applied for large numbers?

Under the assumption that the items are indistinguishable and the people are distinct, let the ordered triple $(a,b,c)$ denote the number of items distributed to the first, second, and third person, respectively. Then determine the number of ordered triples of integers $(a, b, 11-a-b)$ such that $$1 \le a \le b \le 11-a-b < a+b.$$  So the rightmost inequality implies $a+b > 11/2$, which in turn implies $b \ge 3$.  We must also have $a \le 11/3$, or $a \le 3$.  Finally, we must have $b \le 5$ since $2b \le 11 - a \le 11 - 1 = 10$.  So it follows that our candidate triples are $$(a, 3, 8-a), (a, 4, 7-a), (a, 5, 6-a),$$ and the last case can only have $a = 1$ resulting in $(1, 5, 5)$.  The second case can only have $a \in \{2, 3\}$, giving $(2, 4, 5)$ and $(3, 4, 4)$, respectively; and the first case admits $a = 3$, corresponding to $(3, 3, 5)$.  These comprise the only nondecreasing positive integer triples for which the sum is $11$ and the sum of any two elements exceeds the third.  Three of them can be permuted in $3$ ways, and the fourth can be permuted in $6$ ways; hence the total number of triples is $3(3) + 6 = 15$.

Under the assumptions of heropup the sequence is OEIS A008795.  It begins
$$1, 0, 3, 1, 6, 3, 10, 6, 15, 10, 21, 15, 28, 21, 36, 28, 45, 36, 55, 45, $$
and is given by $$a(n)= \begin {cases} {\frac {n-2}2 \choose 2}& n \text { even}\\
{\frac {n-1}2 \choose 2}& n \text { odd}\end {cases}$$


Find $\lim_{m \to \infty} m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right]$ where $N$ is Poisson random variable with mean $m$ Let $N$ be a Poisson random variable with the mean parameter $m$.
We are interested in finding the following limit
\begin{align}
\lim_{m \to \infty} m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right].
\end{align}
Things that I tried:
First, I found an upper bound by using Jensen's inequality
\begin{align}
\lim_{m \to \infty} m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right]\le  \lim_{m \to \infty} m  \log\left( \frac{ E \left[ N  \right]+\frac{1}{2}}{m+1} \right)= \lim_{m \to \infty} m  \log\left( \frac{ m+\frac{1}{2}}{m+1} \right)=-\frac{1}{2}.
\end{align}
For the lower bound I tried to use the CLT argument from a related question in here:
\begin{align}
m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right]= m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) | N \ge \frac{m}{k} \right] P[ N \ge \frac{m}{k} ]+ m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) | N < \frac{m}{k} \right] P[ N < \frac{m}{k} ]
\end{align}
for some $k>0$.  Via the CLT it can be shown that $P[ N < \frac{m}{k} ] \to 0$ for all $0<k <1$, and we have that
\begin{align}
\lim_{m\to \infty}m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) \right]= \lim_{m\to \infty}m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) | N \ge \frac{m}{k} \right] .
\end{align}
Now if we use the lower bound at this point, we get
\begin{align*}
 \lim_{m\to \infty}m E \left[ \log\left( \frac{N+\frac{1}{2}}{m+1} \right) | N \ge \frac{m}{k} \right]  \ge   \lim_{m\to \infty}m  \log\left( \frac{ \frac{m}{k}+\frac{1}{2}}{m+1} \right) =-\infty,
\end{align*}
for all $k \in (0,1)$.

We will show that $lim_m m E\ln \frac{N+ \frac12}{m+1} = -1$.
We have $$m E\ln \frac{N+ \frac12}{m+1} = Em\ln \frac{N+ \frac12}{m+\frac12} + Em\ln \frac{m+ \frac12}{m+1} = Em\ln \frac{N+ \frac12}{m+\frac12} + m \ln (1 - \frac{0.5}{m+1})$$
and thus
$$lim_m m E\ln \frac{N+ \frac12}{m+1} = \lim_m E\ln \frac{N+ \frac12}{m+\frac12}  -\frac12.$$
It's sufficient to show that $\lim_m E\ln \frac{N+ \frac12}{m+\frac12} \to -\frac{1}2$.
Condsider $m \ge 100$, $1+Y_i \sim Pois(1)$, $\xi_m \sim Pois(m - [m])$. Thus $\sum_{i=1}^{[m]} (Y_i + 1) + \xi_m = N$ in distribution.
Fix $x$. $$P(\frac{\xi_m}{\sqrt{m}} \ge x) \le \frac{E \frac{\xi_m}{\sqrt{m}} }{x} \to 0, m \to \infty$$
thus $ \frac{\xi_m}{\sqrt{m}}  \to 0$ in probability.
It follows from CLT that $\frac{ \sum_{i=1}^{[m]} Y_i }{\sqrt{[m]}} \to N(0,1)$. Hence
$$\frac{N-m}{\sqrt{m}} = \frac{ \sum_{i=1}^{[m]} Y_i }{\sqrt{[m]}}\frac{\sqrt{[m]}}{\sqrt{m}}   +  \frac{\xi_m}{\sqrt{m}} \to N(0,1).$$
Put $$\eta_m = \frac{m}{m+\frac12}\cdot \frac{N-m}{\sqrt{m}} = \frac{\sqrt{m}(N-m)}{m+\frac12}$$
Then $\eta_m \to N(0,1)$  and $\eta_m \ge \frac{m}{m+\frac12}\frac{-m}{\sqrt{m}} \ge -\sqrt{m} + \frac{1}{3\sqrt{m}}$.
Further,
$$\lim_m E\ln \frac{N+ \frac12}{m+\frac12} = E m \ln(1 + \frac{\eta_m}{\sqrt{m}}).$$
Put $g(x) = \ln(1+x) - (x - \frac{x^2}2)$. Then
$$ E m \ln(1 + \frac{\eta_m}{\sqrt{m}}) = E m(  \frac{\eta_m}{\sqrt{m}} - \frac12 ( \frac{\eta_m}{\sqrt{m}})^2 + g( \frac{\eta_m}{\sqrt{m}}) ) = - \frac12 + Emg( \frac{\eta_m}{\sqrt{m}})  $$
and it's sufficient to show that $Emg( \frac{\eta_m}{\sqrt{m}})  \to 0$.
Put $A = [-\sqrt{m}+ \frac{1}{3\sqrt{m}}, - m^{\frac{3}{8}}], B= [-m^{\frac{3}{8}}, m^{\frac{3}{8}}], C = [m^{\frac{3}{8}}, \infty)$. Then
$$|Emg( \frac{\eta_m}{\sqrt{m}})| \le Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in A} + Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in B} + Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in C} \le $$
$$\le m \max_{x \in A} |g(\frac{x}{\sqrt{m}})|P(\eta_m  \in A)  + m \max_{x \in B} |g(\frac{x}{\sqrt{m}})|P(\eta_m  \in B) + Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in C}$$
As $m\ge 100$, we know that  $\max_{x \in A} |g(\frac{x}{\sqrt{m}})| = |\ln(\frac{1}{3m})| \le 2 \ln m $.
Further, by Taylor formula
$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} \cdot \frac{1}{(1+\theta_x x)^3}$, $0 \le \theta \le 1$. Thus
$g(x) = \frac{x^3}{3} \cdot \frac{1}{(1+\theta_x x)^3}$
and $\max_{x \in B} |g(\frac{x}{\sqrt{m}})| =  O(\frac{1}{ m^{\frac98 }  })$.
We have $|g(x)| \le 2x^2$ for $x > 0$.
Thus $Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in C} \le E m 2 (\frac{\eta_m}{\sqrt{m}})^2 I_C = 2 E \eta_m^2 I_C$.
Further there will be heuristic arguments.
$2 E \eta_m^2 I_C \approx 2 E N^2(0,1) I_{(N(0,1) > m^{\frac38})} \to 0$ hence $Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in C} \to 0$.
$P(\eta_m \in A) \approx P(N(0,1) \in A) \le P(N(0,1) \le -m^{\frac38})$ and thus
$Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in A}  \le m \max_{x \in A} |g(\frac{x}{\sqrt{m}})|  P(N(0,1) \le -m^{\frac38}) \to 0$.
It's sufficient to show that$Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in B} \to 0$.
But $Em|g( \frac{\eta_m}{\sqrt{m}})| I_{\eta_m \in B} \le m \max_{x \in B} |g(\frac{x}{\sqrt{m}})| = m  O(\frac{1}{ m^{\frac98 }  }) = o(1)$, q.e.d.

This is an respectable answer, what a median in a Poisson distribution is. Your answer targets large, infinite such medians. This gives the formulae: Poisson distribution. This definition also relates the E symbol to the distribution. Expected value makes clear which operation is expected.
$$E[X]=\int_{0}^{\infty} x P(x)dx$$
$X$ is any function of the distributed parameter.
$$E[\log(\frac{N+\frac{1}{2}}{+1})]=\int_{0}^{\infty} \log(\frac{N+\frac{1}{2}}{+1}) P(N)dN$$
Now You named Poisson random variable so,
$$P(\lambda,N)=\frac{\lambda^N\exp(-\lambda)}{N!}$$
Median and $\lambda$ are : $m\approx\lambda-\frac{1}{3}-\frac{0.05}{\lambda}$. Since Your are interested in large medians only:
$$P(m,N)\approx\frac{(m+\frac{1}{3})^N\exp(-(m+\frac{1}{3}))}{N!}$$
For sufficiently high values the Poisson distribution is approximated by the standard distribution. This can only be evaluated numerically or literature.
A plot of the function over m looks like this:

$$\lim_{m->\infty} m E[\log(\frac{N+\frac{1}{2}}{+1})]=0$$
This might give some safety if the median, mean or parameter $\lambda$ has to be estimated. Since $N$ is natural a variation of $\frac{1}{2}$ is rather common as uncertainty. $\lambda$ is real and the variation in first order is estimated $1$. We approximate the entropies roughly and integration the difference between both over the "true" probability distribution with the mean as parameter. In first order the coefficient vanishes and show how good the approximation really is, if alone the Poisson distribution is selected.
This does not compare to other distribution such as the standard distribution. $\frac{1}{2}$ and $1$ might too named information uncertainty. The graph reflects teh general property of the Poisson distribution that goes over into the standard distribution. Mind this plot might have numerical errors. This is crude because it replaces the median for the parameter already $\lambda$. Use Poisson distribution for a good completion of formulae used.


Can infinity be divided by anything? I'm in ninth grade and I've been thinking about this for a while. It's related to a question that came to my mind, namely which is the highest number you can divide 11 by. Is "infinite" a number and is it the highest number by which 11 can divide it?

Thinking and playing with infinity is a popular thing to do, especially for people your age.  To do so correctly however, you need to understand what we actually mean by infinity and what we mean by a "number."

Note: For the remainder of this post, by "division" and "divided by" I will be referring to the usual division operator rather than the related concepts from number theory.

Infinity is not a "real number" (that is not to say it doesn't exist in some contexts, but it does not belong to the set of numbers known as 'the real numbers') so we are not even allowed to refer to operations involving infinity.  It just flat out doesn't make sense in this context.  There is no largest "real number" and so there is no "highest (real) number" that $11$ can be divided by.

Since you are talking about "dividing by infinity" then you are probably working in the extended real numbers rather than the real numbers.  In such a context, yes infinity is the "largest" extended real number and $11$ can indeed be divided by it.  In fact, every real number when treated as an extended real number instead can be divided by infinty and the result would be zero.  In that context, yes... you are correct.

The extended real numbers however is not the usual context to be working in.  If the question was asked "does there exist a largest number that $11$ can be divided by" without any additional clarification about context then the answer would be no.

(Note: infinity and negative infinity can not be divided by infinity)

It is really great to be thinking of stuff like this.

You will find people all around you who will say infinity is not a number and they will stop there, but there are a set of math classes in your future (Calculus) that are all about dealing with numbers that are virtually infinite--they call them limits.

All those people who say that it's absolutely impossible to deal with infinity will find it perfectly fine to deal with "x as x approaches infinity" and you get the same results you would intuit for infinity (Division by infinity is still a bit tricky if I recall correctly)

Anyway, if you have ever noticed that math is often about describing things, calculus is how you describe strange (arbitrary) curves and such--Exactly how much water fits into your strangely shaped sink, or how long will it take to drain said water since the speed it drains is slower as there is less water in the sink.

This kind of thing is calculated by slicing up curves into infinitely (Approaching infinite anyway) thin slices and adding the slices together to get an exact result.

(PS, I haven't done much calculus in decades, if this is factually wrong please feel free to fix it rather than just blindly downvoting.  Thanks)


Interpolation method that does never overshoot for implementing a system that will control hardware, I need an interpolation between points on a graph that does never overshoot.
By overshooting I mean that between two points there may be no y-result that has a higher value than either of thew two points.

First I thought that "monotone cubic interpolation" might be what I need, however after implementation I found that this isn't always the case. After researching, I found that this only reduces overshooting, doesn't prevent it entirely. That can also be seen here, I think: Implementation of Monotone Cubic Interpolation )

So what other algorithms/formulas could fulfill this requirement?

Huge thanks in advance!

One option is use a monotone cubic interpolation but always check the values of the points on both sides, and if the value exceeds one of the two y values, just go with that y value instead.


Order of the centralizer of a permutation Given a permutation $\sigma\in S_{n}$, is there a way to know the order of the centraliser $C_{S_{n}}\left(\sigma\right)=\left\{ \pi\in S_{n},\,\pi\sigma=\sigma\pi\right\}$ , i.e what is $\left|C_{S_{n}}\left(\sigma\right)\right|$?

I would appreciate a proof if the answer is yes.

Also, if the answer above is yes, is there also a way to calcualte the order of the centraliser of a given subset of $S_{n}$, or at least for a pair of permutations?

Let $n_1,n_2,\ldots,n_k$ be the distinct lengths of the cycles of $\sigma$ (including 1 if there are fixed points) and suppose that there are $m_i$ cycles of length $n_i$. Then the centralizer of $\sigma$ can permute the cycles of the same length. Its order is $\prod_{i=1}^k n_i^{m_i}m_i!$.

Calculating the centralizer of a subgroup $H$ of $S_n$ is not difficult, but it is more complicated. The order of the centralizer of a single orbit is equal to the number of fixed points (in that orbit) of the stabilizer of a point in the orbit. But if $H$ has more than one orbit with equivalent actions then the equivalent orbits can be permuted by the centralizer, so the complete centralizer is a direct product of wreath products of centralizers of sets of equivalent orbits.

Factor $\sigma$ into its orbits, and let $m_1, m_2, \ldots, m_i$ be the order of each of those orbits, with $m = \sum_{j=1}^i m_j$. For a permutation $\pi$ to commute with $\sigma$, for each of the orbits of $\pi$ we must have either that it is a power of an orbit of $\sigma$, or that it acts on none of the elements that $\sigma$ does. Thus we have that the centralizer group is isomorphic to

$$
\mathbb{Z}_{m_1}\oplus \mathbb{Z}_{m_2}\oplus \cdots \oplus\mathbb{Z}_{m_i} \oplus S_{n-m}
$$


How does $f$ change with the increasing of $\sigma$ The $f=\int^{+\infty}_{0}x^{\lambda}\frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{(x-m)^2}{2\sigma^2}}dx$, and $\lambda$ is a positive constant value, $m=1000$, how does $f$ change with the increasing of $\sigma$. (if $\sigma^2$ changes from 20 to 200)

How to prove it? Thanks.

If the integral were taken from $-\infty$ to $\infty$, then the integral equals to $\newcommand{\expect}{\mathbf{E}} \expect\left(X^\lambda\right)$ where $X\sim\mathcal{N}(m, \sigma^2)$.

So obviously when $\lambda=1$, it is nothing but the expected value of $X$, hence it stays the same, the value is the mean, $m$.

If $\lambda=2$, then it's $\newcommand{\var}{\mathbf{Var}}\expect X^2 = \var(X) + \left(\expect X\right)^2 = \var(X) + m^2$, hence it's increasing as $\sigma$ increases, since $\var(X) = \sigma^2$.

Taking the derivative

$$f'(\sigma) = \frac{1}{\sigma^4}\int_{0}^{\infty} \frac{x^\lambda}{\sqrt{2\pi}}((x-\mu)^2 - \sigma^2)e^{-\frac{(x-\mu)^2}{2\sigma^2}}dx$$

Now let's take the case of $\mu=0$. This simplifies down to comparing the values of the two integrals
$$I_1=\int_{0}^{\infty} x^{\lambda+2}e^{-\frac{x^2}{2\sigma^2}}dx \hspace{20 pt} I_2=\sigma^2\int_{0}^{\infty} x^{\lambda}e^{-\frac{x^2}{2\sigma^2}}dx$$

$$ I_1 = \sigma^{\lambda+3}2^{\frac{\lambda+1}{2}}\Gamma\left( \frac{\lambda+3}{2}\right) \hspace{20 pt} I_2 = \sigma^{\lambda+3}2^{\frac{\lambda-1}{2}}\Gamma\left( \frac{\lambda+1}{2}\right)$$
$$\implies \frac{I_1}{I_2} = \lambda + 1$$

In other words, this derivative is always positive for $\lambda > 0$, so it is always an increasing function of $\sigma$. I'm not sure yet how this changes in the $\mu\neq 0$ case.


Creating a Tridiagonal matrix in matlab How can I create a tridiagonal matrix that I can use for Crout factorization? And, I don't have any codes on how to create one since I am new to matlab.

Ok, please help me understand what does the sentence "The program should output the $\infty$ norm of the residual of your computed solution and the number of iterations used" mean in this case? I am all confused figuring this out.

You could also use conv2 to create a tridiagonal matrix

B = conv2(eye(5),[-1 2 -1],'same');

B =

 2    -1     0     0     0
-1     2    -1     0     0
 0    -1     2    -1     0
 0     0    -1     2    -1
 0     0     0    -1     2


Is there a function that grows faster than exponentially but slower than a factorial? In big-O notation the complexity class $O(2^n)$ is named "exponential". The complexity class $O(n!)$ is named "factorial".

I believe that $f(n) = O(2^n)$ and $g(n) = O(n!)$ means that $\dfrac{f(n)}{g(n)}$ goes to zero in the limit as $n$ goes to infinity.

Is there any known function between the factorial and exponential complexity classes?

In other words is there any known function $j(n)$  that dominates every function in $O(2^n)$, such that:
$$ (j(n) \neq O(2^n)) \wedge (j(n) = O(n!)) \wedge (n! \neq O(j(n)))
$$
 or, informally, $j(n)$ grows asymptotically strictly faster than $2^n$ but not as fast as $n!$?

Or perhaps it has been proven that no such function can exist?

Note: this may seem like a Computer Science question, but in fact I am attempting to prove that any periodic, convergent power series must have coefficients whose inverses grow asymptotically as fast as $n!$ but not faster. I think I can show they most grow faster than $O(2^n)$, but that does not prove they are in $\Theta(n!)$ unless there is no complexity class between $O(2^n)$ and $O(n!)$.

Hint For exponential you have the growth $$\frac{a_{n+1}}{a_n}=\mbox{constant}$$

For the factorial you have the growth
$$\frac{b_{n+1}}{b_n}=n$$

Take any function $g(n)$ which grows to infinity slower than $n$ and  set
$$\frac{c_{n+1}}{c_n}=g(n)$$

For example, $g(n)=\sqrt{n}$  gives the example $\sqrt{n!}$ given by AntonioVargas.

Another interesting example is $g(n)=\ln(n)$ which gives
$$d_n =\prod_{k=2}^n \ln(k)$$

To add to Especially Lime's answer, there are a couple of common kinds of classes of asymptotic behaviour relative to any given class. As $n \overset{\in\mathbb{N}}\to \infty$, we often naturally encounter not just $O(f(n))$ but also $f(n)^{O(1)}$ and $f(n)^{o(1)}$. (For definition of $o(1)$ see Landau notation.) And there are still asymptotic growth rates between $\log\log(n)^{Θ(1)}$ and $\log(n)^{Θ(1)}$, such as $(\log\log(n))^{\log\log\log(n)}$.


dimension of a von Neumann algebra Is there any dimension on a von Neumann algebra? Is there any relationship between finite von Neumann algebras and finite dimensional von Neumann algebras?

A von Neumann algebra is a vector space, so it has a dimension in that sense, but it is meaningless (actually, it is meaningless for any Banach space as the dimension is either finite or uncountably infinite).

Any finite-dimensional von Neumann algebra is finite. But there are (many) finite, infinite-dimensional von Neumann algebras: precisely those of type II$_1$.


Largest eigenvalue of a simple graph Let $\Gamma$ be a simple graph with $n$ vertices, $e$ edges and largest eigenvalue $\lambda_{max}$.
Show that $\lambda_{max} = \frac{2e}{n}$ iff $\Gamma$ is regular.
I've already shown the if part, which is kinda obvious.
For the converse, I have no idea how to prove it. I've already checked in the internet and even here in MathStackExchange but found almost no clue, and the clue that I found was not useful at all (I will put a link to it). Can someone give me a really useful hint?
Do I need any deep knowledge in Linear Algebra that I don't know because I had not been taught?
I'm currently taking a course in Combinatorics, in a master's degree, and this is part of a suggested exercise by the Professor after a class.
Largest eigenvalue of the adjacency matrix of a graph

We should assume that $A$ is connected/irreducible-- otherwise recurse on each irreducible component.  Mapping this over to quasi-Markov chain territory:  First,  by divide by $\frac{2e}{n}$.  In this setup $A$ being regular is now called being doubly stochastic (and conversely being doubly stochastic implies regularity).
averaging over all $n\times n$ permutation matrices
$A':=\frac{1}{n!}\sum_{k} \big(P^{(k)}\big)A\big(P^{(k)}\big)^T$
(where $A'$ is necessarily doubly stochastic)
there are two ways to finish
(1)  Use sub-additivity of operator 2 norm
$1=\Big \Vert A'\Big \Vert_2=\frac{1}{n!}\Big \Vert\sum_{k} \big(P^{(k)}\big)A\big(P^{(k)}\big)^T\Big \Vert_2\leq \frac{1}{n!}\sum_{k} \Big \Vert\big(P^{(k)}\big)A\big(P^{(k)}\big)^T\Big \Vert_2=\Big \Vert A\Big \Vert_2 = \lambda_1$
with equality iff each graph isomorphic representation of $A$ has a Perron vector in common, i.e. iff the Perron vector is invariant to permutation.
(2) A different way of looking at the equality conditions:  $\mathbf v:= \frac{1}{\sqrt{n}}\mathbf 1$
$1=\Big \Vert A'\Big \Vert_2= \mathbf v^T A'\mathbf v = \frac{1}{n!}\sum_{k}\mathbf v^T\big(P^{(k)}\big)A\big(P^{(k)}\big)^T\mathbf v= \frac{1}{n!}\sum_{k}\mathbf v^TA\mathbf v = \mathbf v^TA \mathbf v\leq \lambda_1$
with equality iff $\mathbf v$ is the Perron vector of $A$, that is iff $A$ is doubly stochastic

By Rayleigh-Ritz, the maximum eigenvalue satisfies:
$$\lambda_1 = \max \frac{v^T A v}{v^T v}.$$
Now, consider the vector $\mathbb{1}$ of all $1.$ Then, $$\mathbb{1}^t A \mathbb{1}=$\frac1n \sum_{j, j} A_{ij} = \frac{2e}n,$$ so $\lambda_1 \geq \frac{2e}n$ (which is the average degree). We have equality if and only if $\mathbb{1}$ is an eigenvector, which means that the graph is regular (notice that a non-negative matrix has at most one non-negative eigenvector).


Prove that $\mathbb{Q}(\sqrt{2},\sqrt[3]{3})=\mathbb{Q}(\sqrt{2}\sqrt[3]{3})=\mathbb{Q}(\sqrt{2}+\sqrt[3]{3})$ Showing that $\mathbb{Q}(\sqrt{2},\sqrt[3]{3})$ has degree 6 over $\mathbb{Q}$ is straighforward:
It contains $\mathbb{Q}(\sqrt{2})$ and $\mathbb{Q}(\sqrt[3]{3})$ which are degree 2 and 3 over $\mathbb{Q}$ and since 2 and 3 are coprime it follows that $[\mathbb{Q}(\sqrt{2},\sqrt[3]{3}):\mathbb{Q}]=6$.
Then since $\mathbb{Q}(\sqrt{2}\sqrt[3]{3})\subset \mathbb{Q}(\sqrt{2},\sqrt[3]{3})$ and $x^6-72$ is the minimal polynomial of $\sqrt{2}\sqrt[3]{3}$ it follows that  $[\mathbb{Q}(\sqrt{2},\sqrt[3]{3}):\mathbb{Q}(\sqrt{2}\sqrt[3]{3})]=1$ and so $\mathbb{Q}(\sqrt{2},\sqrt[3]{3})=\mathbb{Q}(\sqrt{2}\sqrt[3]{3})$.
But I can't figure out how to prove the last equality. Any help is appreciated!

So after some thinking, I came up with two solutions:
Solution 1
Let $\alpha=\sqrt{2},\beta=\sqrt[3]{3}$ and $\gamma=\alpha+\beta$. Let $L=\mathbb{Q}(\alpha, \beta)$, $K=\mathbb{Q}(\gamma)$ and suppose that $\alpha,\beta\not\in K$. Since if one of $\alpha,\beta$ is in $K$, we get the other one for free it follows that $L=K(\alpha)=K(\beta)$. Then the minimal polynomial of $\alpha$ in $K[X]$ is of degree 2 since $\alpha$ is a root of $X^2-2$ and we assumed $\alpha\not\in K$. Since $X^3-3$ has one real root and two non-real roots it follows that it is the minimal polynomial of $\beta$ in $K[X]$. Hence we conclude
$$2=[K(\alpha):K]=[L:K]=[K(\beta):K]=3.$$
Which is a contradiction and so $L=K$ as desired.
Solution 2
We know that $\gamma-\alpha=\beta$ so $(\gamma-\alpha)^3=3$. Expanding the left side we get
$$\gamma^3-3\gamma^2\alpha+6\gamma+\alpha^3=3$$
and so $\alpha=\frac{3-6\gamma-\gamma^3}{2-3\gamma^2}\in K$. Therefore $\beta=\gamma-\alpha\in K$ and so $K=L$.

Show that $x^3-3$ is irreducible over $\Bbb{Q}$ and $x^2-2$ is irreducible over $\Bbb{Q}(\sqrt[3]3)$,
which implies that $[\Bbb{Q}(\sqrt2,\sqrt[3]3):\Bbb{Q}]=6$.
Next show that $\sqrt2\sqrt[3]3$ has $6$ distinct $\Bbb{Q}$-conjugates, which implies that $[\Bbb{Q}(\sqrt2\sqrt[3]3):\Bbb{Q}]=6$.
Same for $\sqrt2+\sqrt[3]3$.


Differention of a First order ODE Integrate the ODE:

\begin{equation}
\frac{d \mu(t)}{dt} = \frac{1}{5}\mu(t)
\end{equation}

Which we solve to arrive:

\begin{equation}
\mu(t)=e^{t/5}
\end{equation}

Can someone please remind me how to integrate this ODE ?

Thanks

$$\mu'-\frac{1}{5}\mu=0$$
the characteristic equation is
$$r-\frac{1}{5}=0$$
$$r=\frac{1}{5}$$
so the solution is
$$y=ce^{\frac{t}{5}}$$


How do I calculate the area $\int_0^2 \frac{1}{\sqrt{2x-x^2}} dx$? I have the integral

$$\int_0^2 \frac{1}{\sqrt{2x-x^2}}dx$$

I know the answer is $\pi$ but I have problems with limits $0$ and $2$. How can I use $\varepsilon$ in this problem?

$$2x-x^2=-(x-1)^2+1\implies \sqrt{2x-x^2}=\sqrt{1-(x-1)^2}\implies$$

$$\int\frac{dx}{\sqrt{2x-x^2}}=\int\frac{dx}{\sqrt{1-(x-1)^2}}=\arcsin(x-1)+C$$

Be careful here: with limits, if one of them is zero or two this is then an improper integral, though it looks nice


For $a,b,c>0$ prove that $abc(a+b+c) \le a^3 b + b^3 c + c^3 a$ if $a,b,c > 0$ then prove that $abc(a+b+c) \le a^3 b + b^3 c + c^3 a$

My attempt
The hint given for this question was to use the Cauchy-schwarz inequality. But if you look at the expression given, the L.H.S is symmetric but the R.H.S is not. If I understand the inequality(Cauchy-Schwarz)  correctly, both the sides of the inequality should be symmetric.
So can we use the Cauchy-Schwarz to prove this?
I also tried applying AM > GM but got stuck.
Any help would be appreciated.

We can write this inequality as $$F=\frac{a^2}{c}+\frac{b^2}{a}+\frac{c^2}{b} \ge (a+b+c)~~~(1)$$
Using Cauch-Schwarts ineq. : $$\frac{A^2}{X}+\frac{B^2}{Y}+\frac{C^2}{Z} \ge \frac{(A+B+C)^2}{X+Y+Z}, X,Y,Z>0~~~~(2)$$
Using (2) it follows that $$F\ge \frac{(a+b+c)^2}{(a+b+c)}=a+b+c.$$

We use following inequality:
$a^4+b^4+c^4 ≥ abc(a+b+c)$
Let $a ≥b ≥c$, we may write:
$a^4 ≥a^3b$
$b^4 ≥b^3c$
$c^4<c^3a$ ⇒ $c^4 ≥ -c^3a$
And we have:
$a^4+b^4+c^4+2c^3a ≥ a^3b+b^3c-c^3a+2c^3a =a^3b+b^3c+c^3a≥ abc(a+b+c)$


Solve trigonometric inequality $ \sin x+2 \cos x<2$ $$ \sin x+2 \cos x<2$$
$$ \dfrac{2t}{1+t^2}+2\dfrac{1-t^2}{1+t^2}<2$$
$$ 4t^2-2t>0$$
$$ 2t(2t-1)>0$$
$$ t(2t-1)>0$$
$$ (t>0 \wedge t>\dfrac{1}{2}) \vee (t<0 \wedge t<\dfrac{1}{2})$$
From this, I  can only find $x<2\pi+2k\pi$, and, $x<2k\pi$, these are good (I think), but I should find another two solutions.

You have used the substitution:
$$
t=\tan \left(\frac{x}{2}\right)
$$ and the solution of the inequality is $ t<0 \lor t>\dfrac{1}{2}$

so you have:
$$
\tan \left(\frac{x}{2}\right)<0 \qquad \lor \qquad \tan \left(\frac{x}{2}\right)> \dfrac{1}{2}
$$

the solution of the first inequality is:

$$
-\dfrac{\pi}{2}+k\pi<\dfrac{x}{2}<0+k\pi \iff (2k-1)\pi<x<2k\pi
$$

and the solution of the second inequality is:

$$
\tan^{-1}\left(\dfrac{1}{2}\right)+k\pi<\dfrac{x}{2}<\dfrac{\pi}{2}+k\pi \iff 2\tan^{-1}\left(\dfrac{1}{2}\right)+2k\pi<x<\pi+2k\pi
$$

The answer is the following set $X$.
$$
X=\{t|2n\pi+2\tan^{-1}(\frac{1}{2})<t<2n\pi+2\pi  {\rm ~for~some~} n \in \mathbb{Z}\}
$$


How to multiply 2 Binary Numbers? Suppose I have two regular numbers $a$ and $b$ in base $10$ like this: (where $N$ is even)
$$a=a_{\frac N2}a_{\frac N2-1}\ldots a_1,\qquad b=b_{\frac N2}b_{\frac N2-1}\ldots b_1$$
So the result of their multiplication is:
$$a\cdot b=a_{\frac N2}a_{\frac N2-1}\ldots a\cdot b_{\frac N2}b_{\frac N2-1}\ldots b_1=\sum_{i=1}^{\frac N2}\sum_{j=1}^{\frac N2}(a_ib_j)10^{i+j-2}$$

Now suppose $a=a_Na_{N-1}...a_1$ and $b=b_Nb_{N-1}...b_1$ are (unsigned) binary numbers, such that each digit is actually 16 bits by its own. For example: $a_1$ can be: $1111010101000000$
So if $a$ has $N$ digits it's actually: $16N$ long binary number.
Given this fact how can I multiply them in a similar way to what is shown above?
Why I need this?
I want to multiply big numbers using the fact that I know only to multiply 16 bits binary numbers.

Small But Different Example:
if $a_M, b_M, a_L, b_L$ are all 16 bits binary numbers then I could use this fact to multiply 32 bits numbers like this:

In base $10$, the digit $a_1$ can take values from $0$ to $9$. Any numbers higher than that can only displayed if you also use $a_2$.
In your setting, $a_1$ can take values from $0$ to $65535$ (which is $1111111111111111$ in binary). Any numbers higher than that can only displayed if you also use $a_2$.
So, in summary, we should consider base $65536$ (instead of base $10$).
The formula becomes
$$a\cdot b=a_{\frac N2}a_{\frac N2-1}\ldots a\cdot b_{\frac N2}b_{\frac N2-1}\ldots b_1=\sum_{i=1}^{\frac N2}\sum_{j=1}^{\frac N2}(a_ib_j)65536^{i+j-2},$$
where instead of $65536^{i+j-2}$
we can also write $(2^{16})^{i+j-2}$, which is the same as $2^{16 (i+j-2)}$.
The number $16 (i+j-2)$ in the exponent of the last expressions
is the number of bits that you need to shift by. to do (because if you multiply by $2^k$ this corresponds to a shift by $k$ bits).

Replace $10$ by the base $b$,
either 2 or 16 or whatever to get
$a\cdot b=a_{\frac N2}a_{\frac N2-1}\ldots a\cdot b_{\frac N2}b_{\frac N2-1}\ldots b_1=\sum_{i=1}^{\frac N2}\sum_{j=1}^{\frac N2}(a_ib_j)b^{i+j-2}
$.
If $b=2$,
$b^{i+j-2}$ does
$i+j-2$ left shifts.
If $b=16$,
$b^{i+j-2}$ does
$4(i+j-2)$ left shifts.


What is the difference between a class and a set? I know what a set is. I have no idea what a class is.

As best as I can make out, every set is also a class, but a class can be "larger" than any set (a so-called "proper class").

This obviously makes no sense whatsoever, since sets are of unlimited size. It's not like you get up to 25 elements and go "oh, hey, sets are limited to a maximum of 20 elements, so we'll have to call this 25-element thing something else".

Does anybody know what the actual difference between a set and a class is?

Several people have pointed out that you can write paradoxes which involve sets. I get that. But where do classes come into the picture?

Your intuition about size limitations is wrong. Think about finite sets: there are sets which are finite but as large as you would want them, $6$ elements, $25$ elements, $216$ elements, whatever. But does that mean that the set of natural numbers is a finite set?

The idea behind the transfinite is what happens after you've gone to infinity and beyond. So there are sets and they grow larger and larger, then they become infinite, and they continue to grow larger and larger... eventually you have gone "all the way". There comes a question - is the collection of everything you have accumulated so far is a set? If so, we can keep on going. Classes tell us that eventually (which is a pretty far eventually) we have to stop somewhere.

In the naive approach to mathematics we think that every collection we can talk about is a set. Simply because in the naive approach there is no definition of a set.

However once axiomatic set theory came into play we have the seemingly circular definition: Sets are elements of a model of set theory.

For example, one of the axioms about sets is that they have power sets. One of the theorems linking a set and its power set is that there is no surjective function from a set to its power set.

Suppose the collection of all sets, $V$ was a set itself, what would its power set be? Well, every subcollection of $V$ is a set and therefore in $V$. This means that $P(V)\subseteq V$. However this means that there is a surjective function from $V$ onto its power set!

Cantor's paradox (as above), as well Russell's paradox (all sets which are not elements of themselves is a collection which is not a set), and so several other paradoxes tell us one thing: not all collections we can define are sets.

In ZFC classes are simply definable collections of sets. What does it mean definable? It means "all sets which have a property which we can describe in the given language".

One simple way to describe the difference between sets and classes in ZFC is that sets are elements of other sets. Classes are not elements of any other class, so if $A\in B$ then $A$ is a set.

To your edit:

The first thing to want from a foundational mathematical theory (one which you hope to later build most of your mathematics on) is that if you have a certain property, then you can talk about all the things in your universe with this property. The various paradoxes tell us that in ZFC (and in its spawns) some of these collections are not sets. The notion of "proper class" tells us that we can still talk about this collection, but it is not a set per se.

For example, we can talk about ordinals (which are a transfinite generalization of the natural numbers in some sense), the collection of all ordinals is a proper class. We can still talk about "all the ordinals" or prove that some property holds for all of them, despite that this is not a set.

This is what I have culled from the sources/references I was fortunate enough to obtain / reach for consultation.

A class is a collection (or, more informally, a putting together) of objects whose origins need not be pre-defined.  A set is a collection of objects whose origins are pre-defined and unambiguous (though not necessarily realizable). Perhaps one similarity between the two notions (though it does not contribute to any enlightenment) is that the process of the putting together of the objects is not an issue at all.

I don't claim this is an all-encompassing distinction- it could be far from it, for all I know.  This struck to me after I went through an assembly line of examples.  Note that I have used a few terms that are open to contest - or, at least an inspection with a critical eye.  For instance: (i) pre-defined (ii) realizable.


Sheffer stroke the most important advance in logic? I think I once read, or heard, that Bertrand Russell once said that the discovery that all logical operators are expressible in terms of the Sheffer stroke was the most significant advance in logic since the publication of Principia Mathematica, and that had he and Whitehead known about it beforehand, they would have proceeded completely differently.

This is a really strange claim, so I imagine that I (or the person who told me) misunderstood, or misheard, or made it up completely.

If Russell really did say something like this, what exactly did he say, and where and when?

I have been reading the recent book The Evolution of Principia Mathematica, by Bernard Linsky.

The quote in the post is in significant ways not quite right. It comes in the Introduction to the 1925 Edition of Principia, first page. Here is what was actually written, in full.

"The most definite improvement resulting from work in mathematical logic is the substitution, in Part I, Section A, of the one indefinable "$p$ and $q$ are incompatible" (or, alternatively, "$p$ and $q$ are both false") for the two indefinables "not-$p$" and "$p$ or $q$." This is due to Dr. H.M. Sheffer. consequently, M. Jean Nicod showed that one primitive proposition could replace the five primitive propositions $\ast$1.2.3.4.5."

So Russell is saying that the greatest improvement in the second edition of Principia that is due to recent advances in mathematical logic, presumably by others, came from using the Sheffer stroke and an axiomatization based on the work of Nicod.  There are also significant changes in the description of type theory, but presumably that is not mentioned here because it is based on ideas of Russell. (By this time, Whitehead had bailed.)

Russell is not saying that Sheffer's work is the most important development in mathematical logic of the previous fourteen years.

The idea of improvement that could have been brought to Principla Mathematica [PM] is quite misplaced! Replacing the primitive team {INCLUSIVE-OR, NEGATION} by Sheffer's (actually Peirce's or even... Chrysippus'!) NAND wouldn't have improved PM, at all.

Because, primo, PM is a piece of kindergaten charabia that should not have been written -- and published -- in the first place: it is nothing but an elementary exercise in formalisation (a rather long one, indeed). As a matter of fact, one can generate it with the help of appropriate software.

Even the idea of formalisation in PM is besides the point: in PM, like in Frege's Begriffsschrift (1879) or in his Grundesetze der Arithmetic (1893--1903) and in Peirce's pioneering 1885-paper, the formalisation concerns only a superficial aspect of logic, namely the (provable) formulas as expressing propositions and / or propositional schemes, not the proofs themselves: the concept of proof -- the central concept of logic, after all -- is left un-formalised in PM. [*]

Second, the handling of quantification (first- and second-order) in PM is seriously flawed and constitutes a considerable step backwards even with respect to Frege. (This is, by the way, one of Goedel's side remarks on Russell, not mine. Cf. Goedel's Russell's mathematical logic, in The Philosophy of Bertrand Russell, edited by Paul Arthur Schilpp, Northwestern University, Evanston and Chicago IL 1944, pp. 123--153.)

Third, the Russellian finding, meant to avoid paradoxes and the like -- namely the so-called ramified theory of types -- is a paradigmatic piece of muddy thinking: in PM, the so-called orders and the type hierarchy are introduced only in order to get rid of them, at a later stage (by the so-called Reducibility Axiom).

Fourth, the actual mathematics `formalised' (so to speak, i.e. in formularian terms) in PM is rather elementary.

Finally, PM (actually Russell and Whitehead) has done more damage than good to the reasearch in (theoretical) logic, during the last hundred years or so.

[*] The proof themselves have been first formalised by Nicolaas Gerrit de Bruijn and his students and collaborators at the Eindhoven Polytechnics (NL), about 40 years ago, within the Automath project (Automated Mathematics), since around 1967--1968. Cf., e.g., https://www.win.tue.nl/automath/ and my Abstract Automath, Mathematical Centre, Amsterdam 1982 [Mathematisch Centrum Tract 160] @ https://www.win.tue.nl/automath/archive/webversion/xaut021/xaut021.html.


Does $\lim_ {(x,y)\to (0,0 )} \frac{x^3+y^3}{x^2 + y^2}$ exist? My solution is the following:

approaching by the y-axis:

$\lim_ {(y)\to (0),(x=0)} =\lim_ {(y) \to (0)}=\frac{0+y^3}{0^2+y^2}=y=0$

approaching by $y=x$

$\lim_ {(y)\to (0),(y=x)} =\lim_ {y=x}=\frac{x^3+x^3}{x^2+x^2}=\frac{2x^3}{2x^2}=x=0$

So I think,that this limit exists.
is it correct in this form?

HINT:
$$
\frac{x^3+y^3}{x^2+y^2}=x\frac{x^2}{x^2+y^2}+y\frac{y^2}{x^2+y^2}
$$

But your method doesn't answer the question.

Try coming at it from any other direction, e.g. y = mx.

Ced


Imaginary numbers - Is there a shortcut? I have recently learned about imaginary numbers and I sat in front of this pattern of -1, -i, 1, i and I asked myself, isn't there an easy way to describe this pattern ? Or: isn't there a really short way to find out about imaginary numbers raised to high powers ?

(Despite the "find a factor of 4"-method) enter link description here

I am not sure if this works really ALL the time and I have only tested it for powers that are positive and negative integers, but haven't found any contradictions and algebraically this works like that power of 4 thing:

When your power divided by 2 gives an integer (so no fraction or decimal):
Just divide the power by 2 : If the result is odd, then the number is = (-1)
If it is even, then the number is =(1) BUT ONLY IF YOUR RESULT IS AN INTEGER !

What if the power of i itself is odd (result is not an integer)? Like 5 or 9999 ?
Then the result is i or -i . To figure out: go back one step and divide the (power-1) by 2 .
If the result is odd => it's -i
If the result is even => it's +i

E.g.: i^4 = ? and i^5= ?
=> 4:2 = 2 ; 2 is even ; => 1
=> 5:2 does not work (for x E Z) so it must be i or -i . So subtract 1 from the power = 4 , 4:2 = even => +i

This also works for negative powers, BUT BE AWARE that, say a power of (-5)-1 is = -6 not -4 !!

Example:
i^9999 ; 9999:2 = Not an integer -> 9999-1 = 9998:2 = 4999 (That is ODD)
so i^9999= -i

i^34; 34:2 = 17 -> Odd number and an integer => i^34=-1

Be aware ! Some numbers like 336 seem to give an odd result "something with 3 at the end" when you are in a hurry - but it's actually even. This is where I often do stupid mistakes :D

Hope you enjoy this. I can not guarantee that this is always true. I just found it out by myself (maybe somewhere on the internet this is already explained :) ))

At my original post at Khan-academy a guy named Ian has well explained why this works out. As also as I mentioned before, you can work this out by algebra since 2 is a factor of 4.Khanacademy post


Intro Math Matrices problem Find all $2 \times 2$ matrices $A= \begin{pmatrix} a & b \\ c & d \end{pmatrix}$ such that $AB= BA $ for all $2\times2 $ matrices B.

I am stumped as to how to approach this question overall. I am not sure if its asking for me to use some theorem or if some easier solution is present.

Let
$$A=\pmatrix{a&b\cr c&d\cr}$$
and suppose that $AB=BA$ for all $B$.  Making a particular choice for $B$ we have
$$\pmatrix{a&b\cr c&d\cr}\pmatrix{1&0\cr0&0\cr}
  =\pmatrix{1&0\cr0&0\cr}\pmatrix{a&b\cr c&d\cr}\ ,$$
so
$$\pmatrix{a&0\cr c&0\cr}=\pmatrix{a&b\cr0&0\cr}$$
which shows that $b=0$ and $c=0$.  Similarly,
$$\pmatrix{a&0\cr 0&d\cr}\pmatrix{0&1\cr0&0\cr}
  =\pmatrix{0&1\cr0&0\cr}\pmatrix{a&0\cr 0&d\cr}$$
gives $a=d$.  Therefore
$$A=\pmatrix{a&0\cr0&a\cr}=aI\ .$$
We have shown that these are the only possibilities for $A$.  This does not mean they actually work for all $B$, but it is easy to check that in fact they do.

In General, matrix multiplication is not commutative.
AB = BA holds only in case if either A or B is an Identity Matrix, or B is the Inverse of A.


If the Goldbach Conjecture is True, does it make it easier to find large primes? I was just reading Is every positive nonprime number at equal distance between two prime numbers? (current hot topic) and was reflecting on the fact that computing security (cryptography) is based around the use of large prime numbers (see: Why are very large prime numbers important in cryptography?).

The answer for the above-mentioned question suggests that the Goldbach Conjecture says that every non-prime positive number is positioned equidistant between two prime numbers.

For the purposes of this question, I'll assume that statement is true (I have no prior knowledge of Goldbach or his/her conjecture).

If the Goldbach Conjecture is true, does it make it easier to find large prime numbers?

For example, I could take any very large number at random and then look at every number below it, find a prime and then work out of the opposite number is also a prime (or something along those lines).  In my mind, it's almost as though the assumption would give me a starting point to find an even larger prime...

I expect I'm not the first person to ask this (and if I am, I've probably missed something somewhere..), but I can't find a similar question here :)

Thanks in advance

Finding primes of the size wanted for cryptography isn't hard.  The prime number theorem says that a "random" number $n$ has one chance in $\ln n$ of being prime.  For a $1000$ bit number, this is about $1$ in $700$.  If you only try numbers congruent to $1$ or $5 \pmod {6}$, you get another factor $3$, so you only have to try a few hundred before you find one.  How to check is described here.  The celebrated prime numbers that are found are much larger.

Yes.  $2n-p=q$, where $p<q$, and $q$ is the larger "flank" of $n$ as $n$ is midway between $p$ and $q$.   You'd probably only have to test the 8 primes less than (but closest to) $p$ to find $q$.

Let's say those primes are called $a,b,c,d,e,f,g,h$.  Then $q$ might be $2n-a$ or $2n-b$ or $2n-c$ or $2n-d$ or $2n-e \ldots$ you get the idea.  Try it with primes you already know.

What if you thought 181 was the largest known prime.  Look for 8 closest primes less than 181, and use $n=200$ (or any number $n$ that is at least 10% greater than your largest known prime).  Find $2n-a$.  Find $2n-b$.  Find $2n-c$.  And so on...

Did you have to go through all 8 to find a prime larger than 181?


Unramification and compositum The background is: a field $K$ complete with respect to a discrete valuation $|\ |$. We write $A$ and $k$ for his discret valuation ring and the residue field of $A$. We assume that $K$ and $k$ are perfect. We have $L$ an algebraic extension of $K$ with $B$ and $\mathfrak{p}$ for the valuation ring and the residue field.

So my question is: let $K'$ and $K''$ unramified (finite) extensions of $K$ in $L$ with the same residue field $k'$. Why do we have that the compositum $K'\cdot K''$ is an unramified extension of $K$ and that $k'$ is his residue field?

Let $A', A''$ be the respective valuation rings of $K', K''$. Then $A'\otimes_A A''$ is unramified over $A$ (used discriminant for example), so it is integrally closed (but not integral in general) because something étale over normal ring is normal. In particular it is a finite product of discrete valuation rings $\prod_i A_i$. Its total ring of fractions is $K'\otimes_K K''=\prod_i \mathrm{Frac}(A_i)$. Consider the canonical map
$$ A'\otimes_A A''\to B, \quad a'\otimes a''\mapsto a'a''.$$
Its image is an integral quotient of $A'\otimes A'''$ hence is one of the factors, say, $A_1$. We then have $K'K''=\mathrm{Frac}(A_1)$. The residue field of $K'K''$ is the residue field of $A_1$, equal to the image of $k'\otimes_k k'$ in the residue field of $B$, and this is just $k'$.

P.S. I don't have enough reputations to comment. This proof is maybe too complicate. You should unaccept this answer, so other people could provide something simpler. I am not convinced by the first proof because we can lift two different roots of  $f(T)\in k[T]$ defining $k'$ to two differents roots $t_1, t_2\in B$ of a lifting $F(T)\in A[T]$ of $f(T)$, and consider $K'=K[t_1]$, $K''=K[t_2]$. If $k'/k$ is not Galois, $K'$ will be different from $K''$ in general.


Covariance of states of a finite Markov chain I know it is possible to construct a covariance matrix for states of a Markov chain but I cannot seem to find a proper way to compute it. I will attach some theories I found from Kemeny and Snell's FMCs which states about limiting covariance.

I just cannot understand how the covariance matrix can be generated using this. I need to know whether there is a much easier way to approach this problem. In the same text in an example the have computed the convariance matrix without any steps showing how it was computed.

Hope someone could help me out on this and give me a better insight to the concept of covarince with respect to states in a Markov chain. Thanks !

Try this (R code)
   a <- A[1,]   # Limiting matrix
   C <- mat.or.vec(m,m)  # mxm covariance matrix
   delta <- function(i,j)  {delta <- 0; if (i==j) delta <- 1; return(delta)}
   for (i in 1:m)   {
      for (j in 1:m)   {
         C[i,j] <- a[i]*Z[i,j]+a[j]*Z[j,i] - a[i]*a[j] - a[i]*delta(i,j)
      }
   }
Result:

  375*C
       [,1] [,2] [,3]
  [1,]  134  -18 -116
  [2,]  -18   36  -18
  [3,] -116  -18  134
  Kemeny's example on page 89.


your proof seems questionable.
Since $g^{\prime}(t)=t^3+2b_2t$ the monotonicity of $g(t)$ will depend on the value of $b_2$.

Why do you assert that

$$g(\sqrt{2})=\dfrac{1}{2}\qquad g(0)=\dfrac{1}{2} g(1)=-\dfrac{1}{2}$$


Euler's totient problem: Prove that the order of $2 \bmod 5^n$ is always equal to $4\cdot 5^{n-1}$ Prove that the order of $2 \bmod 5^n$ is always equal to $4\cdot 5^{n-1}$.

I know that the order must divide $4\cdot 5^{n-1}$ because of Euler's totient, but how do I move forward from here?

There is probably an easier way of proving this, particularly since you are interested in the special case $p=5$, but here is a proof for the general case.

Suppose $p$ is an odd prime and $g$ a primitive root modulo $p$. We first show that either $g$ or $g+p$ is a primitive root modulo $p^2$. (Note that this can be proved more easily using Hensel's Lemma.)

We have for $n\ge 2$ that
\begin{equation*}
  g^{\mathrm{ord}_{p^n}g}\equiv 1\mod{p^n}\quad\Rightarrow\quad
  g^{\mathrm{ord}_{p^n}g}\equiv 1\mod{p} \quad\Rightarrow\quad
  p-1 = \phi(p)\mid \mathrm{ord}_{p^n}g.
\end{equation*}
But also $\mathrm{ord}_{p^n}g\mid \phi(p^n) = p^{n-1}(p-1)$, so that $\mathrm{ord}_{p^n}g = p^k(p-1)$ for some $0\le k\ne n-1$. For $n=2$, if $k=1$ then $g$ is primitive modulo $p^2$. Otherwise, $\mathrm{ord}_{p^2}(g+p) = p-1$ or $p(p-1)$. In the former case, we get
\begin{align*}
  1&\equiv (g+p)^{p-1} \equiv g^{p-1} + (p-1)g^{p-2}p + \cdots + p^{p-1} \\
  &\equiv g^{p-1} + (p-1)pg^{p-2}\mod{p^2} \\
  &\equiv g^{p-2}(g-p)\mod{p^2},
\end{align*}
but this is impossible since $p\nmid g$. Thus $g+p$ is primitive modulo $p^2$. (Note that in your case, $2$ is also primitive modulo $25$, so we can take $g=2$ in what follows.)

Now let $g$ be a primitive root modulo $p^2$ and modulo $p$. Claim $g$ is a primitive root modulo $p^n$ for any positive integer $n$. From before, $\mathrm{ord}_{p^n}g = p^k(p-1)$; we wish to show that $k=n-1$. It suffices to show that $g^{p^{n-2}(p-1)}\not\equiv 1\mod{p^n}$. Using induction on $n$, the base case $n=2$ is our assumption. Assume that the statement holds for $n$. Then from the inductive assumption, we have $g^{\phi(p^{n-1})}\not\equiv 1\mod{p^n}$, while Fermat's little theorem gives $g^{\phi(p^{n-1})}\not\equiv 1\mod{p^{n-1}}$, so that $g^{\phi(p^{n-1})}=1+bp^{n-1}$ with $p\nmid b$. Thus (with $n\ge 2$)
\begin{align*}
  g^{p^{n-1}(p-1)} = (1+bp^{n-1})^p \\
  &= 1 + pbp^{n-1} + p^{n+1}(\ldots) \\
  &\equiv 1+bp^n\mod{p^{n+1}}.
\end{align*}
Since $p\nmid b$, the order of $g$ modulo $p^{n+1}$ must be $p^n(p-1)$ and we are done.

Prove by induction on $n$. Can be shown for $n=1$ and $2$ by computation.

Now assume true for $n=k\ge2$.

As you indicated, noting Euler's totient function, the order of $2$ modulo $5^{k+1}$ divides $4\times5^k$.

To show it is not smaller than $4\times5^k$,

it suffices to show that $2^{2\times5^k}-1$ and $2^{4\times5^{k-1}}-1$ are not divisible by $5^{k+1}$.

But if $5^{k+1}$ divides $2^{2\times5^k}-1$ then certainly $5^k$ divides it, and given that $5^k$ divides $2^{4\times5^{k-1}}-1$,

it would follow that $5^k$ divides $2^{2\times5^{k-1}}-1,$ contradicting the inductive hypothesis.

Likewise, $5^{k+1}|2^{4\times5^{k-1}}-1=(2^{4\times5^{k-2}}-1)(2^{4\times4\times5^{k-1}}+2^{4\times4\times5^{k-1}}+2^{4\times3\times5^{k-1}}+2^{4\times2\times5^{k-1}}+1)$

$\implies5^k|2^{4\times5^{k-2}}-1,$ since $5|(2^{4\times4\times5^{k-1}}+2^{4\times4\times5^{k-1}}+2^{4\times3\times5^{k-1}}+2^{4\times2\times5^{k-1}}+1);$

again, we have a contradiction to the inductive hypothesis.


$\mu^{\ast}(\alpha A) = |\alpha|^2 \mu^{\ast}(A)$ I'm trying to prove that this statement in real analysis:

If $A \subset R^2$ is measurable, then $\alpha A$ is measurable and

$$\mu^{\ast}(\alpha A) = |\alpha|^2 \mu^{\ast}(A)$$

Thanks for your help and time.

My idea is this:

$$
m^*(E)=\inf\left\{\sum_{n≥1}\ell (I_n)\middle|E\subset\cup_{n≥1}I_n\right\}
$$
and we note that $m(E)=m^*(E)$ for measurable sets $E$.

Noe note that any set $A\subseteq(a,b)$ satisfies $\alpha A\subseteq(\alpha a,\alpha b)$ and that $\ell(\alpha(a,b))=\alpha \, \ell(a,b)$.  It follows that for any collection of sets $\{I_n\}_{n≥1}$ whose union contains $E$, we may state that $\{\alpha I_n\}_{n≥1}$ is a collection whose union contains $\alpha E$ and whose total length is $\alpha$ times longer.  For any set $A$ of real numbers, $\inf (\alpha A) = \alpha \inf A$.  The desired statement follows.

Is this correct?

Use the uniqueness of the Lebesgue measure.  The Lebesgue measure $\mu$ on $\mathbf R^n$ is the unique measure on the Borel sets of $\mathbf R^n$ satisfying the following properties:

1 .  $\mu(E + x) = \mu(E)$ for all Borel sets $E$ and all $x \in \mathbf R^n$.
2 .  Every Borel set can be measured from within by compact sets, and measured from without by open sets.
3 .  $\mu( [0,1] \times \cdots \times [0,1]) = 1$.


Find equations of the tangents to a parametric curve that pass through a given point Find equations of the tangents to the curve $x=3t^2+1,\space y=2t^3+1$ that pass through the point $(4,3)$

My attempt:

If we have $x=3t^2+1,\space y=2t^3+1$, then \begin{cases} 4=3t^2+1\\3=2t^3+1 \end{cases}$\implies$\begin{cases} t=1,-1\\ t=1 \end{cases}Therefore we take the parameter of intersection for point $(4,3)$ which is $t=1.$ $$Recall \space \frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{6t^2}{6t}=t$$
So at point $(4,3)$ the slope of the tangent line is $\frac{dy}{dx}=t=1$.

Thus using point slope form: $$y-y_0=\frac{dy}{dx}\big(x-x_0\big)\implies
y-3=x-4\implies y=x-1$$

Our tangent line then is $y=x-1$. However, it can be shown through the elimination of $x$ and $y$ from the tangent line equation that $y=-2x+11$ is also tangent to the curve and passes through $(4,3)$.

My question is:

  How do we know a priori if there exists multiple tangents to a curve through a common point? If it weren't for a solutions manual, I would not think to eliminate $x$ and $y$ to solve for the slope of the tangent. What is the reason there are multiple tangents through a common point? Is this specific to parametric curves?

Tangent line definition is a straight line or plane that touches a curve or curved surface at a point, but if extended does not cross it at that point.
So here even though $(y-3)=-2(x-4)$ satisfies line equation but since it crosses curve at $(4,3)$ its not a tangent line.
There are no multiple tangent lines to the curve which passes through common point


What is the meaning of the notation $A :\Leftrightarrow B$? This is the text from my book:

  To define a statement $A$ so that it is true whenever the statement
  $B$ is true, we write $$A :\Leftrightarrow B$$ and say '$A$ is true,
  by definition, if $B$ is true'.

I don't quite understand it. How does it differ from $A := B$ or $A \Leftrightarrow B$? Please provide some examples if you can? Thanks in advance.

Edit:

Just to check whether I have really understood this notation, is it correct to write $(X \subseteq Y) :\Leftrightarrow (\forall x \in X (x \in Y))$?

$A:=B$ means that $A$ is defined as $B$.

$A \Leftrightarrow B$ is a theorem: it means that $A$ and $B$ are defined independently and now you are saying that indeed $A$ holds if and only if $B$ holds. This should be followed by some proof.

$A :\Leftrightarrow B$ is again a definition. It means that this is not something you have to prove (you may need to prove that this is a good definition, but that's another problem).

Examples:

$f(x):= x^3$ : definition. Nothing to prove

$f(x)>0 \Leftrightarrow x>0$ : theorem, you should provide a proof for it (in this case it's trivial)

$n$ is even $:\Leftrightarrow n = 2k$ for some $k \in \mathbb{N}$: definition of being even. Nothing to prove

Edit: yes, your example is correct as that's actually the definition of being a subset

One possible reading is that A may be true when B is false, in contrast to "A if and only if B" meaning not only that A is true whenever B is, but that A is false whenever B is.


Doubt about uniqueness of limit proof My lecturer gave us a proof of uniqueness of limit by contradiction, the context is: let $X \subset \mathbb{R}$ be an open set, let $x_0$ be a limit point for $X$ and let $f:X \to \mathbb{R}$ a function. If the limit of $f$ exists then it must be unique.
I have a doubt in the proof, which is the following: suppose by the sake of contradiction that $f$ has two limits $y_1 \ne y_2$ as $x \to x_0$. Let $V_1$ be a neighbourhood of $y_1$ and $V_2$ be a neighbourhood of $y_2$, with $V_1 \cap V_2 = \emptyset$. Since $x_0$ is a limit point, we can find two neighbourhoods $U_1$ and $U_2$ of the point $x_0$ such that for all $x \in U_1 \cap X \setminus \{x_0\}$ it is $f(x) \in V_1$ and for all $x \in U_2 \cap X \setminus \{x_0\}$ it is $f(x) \in V_2$. Since $U-1$ and $U_2$ are neighbourhoods of $x_0$ it is $U_1 \cap U_2 \ne \emptyset$ and since $x_0$ is a limit point there exists at least a point $\bar{x} \in X$ such that $\bar{x} \in U_1 \cap U_2$. So if $x \in U_1 \cap U_2$ it is $f(x) \in V_1$ and $f(x) \in V_2$, meaning $V_1 \cap V_2 \ne \emptyset$ and this contradicts the fact that $V_1 \cap V_2 = \emptyset$.
My doubt is: what if $V_1$ and $V_2$ aren't disjoint? I will try to explain my doubt: proofs must be general, and it seems to me that when he says "let $V_1$ and $V_2$ be such that $V_1 \cap V_2 =\emptyset$" we are restricting to a case that it isn't general because it imposes a condition on $V_1$ and $V_2$, that is that they are disjoint. Why this proof works then? I have a suspect that it is because this restriction is "apparent", in the sense that we can always lead back to the case when $V_1$ and $V_2$ are disjoint (for example, if $V_1$ and $V_2$ aren't disjoint I can consider another neighbourhoods $V_1'$ and $V_2'$ with radius, for example, a fraction the radius of $V_1$ and $V_2$ big enough to assure that $V_1'$ and $V_2'$ are disjoint), and so obtain a general proof because the condition we impose can always be imposed and this makes $V_1 \cap V_2 =\emptyset$ an assumption without loss of generality. This has some sense or am I missing the point? Thanks.

I think that you are misunderstanding the meaning of the idea that proofs must be general. We are assuming that both $y_1$ and $y_2$ are limits of $f$ at $x_0$. Asserting that $y_1$ is limit of $f$ at $x_0$ means that for every neighborhood $V_1$ of $y_1$, there is a neighborhood $U_1$ of $x_0$ such that $x\in U_1\implies f(x)\in V_1$. And asserting that $y_2$ is limit of $f$ at $x_0$ means that for every neighborhood $V_2$ of $y_2$, there is a neighborhood $U_2$ of $x_0$ such that $x\in U_2\implies f(x)\in V_2$. Since this is true for every two neighborhoods $V_1$ and $V_2$ of $y_1$ and $y_2$ respectively, then, in particular, it must still be true when we choose those neighborhoods in such a way that they don't intersect. And doing that leads to a contradiction.
A proof being general means that it has to apply to every single case. But, in this context, this does not mean that it must work if we pick any two neighborhoods $V_1$ and $V_2$. If one pair of such neighborhoods leads to a contradiction, that's fine and we have then a proof.


The outside of a $180$-sheet roll of toilet paper is covered by two sheets; the inner cylinder, by one. What's wrong with how I counted the layers? Puzzle: A roll of toilet paper has 180 sheets on it. The outside is covered with exactly two sheets. The inside around the cardboard cylinder is covered by exactly one. Question of the puzzle: how many layers of toilet paper are on the roll of toilet paper?

The given solution: One way to solve this is by saying that the average round is covered by 1.5 sheets, so therefore the answer is $180\times\frac{2}{3}=120$

I tried a similar (but wrong) reasoning: "the average sheet makes an average of $\frac{3}{4}$ rounds (first sheet makes one round and the last sheet makes $\frac{1}{2}$ rounds), so the answer is $180\times\frac{3}{4}=$ 135"

  QUESTION: Apparently my answer is wrong. But since it seems analogical to the given solution I don't understand what error I made.

Possibly the growth of sheets per round is constant? While the (negative) growth of rounds per sheet is not constant? What are the related functions?

Put in another way: if $\frac{dSheets}{dRounds}=Constant$ isn't also $\frac{dRounds}{dSheets}=Constant$?

This question is linked to this question: Using differential equations to determine the number of rolls on a roll of toilet paper

There are twice as many sheets that make half a round than there are that make a single round, so it's inappropriate to simply average the numbers $1/2$ and $1$. A weighted average says the average sheet makes

$${2\cdot1/2+1\cdot1\over3}={2\over3}$$

of a round, which gives the answer $180\cdot2/3=120$ again.

I now understand that one cannot solve this puzzle using continuous functions. We can make an approximation however. This is how that would work:

$S$=total amount of sheets

$n$=number of completed rounds

We know that the growth of the amount of Sheets (S) per Round (n) is constant. Why? Because it is depending om the circumference which grows linearly with the radius with $2\pi r$.

Therefore the second derivative of $S(n)$ is constant: $S''(n)=C_1$

The first derivative is therefore: $S'(n)=C_1n+C_2$

In round $0$ the amount of sheets per round is $1$, so we can approximate: $S'(0)=C_2\approx 1$

In the last round (when we used up all $180$ sheets of toilet paper) the amount of sheets per round is $2$, so we can approximate: $S'(n_{180})=C_1 n_{180}+1\approx 2$

And we can conclude that $C_1=\frac{1}{n_{180}}$

So:

$$\bbox[5px,border:2px solid black]{S'(n)=\frac{1}{n_{180}}n+1}$$

We can integrate to find: $S(n)=\int S'(n)dn=\frac{n^2}{2n_{180}}+n+C_3$

$S(0)=C_3=1$

So:

$$\bbox[5px,border:2px solid black]{S(n)=\frac{n^2}{2n_{180}}+n+1}$$

Now we can take $S(n_{180})=\frac{n_{180}^2}{2n_{180}}+n_{180}+1=180$

To find that after $180$ sheets, the approximate amount of completed rounds is $n_{180}=\frac{358}{3}=119\frac{1}{3}$

Now we can also take the inverse to find that:

$$\bbox[5px,border:2px solid black]{n(S)=\sqrt{\frac{716}{3}S+14001\frac{7}{9}}-\frac{358}{3}}$$

With a derivative of:

$$\bbox[5px,border:2px solid black]{n'(S)=\frac{358}{3\sqrt{\frac{716}{3}S+14001\frac{7}{9}}}}$$

Now I see why $S'(n)$ is lineair, but $n'(S)$ is not lineair! Which was what I was trying to get my head around when asking this question.


Supremum and infimum of $\sum_{k=1}^{n}\frac{a_k}{a_{k}+a_{k+1}+a_{k+2}}$ Let, $n\ge 3$ be a fixed integer. Take all the possible finite sequences $(a_1,a_2,..,a_n)$ of positive numbers. Find, supremum and infimum of
$$\sum_{k=1}^{n}\frac{a_k}{a_{k}+a_{k+1}+a_{k+2}}$$
where, $a_{n+1}=a_1,a_{n+2}=a_2$

My partial progress:
Let, $s=a_1+a_2+..+a_n.$
Since, $$\frac{a_k}{s}\le\frac{a_k}{a_k+a_{k+1}+a_{k+2}}\le 1-\frac{a_k}{s}-\frac{a_{k+1}}{s}$$
which gives, $$1\le\sum_{k=1}^{n}\frac{a_k}{a_k+a_{k+1}+a_{k+2}}\le n-2$$
I guess, $n-2$ and $1$ will be the supremum and infinmum respectively,but I failed to show that.

Put $a_k=m^k$, when $m\rightarrow \infty $ the sum approaches $1$.

Put $a_k=m^{-k}$, when $m\rightarrow \infty $ the sum approaches $n-2$.


Solve for the roots in the following equations I have two equalities:
$$ \alpha x^{2} + \alpha y^{2} - y = 0 $$
$$ \beta  x^{2} + \beta y^{2} - x = 0 $$

Where $$ \alpha, \beta $$ are both known constants.

How can I solve for $x$ and $y$?

The two curves are circles through the origin and they will intersect in at most one other point.

By eliminating $x^2+y^2$, we have

$$\beta y=\alpha x,$$

then

$$\beta(\beta x^2+\beta y^2-x)=(\beta^2 + \alpha^2)x^2 -\beta x = 0. $$

The rest is immediate.

  $$\left(\frac\beta{\beta^2+\alpha^2},\frac\alpha{\beta^2+\alpha^2}\right).$$

We also have the special case $\alpha=\beta=0$, which is trivial.

We have

  $$ \alpha x^{2}+\alpha y^{2}-y=0 \longrightarrow(1) $$ and $\beta
 x^{2}+\beta y^{2}-x=0 \quad \longrightarrow \quad(2)$

Both equations can be written as:

  $\alpha\left(x^{2}+y^{2}\right)=y$ and $\quad \beta\left(x^{2}+y^{2}\right)=x$

Dividing first equation with later one, we get:

  $$
\begin{aligned}
& \frac{\alpha}{\beta}=\frac{y}{x} \\
\Rightarrow & \frac{\alpha x}{\beta}=y
\end{aligned}
$$

Putting the value of $y$ in $(1),$ we get:

  \begin{aligned}
& \alpha x^{2}+\frac{\alpha^{3} x^{2}}{\beta^{2}}-\frac{\alpha x}{\beta}=0 \\
\Rightarrow & \alpha x\left(x+\frac{\alpha^{2} x}{\beta^{2}}-\frac{1}{\beta}\right)=0 \\
\Rightarrow & x+\frac{\alpha^{2} x}{\beta^{2}}-\frac{1}{\beta}=0 \\
\Rightarrow & x\left(1+\frac{\alpha 2}{\beta^{2}}\right)=\frac{1}{\beta} \\
\Rightarrow & x\left(\frac{\beta^{2}+\alpha^{2}}{\beta^{2}}\right)=\frac{1}{\beta} \\
\Rightarrow & x=\frac{\beta}{\beta^{2}+\alpha^{2}}
\end{aligned}

Similarly find the value of $y$, yourself. $^_^$

Note: In this solution it is clear that I have assumed$\alpha \neq0, \beta \neq0,  x \neq0 y, \neq 0$.. For if this was not the case, as Eric Towers states in his answer, solution will be trivial.


Find the volume of a sphere using spherical coordinates Find the volume of the sphere $x^2 +y^2+z^2 \le 4$ that are between planes $z= 1/3$ and $z=-1/4$ using spherical coordinates

My attempt:

I know that $\theta$ would be from  $0\le\theta\le 2\pi $


What's the "dimensionality" of a matrix (not dimension!)? I know that a $m\times n$ matrix has dimension... well, $m\times n$.

However, in a paper (not a mathematics paper!) I encountered the term dimensionality of a matrix, whose definition is nowhere to be found.

Apparently, the dimensionality has to be a scalar, as it says e.g.:

"For all real, skew-symmetric matrices, the rank is 2 times the dimensionality, as eigenvalues come in conjugate imaginary pairs."

So, can anyone tell me what I have to deal with here?

Thanks!

Trying to find information on the 'dimensionality' of a matrix is a bit difficult for me! But I think I figured it out in laymens terms:

The dimensionality of a vector space is the number of unique ways a set of vectors 'point' within the vector space.

Consider a 2x2 vector space  $\vec{V} = \begin{bmatrix}
   a       & b \\
    c       & d \\
\end{bmatrix}$

The dimensionality of $\vec{V}$ is $4$, because it has four independant components that 'point' in different directions and $\vec{V}$ can be represented by a set of four basis vectors:
$A\begin{bmatrix} 1 & 0\\0 & 0\end{bmatrix}  + B\begin{bmatrix} 0 & 1\\0&0\end{bmatrix}+ C\begin{bmatrix} 0 & 0\\1&0\end{bmatrix} +D\begin{bmatrix} 0 & 0\\0&1\end{bmatrix}$. The basis vectors are arbitrary and all kinds of sets can satisfy the basis. But, you get what I mean.

Also consider the vector space $\vec{A}=A_x\hat{x} + A_y\hat{y} + 0\hat{z}$. It has dimensionality $2$ because the components only exist in the $x$-$y$ plane.

Lastly, consider the vector space $\vec{A}=k\hat{x} + k\hat{y} + k\hat{z}$. This one is tricky, because $\vec{A}$ appears to point in three different directions. But since the components are all the same, the vector space is just a one dimensional line from $($-$\infty,$ -$\infty,$ -$\infty)$ to $(\infty, \infty, \infty)$. So the dimensionality is $1$.

Also, I am still confused a bit about vector space dimensionality, so if anyone cares to right my wrongs, please do! I'm really unclear about this kind of stuff and any help is appreciated. Thanks!


Number of points at which $g(x)$ is non-differentiable I could not type the question in latex correctly,so i am posting the pic of my question.

My problem is that in the first part,i think that number of points where function is non differentiable are $3$(i.e at $x=-2,x=0,x=1 )$ but i am not sure about the answer.Is my answer correct.In second part,i am convinced that option (C) is correct.g(x) is continuous and non-differentiable at $x=0$.

For the third part,both $f(x)$ and $g(x)$ are zero at $x=-2$.So how will we find its finite limit as $x$ tends to $-2$.Please help me.

Would rather discard the first answer I have posted, but instead of editing it, I am reposting a more detailed answer, hopefully removing any errors of commission or omission from the first one. Also. OP edited his question in the mean time.

I $\color{blue}{agree}$ with you on the first part - with the following explanation.

We first start with the definition of $|x|$ which is used throughout.

$$
|x| = \left\{
        \begin{array}{ll}
            x & \quad x \ge 0 \\
            -x & \quad x \lt 0
        \end{array}
    \right.
$$

This is now used to simplify $f(x)$ successively

$$ f(x) = |x^2 - 3|x| + 2|$$
$$
 = \left\{
        \begin{array}{ll}
            |x^2 - 3x + 2| & \quad x \ge 0 \\
            |x^2 + 3x + 2| & \quad x \le 0
        \end{array}
    \right.
$$

$$
 = \left\{
        \begin{array}{ll}
            |(x-1)(x-2)| & \quad x \ge 0 \\
            |(x+1)(x+2)| & \quad x \le 0
        \end{array}
    \right.
$$

$$
 = \left\{
        \begin{array}{ll}
            (x-1)(x-2) & \quad x \ge 2 \\
            (x-1)(2-x) & \quad 1 \le x \lt 2 \\
            (x-1)(x-2) & \quad 0 \le x \lt 1 \\
            (x+1)(x+2) & \quad -1 \le x \lt 0 \\
            -(x+1)(x+2) & \quad -2 \le x \lt -1 \\
            (x+1)(x+2) & \quad x \lt -2
        \end{array}
    \right.
$$

Having done it so explicitly, it is easy to see that its roots $\pm 1$ and $\pm 2$

So now, the definition of $g$ becomes easy (there were intervals when maxima had to be found, but leaving that out for brevity).
$$
g(x) = \left\{
        \begin{array}{ll}
            0 & \quad x \le -1 \\
            2 & \quad -1 \le x \lt 1 \\
            0 & \quad 1 \le x \lt 2 \\
            x^2 + 3x + 2 & \quad  x \ge 2
        \end{array}
    \right.
$$

So $-1, 1$ and $2$ are $\color{blue}{3}$ points of discontinuity for $g(x)$, and hence not differentiable. At all other places, it is smooth.

On the $\color{blue}{second}$ part, since $g(x) = 2$ around $0$, so it is both continuous and differentiable.

On the $\color{blue}{third}$ part, the left hand limit is 0, the right hand is infinity, so the limit does not exist.

I read the first part of the definition as
$$\min_{t\in(-\infty,x]}f(t)$$
whenever $x<-1$.  Since $f(t)$ is decreasing on $(-\infty,-2)$, the minimum on the interval is $f(x)$, so $g(x)=f(x)$ if $x<-2$.  For $x\in[-2,-1]$, the interval $(-\infty,x)$ contains $-2$, so $g(x)=0$.


Expand the function as Taylor series Expand given function $f$ as Taylor series around $c=3$
$$f(x) = \frac{x-3}{(x-1)^2}+\ln{(2x-4)} $$

and find out open interval at what that series is convergent. What is radius of convergence?

This is what i have so far. We know that $\ln{(1+x)} = \sum_{n=1}^{\infty} \frac{x^n}{n}$ and $\frac{1}{1-x}=\sum_{n=1}^\infty x^n ,|x| \lt 1$

We can rewrite $\ln{(2x-4)}$ as $\ln{(-\frac{1}{4})}+\ln{(-x/2+1)}$ and then expand last expression, but i don't know what to do with $\ln{(-1/4)}$. We can can split given fraction onto partial fractions as $\frac{x-3}{(x-1)^2} = \frac{1}{x-1}-\frac{2}{(x-1)^2}$. First fraction we can expand easily, but i don't know how to expand fraction with binomial as denominator.

We have that for $|z|<1$,
$$\frac{1}{(1-z)^2}=\left(\frac{1}{1-z}\right)'=\left(\sum_{n\geq 0} z^{n}\right)'=\sum_{n\geq 1}n z^{n-1}.$$
Let $t=x-3$, then for $t\in (-1,1]$
$$\frac{x-3}{(x-1)^2}+\ln{(2x-4)}=\frac{t}{(t+2)^2}+\ln(2)+\ln(1+t)\\
=\frac{t}{4(1-(-t/2))^2}+\ln(2)+\ln(1+t)\\
=\frac{t}{4}\sum_{n\geq 1}n (-t/2)^{n-1}+\ln(2)-\sum_{n\geq 1}\frac{(-t)^n}{n}\\
=\ln(2)+\sum_{n\geq 1} t^n\cdot(-1)^{n-1}\left(\frac{n}{2^{n+1}}+\frac{1}{n}\right).
$$
Note that the radius of convergence is $1$, that is the smaller radius between the radius of the series of $\ln(1+t)$, that is $1$,  and the radius of the series of $\frac{1}{(1-(-t/2))^2}$, that is $2$.


Is $R$ finitely generated? Let $A$ be a commutative ring with identity. Given two submodules $R,S$ of $A^n(n\in\Bbb N)$ and suppose $S$ is finitely generated, if there exists an isomorphism of $A$-modules $A^n/R\simeq A^n/S$, is $R$ finitely generated?

My example (below) was wrong, I misinterpreted what OP was asking for. I'll leave it up so others don't get confused as I did.

No: Let $A = \mathbb{C}[x_1, x_2, \ldots, x_n, \ldots]$ be the polynomial ring in infinitely many variables over $\mathbb{C}$, considered as a module over itself (i.e. the $n$ in your question is just $1$). Let $R$ be the submodule generated the even-indexed variables $x_{2i}$ and let $S = \{0\}$. Then $A/R \cong A/S \cong A$ and $S$ is finitely generated, but $R$ is not finitely generated.


Find the number of ways to choose a countable subset of R I need to find how many ways there are to choose countable subset from $\mathbb R$ the set of real numbers.
This confuse me because $\mathbb R$ is uncountable, although he has countable subsets like Netural Numbers set.

The cardinality is $\mathfrak{c}$. To show this it suffices to find an injective map from the set of infinite sequences of subsets of $\mathbb N$ to subsets of $\mathbb N$. Consider the map
$$ (A_1, A_2, \cdots ) \mapsto \bigcup_{n = 1}^\infty \{p_n^{a} : a \in A_n \}$$
where $p_n$ is the $n$th prime.


Maximum of $\sum_{cyc}x_1x_2x_3\cdots x_k$ given $x_1+x_2+\ldots+x_n=a$ for non-negative $x_1,\ldots,x_n$ for a positive integer $n$ and $1<k<n$ How do I find the maximum value of
$$\sum_{cyc} \prod_{i=1}^{k} x_{i}$$ for $1< k<n$
given that $x_1,x_2,\ldots,x_n$ are non-negative reals satisfying $$x_1+x_2+\ldots+x_n = a$$ for an integer $n\ge1$?

The case where $k=n$ can be solved by straightforward $AM-GM$ hence why I put $k<n$.

If $n=3, k=2$. It's well-known that $x^2+y^2+z^2 \ge xy+xz+yz$ so $3(xy+xz+yx)\leq (x+y+z)^2$.

$k=1$ is trivial hence why I am interested in $1<k$.

$n=3, k=2,3$ is just AM-GM.

Partial Result (when $k$ divides $n$).  The maximum value in this case is $\left(\dfrac{a}{k}\right)^k$.  However, I think the answer is still $\left(\dfrac{a}{k}\right)^k$ even when $k$ does not divide $n$.
Here we assume that $x_1,x_2,\ldots,x_n\geq 0$ (if the variables are strictly positive, then there are cases where the maximum does not exist).  Suppose that $k$ is a divisor of $n$.  Write $n=mk$ for some positive integer $m$.  Let
$$S:=\sum_{i=1}^n\,x_ix_{i+1}x_{i+2}\cdots x_{i+k-1}\,,$$
where the indices are considered modulo $n$.  Observe that
$$S\leq \prod_{r=1}^{k}\,\left(x_r+x_{r+k}+x_{r+2k}+\ldots+x_{r+(m-1)k}\right)\,.$$
By the AM-GM Inequality,
$$S\leq \left(\frac{1}{k}\,\sum_{r=1}^k\,(x_r+x_{r+k}+x_{r+2k}+\ldots+x_{r+(m-1)k})\right)^k\,.$$
Consequently,
$$S\leq \left(\frac{1}{k}\,\sum_{i=1}^n\,x_i\right)^k=\left(\frac{a}{k}\right)^k\,.$$
The maximum value is attained, for example, when $$x_1=x_2=\ldots=x_k=\dfrac{a}{k}\,,$$ and $$x_{k+1}=x_{k+2}=\ldots=x_n=0\,.$$

Another Partial Result (when $k=2$ and $n\geq 5$ is an odd integer).  The maximum value is also $\left(\dfrac{a}{k}\right)^k=\left(\dfrac{a}{2}\right)^2$.
To show this, suppose without loss of generality that $x_1$ is the smallest value among $x_1,x_2,\ldots,x_n$.  Observe that
$$S\leq S_1S_2\,,$$
where
$$S:=x_1x_2+x_2x_3+\ldots+x_{n-1}x_n+x_nx_1\,,$$
$$S_1:=x_1+x_3+x_5+\ldots+x_{n}\,,$$
and
$$S_2:=x_2+x_4+x_6+\ldots+x_{n-1}\,.$$
This is because all terms $x_1x_2,x_2x_3,\ldots,x_{n-1}x_n$ in $S$ appear in $S_1S_2$, and the term $x_nx_1$ in $S$ is less than or equal to the term $x_nx_2$ in $S_1S_2$.  Therefore,
$$S\leq \left(\frac{S_1+S_2}{2}\right)^2=\left(\frac{a}{2}\right)^2\,.$$
The maximum value is attained, for example, when $$x_1=x_2=\dfrac{a}{2}\text{ and }x_3=x_4=\ldots=x_n=0\,.$$

Whenever you have an expression that is symmetrical over all variables, if you have only one point with extreme value it is achieved when all variables are equal.
Therefore max or min the sum of $k$-cyclic products of $n$ elements is $n(\frac{a}{n})^k$ and if you take $n>1, x_1=x_2=x_{n-1}=0, x_n=a$ you have the sum equal to $0 < n(\frac{a}{n})^k$, that means that $n(\frac{a}{n})^k$ is the maximum value.


Why are the laws of sines/cosines "laws" and not "theorems"? So in logic we have every line of a proof being either an axiom or a theorem -- but then why do we have concepts like the "law of sines" and the "law of cosines"? Are these technically "theorems" as well? How is "law" being used here? Is there a mathematical reason? A historical one?

One could be forgiven for thinking the terminology is a mess. There are other examples of laws/rules in mathematics, some trigonometric, but not in general. The term "identity" is also common (in fact, some of those links came from here). I could find nothing anywhere indicating when to use each term. However, based on my experience, the way people use the terms in the names of named results seems to go as thus (no doubt with plenty of exceptions):

Any result of the form $f(x)=g(x)$ with the same variables $x$ on each side tends to be called an identity (whereas $f(x)\ne g(x)$ would be an inequation, and $f(x)\ge g(x)$ and $f(x)>g(x)$ would be inequalities), unless it governs the arithmetical operations on such variables, e.g. $ab=ba$, in which case "law" is typically used;
Any equally simple to use result for calculations, which may use different variables on each side, is called a law or rule, sometimes interchangeably (e.g. $a=2R\sin A$ has been called both the law of sines and the sine rule, although for some reason you only seem to see "law" in trigonometry, e.g. no-one would refer to Cramer's law);
A result you can use procedurally to verify a claim tends to be called a criterion;
Other results tend to be called theorems: in particular, "for all $x$ there exists a $y$ such that..." would be a theorem.

Technically, all of these are theorems, but even if people acknowledge them as such, the name of the result might not reflect that.

This is a law because no other law can prove this. Though using basic mathematics can prove this. Like pythagoras theorem is a special case of cosine law hence it is called a theorem


Investment question using logarithm 8,000 dollars is invested in an account that yields 6% interest per year. After how many years will the account be worth $14 000, to the nearest half year, if the interest is compounded monthly?

Progress
$A = P (1 + i)^n$, so $14 000 = 8 000 (1 + 0.06)^n$, $14 000 = 8 000 (1.06)^n$, $1.75 = (1.06)^n$, $\log 1.75 / \log 1.06 = n$, and so $n = 9.6$ years. Am I correct?

OK.  If you have an annual percentage rate of $6\%$ compounded monthly, that means each month you add $\frac{6\%}{12}$ of the investment to itself.

So each month the investment is multiplied by by $1.005$.  ($\text{Investment plus Investment}\times0.005$).

So after $x$ months, your investment is $8000\times(1.005)^x$  Therefore, you need to solve the following equation:$$8000\times(1.005)^x=14000$$

to get the number of months it will take to get to \$14000.  I see in your comment above that you know how to solve this. You will have to divide by 12 to get years...

Halfway between 9 and 10 years. Set a formula multiplying the base by 1.06 per year, on the compounding amount.


Is there any shortcut to find if a number is a perfect cube? Is there any shortcut to find if a number is a perfect cube?

I am taking for instance finding if a number is a perfect square. So , if a number ends with $2,3,7,8$. It cannot be a square. But if it does not end then it is not compulsory that the number is a perfect square. So we add up the digits to find if the sum is $1,4,7$ or $9$. If it is then it is a perfect square.

Example: $13689 =$ not ending with $2,3,7$ or $8$. We add up the digit $= 9$. So it's a perfect square $= 117$. But for $44$ does not end with $2,3,7,8$. We add up the digits $= 8$, hence not a perfect square.

So, I was wondering if there is any such rule to find if a number is a perfect cube. Please, no computing or algorithmic solutions. Just need a theoretical answer.

I have been making something that needs to check a bunch of numbers and see if they are cubes or not.  This is what I found out in the process:

If you take your number $n$ module $7$ and the result is not $0, 1$, or $6$ then it's not a cube.
If $n$ module $13$ is not $0, 1, 5, 8$, or $12$ then $n$ is not a cube.
There are many more like the first 2 examples which I have a list of at the bottom.
If you know any of it's prime factors, like $2$ for example, then you can divide your number by $2^3$ (If it can't divide by $2^3$ then it's not a cube, but if it can then...)
check if that is a cube.
The best thing (that I know of) to do after those tests is to make guesses and then cube them and then pick higher or lower guesses and repeat.
EXAMPLE: $1718$.  Well maybe to start you guess $10$ which cubed is $1000$, since $1718 > 1000$ then we pick a higher number.  Say we pick $14$.  Well, $14^3 = 2744$ and since $1718 < 2744$ then next time we pick a lower guess... and repeat until we have it between $11$ and $12$ then since it is in between $1718$ isn't a cube.

LIST OF MODULE CHECKS:

if $n$ module $7$ is not one of these: $0, 1, 6$ then $n$ is not a cube
if $n$ module $9$ is not one of these: $0, 1, 8$ then $n$ is not a cube
if $n$ module $13$ is not one of these: $1, 8, 12, 5, 0$ then $n$ is not a cube
if $n$ module $19$ is not one of these: $1, 8, 7, 11, 18, 12, 0$ then $n$ is not a cube
if $n$ module $31$ is not one of these: $1, 8, 27, 2, 30, 16, 29, 23, 4, 15, 0$ then $n$ is not a cube
if $n$ module $37$ is not one of these: $1, 8, 27, 14, 31, 10, 26, 36, 6, 29, 23, 11, 0$ then $n$ is not a cube

Hope that gives you what you were looking for. :)

This method will work for 1 followed by any number of zeros;

Let's take 1000 for example.
1000 has three zeros, and 3 is divisible by 3, so 1000 is a perfect cube
Here:
 The prime factorisation of 1000 will give us 2*2*2*5*5*5. Since this is cube roots, we group the 2's and 5's to get 2*5=10, and 10 is the cube root of 1000.

Next example, 10000, which has 5 zeros. 5 can't be divided by 3 to get equal whole numbers, so 10000 is not going to be a perfect cube number.
 Here,
the prime factorisation of 10000 will give us 2*2*2*2*2*5*5*13. We can't group the 5's and the 13, so this shows that 10000 is not a perfect cube number.

This method can work for only a 1 followed by how many zeros it has. Pretty much, if the number of zeros can be divisible by 3 to get equal whole numbers, it's a perfect cube number!

Something else you might want to know,

Take 89 cubed
89 cubed - 89*89*89 = 684909
The number of digits of the cube are equal to the number of 89's that are their, which is 3 89's. This shows that the number of digits that are there for the extended form of 89 cubed, ( which is 6 here) gives us how many digits there will be in the answer, which is 6. This is for odd numbers, but for even numbers it would change.
 Now let's take 34
34 cubed= 34*34*34. The extended form has 6 digits. But for this, we have to actually minus 1 form the total number of digits. 6-1=5. So, the answer should have 5 digits. Let's check. The cube of 34 gives us( drumroll please) 39304, which has ( drumroll) 5 digits! This method works for whole number cubing.
 Sorry if this is kind of complicated, and again if this does not satisfy you.
By the way, I also had the same question, and did some research on the internet. I came up with nothing! The stuff I gave you above is pure observation and multiple testings! If you're not sure if they are not correct, try out those methods on some numbers and see if they are actually accurate ( I've already tried it, that's why I'm posting this here, duh)! Have fun!

-TheDRAGon


Calculating Combinations with Duplicates and Replacements Hi I'm stuck at doing my student project work, and this is my first time doing Combinatorics, so any help would be useful.
I'm attempting to calculate the probability that given $k$ blocks of ice falling and filling $n$ empty blocks on the ground, that all empty blocks are filled.
So I've hit a bit of a snag. The way I'm approaching this is to visualize the "ice falling" as a random bag that are filled with $1 - n$ numbers (denoting the spot the ice would fall in). It is possible for ice to fall on the same spot (meaning with replacement).
So what I've know is that I have to calculate the number of ways ice fall using combinations with replacement (combinations because as long as all holes are filled it does not matter). What I don't know is how to handle the duplication.
Furthermore, I'm also a little lost on how to calculate the numerator. I figure I need to find the excess number of duplicates, and then find the number of ways with only that amount of duplicates.
So given $k = 3$, and $n = 2$, I need to find:

the total ways it can fill the area, and
the number of ways it can fill the area given x duplicates.

I saw this post, and it seems pretty close to what I have to do, but I don't really understand why its doing $(n-2)$ and then $(n-4)$.
Example: Given 3 ice blocks and 2 holes on the ground, there are 4 ways to fill it up:
$$
\newcommand{\set}[1]{\{#1\}}
\set{1,1,1}\\\set{1,1,2}\\\set{2,2,1}\\\set{2,2,2}
$$
and two of these methods would work.

I don't want to post a complete answer, because I'm afraid I'd end up doing your project, but I'll get you started.  There are $n^k$ possible draws you can make.  Let's subtract the ones that don't include every number from $1$ to $n$.  There are $n$ choices for the omitted number, and $(n-1)^k$ ways to make all selections from the other other numbers, so that gives $$n^k-n(n-1)^k$$
The problem with this computation is that draws that miss two numbers are subtracted twice, once for each excluded number.  At this point, you need to look at the principle of inclusion and exclusion.
EDIT
In the case $k=3, n=2$ it is impossible for two numbers to be omitted, as then we wouldn't choose any numbers at all, so the inclusion-exclusion adjustment is not required.  The probability is$$
\frac{n^k-n(n-1)^k}{n^k}=\frac{2^3-2\cdot1^3}{2^3}=\frac{8-2}{8}=\frac68$$


How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression? How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression? For example here is the sum of $\cos$ series:

$$\sum_{k=0}^{n-1}\cos (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \cos \biggl( \frac{ 2 a + (n-1)\cdot d}{2}\biggr)$$

There is a slight difference in case of $\sin$, which is:
$$\sum_{k=0}^{n-1}\sin (a+k \cdot d) =\frac{\sin(n \times \frac{d}{2})}{\sin ( \frac{d}{2} )} \times \sin\biggl( \frac{2  a + (n-1)\cdot d}{2}\biggr)$$

How do we prove the above two identities?

Let $$ S = \sin{(a)} + \sin{(a+d)} + \cdots + \sin{(a+nd)}$$ Now multiply both sides by $\sin\frac{d}{2}$. Then you have $$S \times \sin\Bigl(\frac{d}{2}\Bigr) = \sin{(a)}\sin\Bigl(\frac{d}{2}\Bigr) + \sin{(a+d)}\cdot\sin\Bigl(\frac{d}{2}\Bigr) + \cdots + \sin{(a+nd)}\cdot\sin\Bigl(\frac{d}{2}\Bigr)$$

Now, note that $$\sin(a)\sin\Bigl(\frac{d}{2}\Bigr) = \frac{1}{2} \cdot \biggl[ \cos\Bigl(a-\frac{d}{2}\Bigr) - \cos\Bigl(a+\frac{d}{2}\Bigr)\biggr]$$ and $$\sin(a+d) \cdot \sin\Bigl(\frac{d}{2}\Bigr) = \frac{1}{2} \cdot \biggl[ \cos\Bigl(a + d -\frac{d}{2}\Bigr) - \cos\Bigl(a+d+\frac{d}{2}\Bigr) \biggr]$$

Then by doing the same thing you will have some terms cancelled out. You can easily see which terms are going to get Cancelled. Proceed and you should be able to get the formula.

I tried this by seeing this post. This has been worked for the case when $d=a$. Just take a look here:

http://mathforum.org/library/drmath/view/72777.html

There's also a nice proof of the cosine formula by Michael Knapp in Mathematics Magazine (vol 82, number 5, December 2009, p. 371-372). Download a free version here: https://www.maa.org/sites/default/files/Knapp200941575.pdf
Knapp mentions in his article that a proof for the sine formula was given by Samuel Greitzer in the obscure student-oriented math journal Arbelos from 1986.


How can I prove that $2^n\ge n^4$? I'm trying to prove the following inequality.
$$2^n\ge n^4 (n>n_0)$$
I noticed that from $n = 20$, the following inequality holds, and I want to use induction for the proof. However I simply cannot find any way to prove whenever $n$ holds, $n+1$ also holds as well. Any help would be appreciated. Thanks!

You can use induction to show that $2^n \geq n^4$ for all $n \geq 16$. When $n = 16$, we have $2^{16} = 16^4$. Suppose that $2^n \geq n^4$ for $n \geq 16$. We will show that $2^{n+1} \geq (n+1)^4$:
$$
(n+1)^4 = \left(1 + \frac{1}{n}\right)^4 \cdot n^4 \leq \left(1 + \frac{1}{16}\right)^4 \cdot 2^n < 2 \cdot 2^n = 2^{n+1}.
$$


Synthetic homotopy theory lecture question In this lectureLogic and topology at 20:21 there is a statement that type isContr which means "is contractible" means that for a given type A and every element x and pointed  element a of type A, there is a path from a to x ( first formula on the slide).
As it of course is definition, it just means everything author may want to, but as ,,"is contractible" has well topological meaning, I would like to ask if it is true?
Because at topological level "is contractible" means that every closed path ( loop ) has to be contractible and it is quite different. From other HTT lectures I know that it is a notion of homotopic equivalence to trivial path ( path(A,a,a) = Id) should be used so it looks like path operation is not well defined on this lecture or presenter at least forgot to mention it is a loop in fact...

Or do I do not understand anything?

In classical topology, a contractible space is one that's equipped with a basepoint and a homotopy from the identity map to the constant map at that basepoint, i.e. a point $a$ and for every point $x$ a path from $x$ to $a$, varying continuously with $x$.  This is exactly what the HoTT definition of IsContr says, except that the "varying continuously" doesn't have to be said explicitly since inside of type theory everything is "continuous".

I suppose Voevodsky use isContr as "is Contractible" in a meaning that there is at least one element of that type. In fact it would be equivalent to saying that type is not empty. It became clear at the univalence axiom level where there's consequence that "unique existence equals to real excistence". So if some object is unique ( up to isomorphism which is equivalent to up to contractible fibration of type) then there is actually an element of it ( which is the point of we contact to) .


Proving $\int_{0}^{\infty} \mathrm{e}^{-x^2} dx = \frac{\sqrt \pi}{2}$ How to prove
 $$\int_{0}^{\infty} \mathrm{e}^{-x^2}\, dx = \frac{\sqrt \pi}{2}$$

This is an old favorite of mine.
Define $$I=\int_{-\infty}^{+\infty} e^{-x^2} dx$$
Then $$I^2=\bigg(\int_{-\infty}^{+\infty} e^{-x^2} dx\bigg)\bigg(\int_{-\infty}^{+\infty} e^{-y^2} dy\bigg)$$
$$I^2=\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}e^{-(x^2+y^2)} dxdy$$
Now change to polar coordinates
$$I^2=\int_{0}^{+2 \pi}\int_{0}^{+\infty}e^{-r^2} rdrd\theta$$
The $\theta$ integral just gives $2\pi$, while the $r$ integral succumbs to the substitution $u=r^2$
$$I^2=2\pi\int_{0}^{+\infty}e^{-u}du/2=\pi$$
So $$I=\sqrt{\pi}$$ and your integral is half this by symmetry

I have always wondered if somebody found it this way, or did it first using complex variables and noticed this would work.

No entender mucho el idioma pero aqui dejo mi solucion en una imagen didactica.


complex numbers - argand diagram How do you draw on an argand diagram:
$\{z\in{\mathbb C}: \arg(z-1) < \arg(z-i)\}$?
I can plot both points but I don't know what to do with arguments and inequalities.

If you simply interpret what it means to say that $\arg(z-1)<\arg(z-i)$ geometrically, then it shouldn't be too hard to visualize the region.  In each of the regions below, the argument of the point ($z$) minus the number (either $1$ or $i$) simply the angle from the dashed line to the arrow.  It's easy to see that this angle is smaller from the complex number $1$ in exactly the blue regions.

Note that I'm assuming a branch cut at $\pi$ in the argument function so that, in the horizontal strip, $\arg(z-1)$ is definitely negative while $\arg(z-i)$ is positive.

Setting $\displaystyle z=x+iy,$ we need $$\arctan\frac y{x-1}<\arctan\frac{y-1}x$$

$$\iff\arctan\frac y{x-1}-\arctan\frac{y-1}x<0 $$

$$\iff\arctan\left(\frac{\dfrac y{x-1}-\dfrac{y-1}x}{1+\dfrac y{x-1}\cdot\dfrac{y-1}x}\right)<0 $$

$$\iff\arctan\left(\frac{x+y-1}{x^2+y^2-x-y}\right)<0 $$

$$\iff\frac{x+y-1}{x^2+y^2-x-y}<0 $$

So, we need the sign of $\displaystyle  x+y-1$ and $\displaystyle x^2+y^2-x-y=\left(x-\frac12\right)^2+\left(y-\frac12\right)^2-\frac12$ to be opposite

We know, $\displaystyle \left(x-\frac12\right)^2+\left(y-\frac12\right)^2-\frac12=0$ is a circle with radius $\frac1{\sqrt2}$ with center at $\left(\frac12,\frac12\right)$.

Edit

As can easily be plotted on WolframAlpha, this region looks like so:

Unfortunately, the arguments can easily be read off of the image as the angles between the vectors and the dashed lines.  It certainly appears that $\arg(z-1)>\arg(z-i)$ for certain points in the region.