probability question (dice) Suppose you toss three fair six-sided dice simultaneously until you observe a six. So if none of the three
  dice are a six, then you toss them simultaneously again. Find the probability that the first occurrence
  of a six requires n tosses.

Since P(a six first toss) =$1/6+1/6+1/6=1/2$ shouldn't it be $n/2$ after $n$ tosses?

The probability of getting a six on the first toss is not $1/2$. If the reasoning that led you that conclusion were valid, you could argue that with six dice you’d be certain to get a six on the first toss, which clearly isn’t the case.

The easiest way to calculate the correct probability of getting a six on the first toss is to calculate the probability of not getting a six and subtract that from $1$. In order not to get a six, you must roll a not-$6$ on each of the three dice. The probability of rolling a not-$6$ on one die is $5/6$, and the dice rolls are independent, so the probability of rolling a not-$6$ on all three dice is $(5/6)(5/6)(5/6)=125/216$, and the probability of getting at least one six is therefore $1-125/216=91/216$, or a little over $0.42$.

Now in order to get a six for the first time on the $n$-th toss, you must get no sixes on each of the first $n-1$ tosses, and then you must get at least one six on the $n$-th toss.  The probability of getting no six on one toss is, as we just saw, $125/216$, so the probability of getting no six on each of $n-1$ tosses is $$\left(\frac{125}{216}\right)^{n-1}\;.$$ This needs to be followed by a successful toss, an event whose probability is $91/216$, so the probability of getting your first six on the $n$-th toss is $$\left(\frac{125}{216}\right)^{n-1}\left(\frac{91}{216}\right)=\frac{91\cdot125^{n-1}}{216^n}\;.$$

The first toss:

$$P(\text{no 6}) = \left(\frac{5}{6}\right)^3 = \frac{125}{216}$$

We generalize. The chance you throw a six after $n$ turns is:

$$P (\text{six in }n\text{th throw}) = \left(\frac{125}{216}\right)^{n-1} \frac{91}{216} = \frac{91 \cdot 125^{n-1}}{216^n}$$


How would you solve this differential equation: $\frac{d^2y}{dx^2} = \frac{100}{y}$? The equation is:

$$\frac{d^2y}{dx^2} = \frac{100}{y}$$

Also if $y = f(x)$,

$f'(0) = 0$ and $f(0) = 10$

How would you solve this for $y$?

$$y''=\frac {100} y$$
is autonomous (ie $x$ is not present in the equation). The standard procedure therefore is to substitute $u(y)=y'$ with
$$ y'' = \frac{du}{dx} = \frac{du}{dy}\frac{dy}{dx} = u'u $$
solve
$$ u'u = 100/y $$
with $u'u = \frac{1}{2}(u^2)'$ we get
$$ u(y)^2 = 2\left(c_1 + 100\ln(y)\right). $$
This equation "can be integrated"
$$ y' = \pm\sqrt{2\left(c_1 + 100\ln(y)\right)} $$
$$ \Rightarrow \int\frac{1}{\sqrt{2\left(c_1 + 100\ln(y)\right)}}dy = \mp\int 1\,dx $$
This is the point where it gets ugly...

The integral can be evaluated to
$$ \sqrt{\frac{\pi}{200}} e^{-c_1/100}\operatorname{erfi}\left(\sqrt{\frac{c_1}{100}+\ln(y)}\right) + c_2 = \mp x $$
where $\operatorname{erfi}$ is the imaginary error function. Solve for $y$ and you are done...

Wolfram alpha gives this result which shows it is not easy to find.


Determine whether the sequence is convergent or divergent Here is the sequence:

$$a_n = \frac{\sqrt{n^4+1}}{n^2+1}$$

How to do the limit?

Hint. Use equivalents, $\sqrt{n^4+1}\sim n^2$ and $n^2+1\sim n^2$.

Observe that
$$
\lim_{n\to\infty} a_n^2=\lim \frac{n^4+1}{n^4+2n^2+1}=1.
$$
Since $a_n>0$ and the function $\sqrt{x}$ is continuous for $x>0$, it follows that $a_n$ converges to $1$ as well.

Furthermore, the series $\sum_{n=1}^{\infty} a_n$ is divergent (but this might not have been what you were asking for?)


proving via binomial theorem that $6^n-5n \equiv 1 \pmod{25}$ Use the binomial theorem to prove that remainder is always $1$ when $6^n-5n$ is divided by $25$

I haven't tried anything yet. I know only mathematical induction theorem to prove this type of question, but I can solve it with mathematical induction.

$$\begin{align}
6^n-5n
&= (1+5)^n -5n \\
&=((n\mathop{\mathsf{C}}0)+(n\mathop{\mathsf{C}}1) \cdot 5+(n\mathop{\mathsf{C}}2) \cdot 5^2+ \cdots +(n\mathop{\mathsf{C}}n) \cdot 5^n) -5n \\
&= 1+5n-5n+((n\mathop{\mathsf{C}}2) \cdot 5^2+(n\mathop{\mathsf{C}}3) \cdot 5^3 + \cdots + (n\mathop{\mathsf{C}}n) \cdot 5^n) \\
&= 1+5^2((n\mathop{\mathsf{C}}2)+(n\mathop{\mathsf{C}}3)\cdot 5 +\cdots+(n\mathop{\mathsf{C}}n)5^{n-2})
\end{align}$$
Hence $1$ will always be the remainder when divided by $25.$


Finding the Moment Generating function of a Binomial Distribution Suppose $X$ has a $\rm{Binomial}(n,p)$ distribution. Then its moment generating function is

  \begin{align}
M(t)
&= \sum_{x=0}^x e^{xt}{n \choose x}p^x(1-p)^{n-x} \\
&=\sum_{x=0}^{n} {n \choose x}(pe^t)^x(1-p)^{n-x} \\
&=(pe^t+1-p)^n
\end{align}

Can someone please explain how the sum is obtained from lines (2) to (3)?

The moment generating function for the binomial distribution $B_{n,p}$, whose discrete density is $\binom{n}{k}p^k(1-p)^{n-k}$, is defined as
$$
\begin{align}
M_{B_{n,p}}(t)
&=\mathrm{E}(e^{tk})\\
&=\sum_{k=0}^n\binom{n}{k}p^k(1-p)^{n-k}e^{tk}\\
&=\sum_{k=0}^n\binom{n}{k}\left(pe^t\right)^k(1-p)^{n-k}\\
&=\left(pe^t+(1-p)\right)^n
\end{align}
$$
The last step is simply an application of the binomial theorem.

$\phi(t) = \mathbb{E}[e^{tX}]
       \Rightarrow \mathbb{E}[e^{t\cdot(\Sigma x)}]
       \Rightarrow \mathbb{E}[e^{tx_1}\cdot e^{tx_2}\cdot...\cdot e^{tx_n}]
       \Rightarrow \mathbb{E}[e^{tx_1}]\cdot \mathbb{E}[e^{tx_2}]\cdot...\cdot \mathbb{E}[e^{tx_n}]$
Since all individual events are independent,
$\Rightarrow [pe^t + (1-p)].[pe^t + (1-p)].[pe^t + (1-p)]...[pe^t + (1-p)]$ n times, since all n random variables are Bernoulli random variables
$\Rightarrow[pe^t + (1-p)]^n=[pe^t + q]^n$, where $q=1-p$.


The shape of a graph of a function with $n$th-roots? Not just these type of functions:

$$\sqrt[3]{x}=x^{1/3} \;\;\;\text{and} \;\;\; \sqrt[8]{x}=x^{1/8}$$

But also more complicated expressions, like expressions that have $n$th roots inside of $n$th roots, or exponentiation inside of $n$th roots:

$$ \sqrt[5]{\sqrt{x}} \;\;\;\;\;\text{or}\;\;\;\;\; \sqrt[9]{x^{12}} \;\;\;\;\;\text{or}\;\;\;\;\; \sqrt[13]{\sqrt[7]{\sqrt{x}}}.$$

Is there a general idea about the shapes of graphs of such functions? What shapes does a graph take on if you keep adding $n$th roots to the inside of the function, for example going from $\sqrt[7]{x}$ to $\sqrt[7]{\sqrt{x}} \dots$ etc?

Thank you.


Prove $\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \cdots$ converges to $\frac 1 2 $ Show that
$$\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \cdots = \frac{1}{2}.$$

I'm not exactly sure what to do here, it seems awfully similar to Zeno's paradox.
If the series continues infinitely then each term is just going to get smaller and smaller.

Is this an example where I should be making a Riemann sum and then taking the limit which would end up being $1/2$?

Solution as per David Mitra's hint in a comment.

Write the given series as a telescoping series and evaluate its sum:

$$\begin{eqnarray*}
S &=&\frac{1}{1\cdot 3}+\frac{1}{3\cdot 5}+\frac{1}{5\cdot 7}+\cdots  \\
&=&\sum_{n=1}^{\infty }\frac{1}{\left( 2n-1\right) \left( 2n+1\right) } \\
&=&\sum_{n=1}^{\infty }\left( \frac{1}{2\left( 2n-1\right) }-\frac{1}{
2\left( 2n+1\right) }\right)\quad\text{Partial fractions decomposition}  \\
&=&\frac{1}{2}\sum_{n=1}^{\infty }\left( \frac{1}{2n-1}-\frac{1}{2n+1}
\right) \qquad \text{Telescoping series} \\
&=&\frac{1}{2}\sum_{n=1}^{\infty }\left( a_{n}-a_{n+1}\right), \qquad a_{n}=
\frac{1}{2n-1},a_{n+1}=\frac{1}{2\left( n+1\right) -1}=\frac{1}{2n+1} \\
&=&\frac{1}{2}\left( a_{1}-\lim_{n\rightarrow \infty }a_{n}\right) \qquad\text{see below} \\
&=&\frac{1}{2}\left( \frac{1}{2\cdot 1-1}-\lim_{n\rightarrow \infty }\frac{1
}{2n-1}\right)  \\
&=&\frac{1}{2}\left( 1-0\right)  \\
&=&\frac{1}{2}.
\end{eqnarray*}$$

Added: The sum of the telescoping series $\sum_{n=1}^{\infty }\left( a_{n}-a_{n+1}\right)$ is the limit of the telescoping sum $\sum_{n=1}^{N}\left( a_{n}-a_{n+1}\right) $ as $N$ tends to $\infty$. Since

$$\begin{eqnarray*}
\sum_{n=1}^{N}\left( a_{n}-a_{n+1}\right)  &=&\left( a_{1}-a_{2}\right)
+\left( a_{2}-a_{3}\right) +\ldots +\left( a_{N-1}-a_{N}\right) +\left(
a_{N}-a_{N+1}\right)  \\
&=&a_{1}-a_{2}+a_{2}-a_{3}+\ldots +a_{N-1}-a_{N}+a_{N}-a_{N+1} \\
&=&a_{1}-a_{N+1},
\end{eqnarray*}$$

we have

$$\begin{eqnarray*}
\sum_{n=1}^{\infty }\left( a_{n}-a_{n+1}\right)  &=&\lim_{N\rightarrow
\infty }\sum_{n=1}^{N}\left( a_{n}-a_{n+1}\right)  \\
&=&\lim_{N\rightarrow \infty }\left( a_{1}-a_{N+1}\right)  \\
&=&a_{1}-\lim_{N\rightarrow \infty }a_{N+1} \\
&=&a_{1}-\lim_{N\rightarrow \infty }a_{N} \\
&=&a_{1}-\lim_{n\rightarrow \infty }a_{n}.\end{eqnarray*}$$


Do 'Sample from' and 'insert parameter' commute? The Setting is as follows:

We are given random variables $X$ and $\Theta$ but we are not so much interested into $X$ itself as its Distribution needs a Parameter $\theta$ which is produced by $\Theta$. Concrete example:

$X$ is a binary variable mapping to $\{0,1\}$ and is Bernoulli distributed with Parameter $\theta$. $\Theta$ is Beta-distributed with fixed Parameters $\alpha_0, \beta_0$. This means that we actually model the conditional density

$$f_{X|\Theta}(x, \theta) = \theta^x (1-\theta)^{1-x}$$
and
$$f_\Theta(\theta) = \frac{\Gamma(\alpha_0)\Gamma(\beta_0)}{\Gamma(\alpha_0 + \beta_0)} \theta^{\alpha_0} (1-\theta)^{1-\beta_0}$$

Now let us assume that I want to sample from $X$. People do the following:

(A):

1) Sample a value $\theta$ from the Beta distribution

2) Insert this value into the Bernoulli distribution and sample from it with this concrete Parameter

We could equivalently do the following:

(B):

Compute the Distribution of $X$ and then sample from it directly.

QUESTION: DO THESE METHODS GIVE THE SAME RESULT?

I guess that one should ask the question more precisely: Do I really get the Distribution of $X$ if I do (A) often enough? I dont see why an ''equation'' of the form

$$\text{Sample}(X|(\Theta|(\alpha_0, \beta_0))) = \text{Sample}(X|\text{Sample}(\Theta|(\alpha_0, \beta_0)))$$
should be true...

Note: (dunno whether it helps or not)
In the example given it is easy to compute the Distribution of $X$ as

$$f_X(x) = \int_\theta f_{X|\Theta}(x, \theta) f_\Theta(\theta) d\theta$$

and it turns out that

$$P[X=x] = \frac{B(\alpha_0+x, \beta_0+(1-x))}{B(\alpha_0, \beta_0)}$$

where $B$ is the Beta function.

Schemes A and B don't return the exact same samples, but each item returned has the same distribution under A as under B. The reason is precisely the equation
$$
f_X(x) = \int_\theta f_{X|\Theta}(x, \theta) f_\Theta(\theta) d\theta.\tag1
$$
Under scheme A, the joint density of $(X,\Theta)$ is
$$
f_{X,\Theta}(x,\theta)=f_{X|\Theta}(x,\theta)f_\Theta(\theta)\tag2
$$
so the marginal density of $X$ is obtained by integrating out $\Theta$, as in equation (1).  What (1) is saying is that if you generate $\Theta$ from $f_\Theta$, and then generate $X$ conditional on $\Theta$ from $f_{X|\Theta}$, and then throw away $\Theta$, the resulting $X$ has the same density as if you generated $X$ from $f_X$ directly.

The reason why scheme A is often used is that there may not be a closed form for the marginal density $f_X$ in (1), so scheme B is not possible.

EDIT: An empirical simulation might help. Here's some code in R to apply scheme A to a situation where $\Theta$ has a normal distribution with mean $2$ and SD $1$, and having observed the value $\Theta=\theta$, we generate $X$ as normal with mean $\theta$ and SD $1$.

> set.seed(17)
> m <- 2
> theta <- rnorm(5000, m, 1)
> x <- rnorm(5000, theta, 1)
> mean(x)
[1] 2.014371
> sd(x)
[1] 1.41051
> mean(theta)
[1] 1.985225
> sd(theta)
[1] 0.9878088

When we simulate 5000 times, we find that $X$ has approximately mean $2$ and SD $\sqrt 2$. This agrees with (1), which gives the marginal density of $X$ as normal with mean $2$ and variance $2$.

You can compute the joint distribution of $(X,\Theta)$ using (2): It is bivariate normal with mean vector $\begin{pmatrix}1\\1\\\end{pmatrix}$ and covariance matrix $\begin{pmatrix}2&1\\1&1\\\end{pmatrix}$. The simulation also agrees with this:


Test the hypothesis .... A random sample of size 20 from a normal population give a sample mean of 42 and sample
standard deviation of 6. Test the hypothesis that the population mean is 44 at 5 % level of
significance.

First calculate the statistic
$$t=\frac{44-42}{6}\sqrt{20}=1.49$$
then decide, using the fact that $t\sim \mathcal{T}_{(19)}$


$\mathbb{Q}[X,Y]/(X^2+Y^2-1)$ is integrally closed I'm was browsing this question, where it is proven the quotient field of $\mathbb{Q}[X,Y]/(X^2+Y^2-1)$ is isomorphic to the rational function field $\mathbb{Q}(t)$ under the isomorphism
$$
(x,y) \mapsto \left( \frac{2t}{t^2+1}, \frac{t^2-1}{t^2+1} \right).
$$
How can this isomorphism be used to show $\mathbb{Q}[X,Y]/(X^2+Y^2-1)$ is normal in its quotient field?

  I identified $\mathbb{Q}[X,Y]/(X^2+Y^2-1)$ with $\mathbb{Q}\left[\frac{2t}{t^2+1}, \frac{t^2-1}{t^2+1} \right]$, and was trying to show this is integrally closed in $\mathbb{Q}(t)$.

If $f/g\in\mathbb{Q}(t)$, with $f,g$ coprime, is integral over $\mathbb{Q}[2t/(t^2+1),(t^2-1)/(t^2+1)]$, then
$$
(f/g)^n+h_{n-1}(f/g)^{n-1}+\cdots+h_1(f/g)+h_0=0,\qquad h_i\in \mathbb{Q}[2t/(t^2+1),(t^2-1)/(t^2+1)].
$$
Multiplying through by $g^n$, I found the relation
$$
f^n=g(-h_{n-1}f^{n-1}-\cdots-h_0g^{n-1})
$$
so any prime dividing $g$ divides $f$, so $g$ is constant, and in fact $f/g\in\mathbb{Q}[t]$. I was hoping to see that is is in fact in $\mathbb{Q}[2t/(t^2+1),(t^2-1)/(t^2+1)]$, but I think I've made a mistake somewhere.

Let's prove that $R=\mathbb{Q}\left[\frac{2t}{t^2+1},\frac{t^2-1}{t^2+1}\right]$ is integrally closed.

We have $\mathbb{Q}\left[\frac{2t}{t^2+1},\frac{t^2-1}{t^2+1}\right]=\mathbb{Q}\left[\frac{1}{t^2+1},\frac{t}{t^2+1}\right]$. It's obvious that $\mathbb Q(t)$ is its field of fractions.  Let $f(t)/g(t)\in\mathbb Q(t)$ integral over $R$ with $\gcd(f(t),g(t))=1$. For any element $h\in R$ there is $r\in\mathbb N$ such that  $(t^2+1)^rh\in\mathbb Q[t]$. Consider an integral dependence relation $$(f/g)^n+h_{n-1}(f/g)^{n-1}+\cdots+h_1(f/g)+h_0=0$$ with $h_i\in R$ and multiply this by $(t^2+1)^{nr}$ with $r$ large enough in order to get an integral dependence relation for $(t^2+1)^rf(t)/g(t)$ over $\mathbb Q[t]$. Since $\mathbb Q[t]$ is integrally closed we get $(t^2+1)^rf(t)/g(t)\in\mathbb Q[t]$, and therefore $g(t)\mid(t^2+1)^r$. Since $t^2+1$ is irreducible over $\mathbb Q$ we have $g(t)=(t^2+1)^s$. Now it's clear that $f(t)/g(t)\in R$.

Due to both denominator's being the same that ring  is isomorphic to the ring $\mathbf{Q}[2t, t^2-1]$. That is, the subring of $\mathbf{Q}[t]$ generated by $2t$ and $t^2-1$. But in the presence of $2t$ we do not need $t^2-1$. So this subring can be  generated by just $2t$. So this is isomorphic to the PID $\mathbf{Q}[2t]$, hence normal.


How to prove that the number of diagonals of a $M \times N$ matrix is $M+N-1$? Given a matrix $M \times N$, how to prove that the number of diagonals that can be drawn, like in the figure below, is equal to $M + N - 1$?

My idea would be to prove it with induction and consider the 3 possible cases: $M=N$, $M>N$ and $M<N$. But I don't know how to do it.

This is a counting problem.

Start at the first row and count diagonals passing through elements on the first row. You get $N$ of those.

Now go down through the last column and you get $M$ of those.

There is one which is counted twice, so the total is $M+N-1$

Each entry of the matrix is given by the row number $i$ and the column number $j$, and we say that $(i,j)$ is the index of this entry.  Notice that the $k$-th diagonal is defined by $(i,j)$-entries with $i+j=k+1$.  Since $i\in\{1,2,\ldots,M\}$ and $j\in\{1,2,\ldots,N\}$, we see that
$$k=i+j-1\in\{1,2,\ldots,M+N-1\}\,.$$

Each $k\in\{1,2,\ldots,M+N-1\}$ corresponds to a nonempty diagonal line.  For $k=1,2,\ldots,M$, the $k$-th diagonal contains the entry indexed by $(k-1,1)$.  For $k=M+1,M+2,\ldots,M+N-1$, the $k$-th diagonal contains the entry indexed by $(M,k+1-M)$.


Examples of mathematical induction What are the best examples of mathematical induction available at the secondary-school level---totally elementary---that do not involve expressions of the form $\bullet+\cdots\cdots\cdots+\bullet$ where the number of terms depends on $n$ and you're doing induction on $n$?

Postscript three years later: I see that I phrased this last part in a somewhat clunky way. I'll leave it there but rephrase it here:

      --- that are not instances of induction on the number of terms in a sum?

Here is the first example I saw of induction, and I still think it's a beautiful one.

Problem: Prove that a $2^n \times 2^n$ sheet of graph paper with one box removed, can be tiled with L-shaped trominos.

Proof: For the $n=1$ case, there is nothing to prove: a $2 \times 2$ grid with one box removed is exactly a L-tromino.

For $n=2$, consider the $4 \times 4$ grid. You can divide it into four $2\times 2$ grids. The removed box is in one of those four sub-grids, so that sub-grid  can be covered with an L-tromino (is an L-tromino, rather). What about the other 3 sub-grids? Put an L-tromino right in the center of the huge grid, which covers them!

Now the remaining part of each of them is a $2\times2$ grid with one box removed. I leave it to you to complete the proof, and teach it to the students as you best see fit.

The figures above are from Mathematical Circles: Russian Experience by Dmitri Fomin, Sergey Genkin, and Ilia Itenberg, specifically the chapter on Induction which is written by I.S. Rubanov. The book actually doesn't use a variable $n$, but asks for a $16\times 16$ square, then in the form of a discussion between a teacher and a student works through the $2\times 2$ and $4\times 4$ and $8\times 8$ cases, until it is obvious that we have in fact proved a theorem for any $2^n \times 2^n$ ('It looks like we have a "wave of proofs running along the chain of theorems $2\times2 \longrightarrow 4\times4 \longrightarrow 8\times8 \longrightarrow$ It is quite evident that the wave will not miss any statement in this chain.')

As an aside, this is a lovely book with quite a bit of non-trivial mathematics suitable for elementary school and high-school students (though I read in late high school).

This theorem and proof are also on the cut-the-knot website: Tromino Puzzle and Golomb's Inductive Proof.

Maybe it would be good to start with something that the students already know is obviously true. For example, you could develop an inductive proof that $2n$ is always even.

I think this would make it easy for them to see what the base case, inductive hypothesis and inductive steps mean.

For the base case we show our statements holds for the smallest possible $n$. In this case it is $0$ and $2(0)=0$, which is even.

For the inductive step we have to show that $2(n+1)$ is even but we get to use the inductive hypothesis: $2n$ is even. Then it's just a matter of the distributive property to get $2n+2$ and the common knowledge that adding $2$ to an even number gets another even number.

Other candidates could be anything shown with straightforward structural induction on Peano numbers.


How to prove that the exponential function is the inverse of the natural logarithm by power series definition alone The exponential function has the well-known power series representation/definition:

$e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$

And the natural logarithm has the less well-known power series representation/definition:

$\ln(x) = \sum_{k=1}^\infty \frac{(x - 1)^k}{k}(-1)^{k + 1}$

Let's for the purpose of this question pretend we have no previous concept of exponentials and logarithms: all we have are two power series called $e^x$ and $\ln(x)$ and, just for the fun of it, we want to see if they are the inverse of one another. So we put one power series into the other:

$\ln(e^x) = \sum_{k=1}^\infty \frac{(\sum_{n=0}^\infty \frac{x^n}{n!} - 1)^k}{k}(-1)^{k + 1}$

How does one prove, using only the above power series, any algebraic manipulation of the expression and any theorems pertaining to power series that do not require the anterior knowledge of $e^x$ or $\ln(x)$, that this is x?

I enjoyed Christian's answer, and if you want the shortest proof starting from the series definitions, then that is the way to go.  But, like you, I was curious about this question because I like to manipulate power series, and because this problem ended up being harder than I thought it would be.
Here is a solution that meets your requirements.  The only thing I have omitted is proving that exchanging the order of summation is valid when I do it below, which is probably not the part of the proof you are interested in.
A few lemmas are needed, which I will list first.  The first two are well-known combinatorial identities, and I will omit the proofs.  I'll follow your approach, but I'll delay the expansion of $e^x$ a little bit longer than you did, because otherwise you get sums over an arbitrarily large number of variables, which end up being difficult to handle.
Lemma 1: For $m,n \in \mathbb{Z}$ with $0 \le m < n$,
$$ \sum_{k=0}^n (-1)^k \binom{n}{k} k^m = 0 $$
The easiest way to prove this is with calculus, but it can be proven by induction on $m$ from the result that for $n \ge 1$,
$$ \sum_{k=0}^n (-1)^k \binom{n}{k} = 0 $$
which itself can be proven by induction on $n$.  So, no calculus required.
Lemma 2:
$$ \sum_{n=k}^m \binom{n}{k} = \binom{m+1}{k+1} $$
This can also be proven by induction.
Lemma 3: $ e^x e^y = e^{x+y} $
Proof: This one is a lot easier than the main result.  Expand the right hand side using the series definition and the binomial theorem to get
$$ e^{x+y} = \sum_{m=0}^\infty \frac{(x + y)^m}{m!} = \sum_{m=0}^\infty \sum_{i=0}^m \frac{1}{m!} \binom{m}{i} x^i y^{m-i} = \sum_{i=0}^\infty \sum_{m=i}^\infty \frac{x^i y^{m-i}}{i! (m-i)!} $$
Let $j=m-i$ and the sums become independent:
$$ \sum_{i=0}^\infty \sum_{m=i}^\infty \frac{x^i y^{m-i}}{i! (m-i)!} = \sum_{i=0}^\infty \sum_{j=0}^\infty \frac{x^i y^j}{i!j!} = \left( \sum_{i=0}^\infty \frac{x^i}{i!} \right) \left( \sum_{j=0}^\infty \frac{x^j}{j!} \right) = e^x e^y $$
Lemma 4: $ (e^x)^y = e^{xy} $ when $y$ is a nonnegative integer.
Proof: This is obvious using Lemma 3.
Main result: $ \log(e^x) = x $
Proof: Apply the power series for the logarithm, then a binomial expansion, and then the power series for the exponential.
$$ \begin{align}
\log(e^x) &= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} ( e^x - 1 )^n \\
&= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \sum_{k=0}^n \binom{n}{k} (-1)^{n-k} e^{kx} \\
&= \sum_{n=1}^\infty \frac{(-1)^{n+1}}{n} \sum_{k=0}^n \binom{n}{k} (-1)^{n-k} \sum_{m=0}^\infty \frac{k^m x^m}{m!} \\
&= -\sum_{m=0}^\infty \frac{x^m}{m!} \sum_{n=1}^\infty \frac{1}{n} \sum_{k=0}^n \binom{n}{k} (-1)^k k^m \\
&= -\sum_{m=0}^\infty \frac{x^m}{m!} a_m \tag{1} \label{1}
\end{align} $$
where $a_m$ has been defined as
$$ a_m = \sum_{n=1}^\infty \frac{1}{n} \sum_{k=0}^n \binom{n}{k} (-1)^k k^m $$
Split the sum over $n$ into two parts, from $1$ to $m$ and from $m+1$ to $\infty$.
$$ a_m = \sum_{n=1}^m \frac{1}{n} \sum_{k=0}^n \binom{n}{k} (-1)^k k^m + \sum_{n=m+1}^\infty \frac{1}{n} \sum_{k=0}^n \binom{n}{k} (-1)^k k^m $$
On the right, the summation over $k$ is zero by Lemma 1 because $n > m$.  This gives us
$$ a_m = \sum_{n=1}^m \frac{1}{n} \sum_{k=0}^n \binom{n}{k} (-1)^k k^m \label{2} \tag{2} $$
In this expression, when $k \ge 1$, we can apply the identity
$$ \frac{1}{n} \binom{n}{k} = \frac{1}{k} \binom{n-1}{k-1} \label{3} \tag{3} $$
Fortunately, the $k=0$ terms in \eqref{2} are all zero because of the $k^m$ term (since we are only considering $a_m$ where $m \ge 1$).  Dropping the $k=0$ terms, moving the summation over $n$ to the right, and applying \eqref{3} in \eqref{2}, we get
$$ \begin{align}
a_m &= \sum_{k=1}^m (-1)^k k^m \sum_{n=k}^m \frac{1}{k} \binom{n-1}{k-1} \\
&= \sum_{k=1}^m (-1)^k k^{m-1} \sum_{n=k}^m \binom{n-1}{k-1}
\end{align} $$
Applying Lemma 2 to the summation over $n$ gives
$$ a_m = \sum_{k=1}^m (-1)^k k^{m-1} \binom{m}{k} $$
This can be simplified by again applying Lemma 1.  However, note that Lemma 1 requires the $k=0$ terms, which we dropped earlier so that we could apply \eqref{3}.  Now the $k=0$ term in the sum must be replaced by including $k=0$ in the sum and subtracting it outside the sum.
$$ a_m = \left( \sum_{k=0}^m (-1)^k k^{m-1} \binom{m}{k} \right) - 0^{m-1} $$
Obviously this term on the right is zero when $m>1$, but the $m=1$ case results in $0^0$.  Here the appropriate convention is that $0^0=1$.  (For example, when we expanded the exponential as $e^{kx} = \sum_m k^m x^m/m!$, the $m=0$ term is interpreted as $1$ even when $k=0$.)  Since the $0^{m-1}$ term yields 0 for $m \ne 1$ and 1 for $m=1$, we can write this using the Kronecker delta as $\delta_{m,1}$.
$$ a_m = \left( \sum_{k=0}^m \binom{m}{k} (-1)^k k^{m-1} \right) - \delta_{m,1}  $$
Now Lemma 1 can be applied again.  (Note that the conditions of the lemma are satisfied because $m-1 < m$.)  This gives $ a_m = -\delta_{m,1} $.  Inserting this into $\eqref{1}$ gives the final result,
$$ \log(e^x) = x $$

I like the series approaches above, quite interesting.

I had a much more mundane approach that involves using the chain rule and inverse functions.  Suppose that g(x) is the inverse of f(x), whereby

g(f(x))=x

Take the derivative of both sides (remember the chain rule):

(dg/df)(df/dx)=1

Take the special case that f(x)=exp(x).

(dg/df)exp(x)= (dg/df)f = 1
or
dg/df = 1/f

Integrating both sides, you'll get that g(f) is whatever the integral of 1/f comes to.  Calling that the logarithm,

g(f)=ln(f)

That works with x too:  g(x)=ln(x) which is the inverse of f(x)=exp(x)


Population dynamics, square root and zero derivative Hy everybody !

I'm studying population dynamics for my calculus exam, and I don't understand something that seems really easy, so I thought you might be able to help me out ;)

Here's the thing. I have this differential equation $\frac{dN}{dt} = \sqrt{N}$.

Our book makes us realize that both $N(t) = 0$ and $N(t) = \frac{t^2}{4}$ are solution, which makes sense so far, simply by replacing in the original equation.

Now suppose we start at $N(0) = 0$. How can $N$ start growing like $\frac{t^2}{4}$ if it's derivative is $0$ at $t = 0$ ? Because zero derivative should mean no growth, so $\sqrt{N}$ should remain zero, which means still no growth, and so on. My brain is melting right now.

Zero derivative at one point does not mean no growth at other points.

For example the  derivative of function $$N(t)=\frac {t^2}{4}$$ is $$N'(t)=t/2$$ That is the derivative is zero at $t=0$ but it is not zero at other points. The point of the problem is that in this case the solution is not unique.

You are correct. There are indeed two solutions. If the population is zero it will stay zero. The only way it will grow is if the initial conditions are positive. However, the population zero solution is "unstable" in the sense that even if the initial condition is slightly positive, the $t^2$ solution will kick in and push the population away from zero.


In most geometry courses, we learn that there's no such thing as "SSA Congruence". In most geometry courses, we learn that there's no such thing as "SSA Congruence". That is, if we have triangles $ABC$ and $DEF$ such that $AB = DE$, $BC = EF$, and $\angle A = \angle D$, then we cannot deduce that $ABC$ and $DEF$ are congruent.

However, there are a few special cases in which SSA "works". That is, suppose $ABC$ is a triangle. Let $AB = x$, $BC = y$, and $\angle A = \theta$. For some values of $x$, $y$, and $\theta$, we can uniquely determine the third side, $AC$.

(a) Use the Law of Cosines to derive a quadratic equation in $AC$.

(b) Use the quadratic polynomial you found in part (a) in order to find conditions on $x, y,$ and $\theta$ which guarantee that the side $AC$ is uniquely determined.

I have solved part (a) and gotten $z^2-2xz\cos (\theta) + (x^2-y^2)=0$.

I don't quite get part (b). It is obvious $90^{\circ}$ is a solution but do I have to prove that this is right? Or do I need to prove that this is the only solution? Or is there other answers?

My work: All I know is that the discriminant would be $4x^2z^2\cos^2 \theta-4(-x^2+y^2)$

P.S. My answer for part(a) is correct, right?

In the generic case, there are two triangles (up to congruence).  To see this, suppose we are given AB, BC, and the angle at A.  Fix the segment AB and let BC rotate through a circle (centered at B).  Form the line from A at the given angle from AB.  That line can intersect the circle at two points.

True, it can be tangent to that circle but that's precisely the case in which the leg BC is perpendicular to AC.

You have a quadratic equation for $z$. Most quadratic equations have either two solutions. If you want to uniquely determine z, you need to have only ONE solution, right?

My advice (I'm not going to solve the problem, but this is a pretty good hint!) is to use your knowledge of solving quadratic equations to solve for z, and then ask yourself when the two solutions you get will be equal. For example, if I'm solving $z^2-5z+6$, I could use the quadratic formula to find $z=2$ or $z=3$, so there's two solutions, and I haven't uniquely determined $z$.

If you get stuck, look up "discriminant".


Completeness of the Hausdorff distance. THEOREM
Let us define $$F(X) := \{ A \subseteq X | A \neq \emptyset, A - \text{closed and bounded} \} $$
Then if $(X, \rho ) $ is complete, then $(F(X), d_h )$ is complete.

Are there any interesting proofs of this theorem other than the ones with long computations with $\epsilon$? I have been searching the internet for a while, but could not find anything.

HINT:

Step 1. Reduction to a simpler case. Assume that we have $A_n$ bounded and closed such that $d(A_m, A_n) \to 0$. Consider $A'_n \colon = \overline{\cup_{m \ge n} A_m}$.  Check that $d(A_n, A'_n) \to 0$ ( simple). Now we have $(A'_n)$ decreasing and Cauchy.

Step2: Proof in the simpler case : if $A_n$ is a decreasing sequence of nonvoid, closed and bounded subsets of $X$ such that $d(A_n, A_m) \underset{m, n \to \infty}{\longrightarrow} 0$, then $A\colon =\cap A_n$ is nonvoid and $A_n \to A$.

A bit of $\epsilon$ work : take $\epsilon > 0$ and consider $(n_k)$ a subsequence of the naturals so that $d(A_m, A_n) \le \epsilon/2^k$ for $m,n \ge n_k$. Every point in $A_{n_1}$ is at distance at most $\epsilon$ from a point in $A$. Indeed, take $x_1$ in $A_{n_1}$ arbitrary, then $x_2 \in A_{n_2}$ , $d(x_2, x_1) \le \epsilon/2$, then $x_3 \in A_{n_3}$, $d(x_3, x_2) \le \epsilon/2^2$, and so on. The sequence $x_n$ converges to a point in $x$ in $A$ and we have $d(x, x_1) \le \epsilon/2 + \epsilon/2^2 + \cdots = \epsilon$.

Basically we show that $A_n$ converges to $\lim \sup A_n$ ( limsup in the lattice of closed subsets).

Hint: You know that the Hausdorff distance is equivalently defined by $d_h(A, B)=\max\{\sup\limits_{a\in A}\rho(a, B),\sup\limits_{b\in B}\rho(b, A)\}.$ Pick a Cauchy sequence $A_n\subset F(X)$ and show that it converges with this metric.


groups of order $p^5$ and exponent p We know that a group of order $p^5$ is an extra special group. How we can show it doesn't have any abelian subgroup of order $p^4$? Also what is the presentation of this group if its exponent is equal to $p$ ?

If $P$ is an extraspecial non-Abelian $p$-group of order $p^{5},$ then $Z(P) = [P,P] = \Phi(P)$ has order $p,$ so is cyclic, say equal to $\langle z \rangle.$ Then $P/Z(P)$ is an elementary Abelian group of order $p^{4},$ and may be viewed as a $4$-dimensional vector space $V$ over $F,$  the field with $p$-elements.

The vector space $V$ admits a well defined non-degenerate alternating $F$-bilinear form $\langle, \rangle :V \times V \to F$ by setting $\langle xZ(P),yZ(P) \rangle = j1_{F}$ if $[x,y] = z^{j},$ (where $0 \leq j \leq p-1$).

If $A$ is an Abelian subgroup of $P$ containing $Z(P),$ then $A/Z(P) \leq \left( A/Z(P) \right)^{\perp},$ so $A/Z(P)$ is at most $2$-dimensional, as the form is non-degenerate.
Hence $|A| \leq p^{3}.$

An alternative proof may be constructed with some knowledge of character theory. If $P$ had an Abelian subgroup $A$ of order $p^{4},$ then by Clifford's theorem, every irreducible character of $P$ of degree greater than $1$ is induced from a degree $1$ irreducible character of $A,$ so has degree at most $p.$ However, $P$ has $p-1$ complex irreducible characters of degree $p^{2}$ and $p^{4}$ irreducible characters of degree $1,$ so there can be no such $A.$

Whether, in the above example, the equation C_P(x)=Z(P) , (x is not in Z(P)), holds or not?
In fact, I am looking for a 2-group such that P/Z(P) is an elementary abelian, but
C_P(x)≠Z(P) .


Two normal operator that commutes Suppose $N\in B(H)$ is normal, and $T\in B(H)$ is invertible. Prove that if $TNT^{-1}$ is normal then $N$ commutes with $T^*T$.

I can not any idea to prove it, just I know  $T^*TNT^{-1}{T^*}^{-1}N^*T^*T =N^*T^*TN$, because $TNT^{-1}$ is normal.

Please just give me a hint .

Thanks in advance.

Because $V = TNT^{-1}$ is normal and $N$ is normal, then
$$
           V = \int \lambda dE_{V}(\lambda),\;\;\; N=\int\lambda dE_{N}(\lambda),
$$
and $E_{V}(S) = TE_{N}(S)T^{-1}$ for all Borel subsets $S$. Hence,
$$
  TE_{N}(S)T^{-1} = E_{V}(S)=E_{V}(S)^{\star}=(T^{-1})^{\star}E_{N}(S)T^{\star}.
$$
Therefore,
$$
                 T^{\star}TE_{N}=E_{N}T^{\star}T.
$$
There are details for you to fill in, as requested.

Here is an interesting observation:

if $v \in \ker(N)$ then $T^*T v \in \ker(N)$

Proof:  keep in mind that $\ker(N)=\ker(N^*)$ and that this holds for all normal operators. Let $v \in \ker(N)$ be any vector. Since $T$ is invertible there is $\tilde{v}$ such that $v = T^{-1}\tilde{v}$. Then $T N T^{-1} \tilde{v} = 0$. But being $T N T^{-1}$ normal we get $(T N T^{-1})^* \tilde{v} = 0$ hence $$(T^*)^{-1} N^* T^*\tilde{v} = 0$$ so $$T^*\tilde{v} \in \ker(N)$$ that is $T^*Tv \in \ker(N)$ as we wanted to show.

This solves the problem for finite dimensional spaces. Indeed, for any $\lambda \in \mathbb{C}$ consider the normal operator $N_{\lambda} := N - \lambda Id$. Then $T N_{\lambda} T^{-1}$ is normal hence (by the above applied to $N_{\lambda}$) we get $T^*T$ preserves $\ker(N_{\lambda})$ which are (for the adequated $\lambda$'s) the eigenspaces of $N$. This implies that $N$ commutes with $T^*T$. Indeed, the vectors space $H$ is a finite sum of eigenspaces $V_{\lambda_j}:= \ker(N_{\lambda_j})$, $j=1,\cdots,k$. i.e.
$$H = \sum_{j=1}^k V_{\lambda_j} \,.$$
Let $v \in H$ arbitrary and $v = \sum_{j=1}^k v_j$ its decomposition according to the above decomposition of $H$. Then $$ \begin{aligned} T^*T N v &= \sum_{j=1}^k T^*T N v_j = \sum_{j=1}^k T^*T \lambda_j v_j\\ &= \sum_{j=1}^k \lambda_j T^*T v_j = \sum_{j=1}^k N T^*T v_j \\ &= N T^*T \sum_{j=1}^k v_j = N T^*T v \end{aligned}$$
since $v \in H$ is arbitrary we get $T^*T N = N T^*T $  .


Why is discriminant less than zero? the question is find the range of values of $c$ for which the expression $4x^2-4x+4c^2-8$ is non-negative for all real values of $x$.

I got the discriminant but I don't understand why it has to be less than zero. If it is non-negative, shouldn't it be positive and more than zero?

Can someone help me explain this, thanks :)

Here it is easy to complete the square to obtain

$$(2x-1)^2+4c^2-9\ge 0$$

You ought to recognise your discriminant here. The minimum value of the left-hand side clearly occurs with $2x=1$ and you should be able to finish from here.

With $ax^2+bx+c$ you can complete the square cleanly by multiplying by $4a$ to obtain $$4a^2x^2+4abx+4ac=(2ax+b)^2+4ac-b^2$$

You just need to bear in mind that if $a$ is negative, multiplying by $4a$ changes the sign.

The form of the equation in this question meant that no fiddling was necessary. But the general case shows you where the discriminant comes in, and might help you to understand the sign.

The Parabula is "smiling" and hence in order it will be positive, you need that there will be no intersection with the $x$ axis, that is there are no solutions to the equation
$$4x^2-4x+4c^2-8=0.$$
That happens if and only if the discriminant is negative.


How to find the number of trousers sold by a merchant on a specific day of the month? The problem is as follows:

The owner of a bazaar in Daegu obtained for the sale of pants of the
same price, $6804$ usd in September, $10800$ usd in October and $7938$
usd in December. If the price of each pants is greater than $14$ usd
and less than $21$ usd, how much did he earned on the first day of
January, assuming that he sold a number of pants equal to the product
of the digits of its price?.

The alternatives given in my book are as follows:
$\begin{array}{ll}
1.&\textrm{360 usd}\\
2.&\textrm{252 usd}\\
3.&\textrm{486 usd}\\
4.&\textrm{144 usd}\\
\end{array}$
The official solution according to my precalculus workbook is as follows:
First, find the $\operatorname{gcd}$ of $6804$, $10800$ and $7938$.
Therefore:
$\operatorname{gcd}(6804, 10800, 7938)=54$
This can be obtained from the prime factorization of each number:
$6804=2^2\cdot 3^5\cdot 7$
$10800=2^4 \cdot 5^2 \cdot 3^3$
$938=3^4\cdot 7^2 \cdot 2$
Then $\operatorname{gcd}(6804, 10800, 7938)=2\cdot 3^3 = 54$
Therefore the pants price is $18$ usd.
Because the quantity sold on January 1st coincides with the product of the digits of the price then:
$\textrm{number of pants}=1\times 8=8$
Since the price per pant is $18$ usd. Then the quantity he earned is:
$8\times 18 =144\,\textrm{usd}$
Therefore the answer is option 5.
This is the part where it ends the official solution.
Now my problem with this answer is I don't know why did the author used the greatest common divisor? I mean at some point he does obtain $54$, but what is it used for? Can someone help me here?.
There is a very imporant aspect and missing step here and that is how did he got $18$ usd the price per pant?
I'm not very savvy with the correct usage of the greatest common divisor, but I believe the intented approach is that to get the price I must find a common divisor for the three. But why should this be the maximum?
The numbers which $54$ multiplies, with those being 126, 200 and 147 I assume that these must be the number of pants sold on the months of september, october and december. But from this logic I'm getting that the price of the pants is 54 usd.
This itself causes a contradiction as the boundaries set by the problem constrain the value of the price of the pants between:
$14<p<21$
If this were the case, then I believe that the price of the pants might be a sub multiple of 54?. But which?, 2?, 3?, 6?, 9?.
Because if any of those were the case the price could be, 15,16,18 and 20. This part confuses me the most.
So in brief the sort of answer which I'm aiming to get is a one which first and foremost explain why was it used the greatest common divisor then how was it obtained the price per pant and on which cases does the greatest common denominator is used? and how should it be avoided its confusion with the least common multiple?. In other words why using the least common multiple isn't used here. By the way I've read the wikipedia entry on the subject but it did not helped me much as it only covers two applications none of which seem to properly explain its usage in word problems.
Since what I'm lacking is the conceptual explanation it would really help me a step by step explanation for this problem. So can someone help me here?.

The price per pants must be a divisor of each of the amounts of money made in each of the months, as presumably he sold an integral number of pants. Thus, the price per pant must divide the greatest common common divisor of 6804 (money made in September), of 10800 (money made in October), and of 7938 (money made in December). Since the greatest common divisor is 54, the price per pant must be a divisor of 54.
(The defining property of the greatest common divisor of a set of numbers is that (i) it divides each of the numbers in the set; and (ii) any number that divides all numbers in the set divides the greatest common divisor. So the price per pant, which must divide each of the three amounts, must also divide $54$)
Now, what are the divisors of $54$? As $54=2\times 3^3$, the divisors are: $1$, $2$, $3$, $6$, $9$, $18$, $27$, and $54$. The price must be one of them.
We are told the price is “more than $14$ and less than $21$”. Well, the only possibility then is that the price per pant is $18$.

The way via the greatest common divisor is absolutely not necessary. Actually the revenues for the given three months aren't necessary at all to solve the problem. So, I add this short answer, even if it does not address the OP's question:
We know for the price $p=10a+b$ and $a\cdot b \in \{5,6,7,8,9\}$ since a price of $20$ would mean no pants sold.
Given the options, the revenue $p\cdot a\cdot b$ must be even. So only $16$ or $18$ are to be considered. Checking these, $p= 18$, hence, a revenue of $144$ is the only possible answer.


Find $\lim_{x\to 0} \frac {\ln( \sin 3x)}{\ln (\sin x)}$ Find $$\lim_{x\to 0} \frac {\ln (\sin 3x)}{\ln (\sin x)}$$ without using L'Hospital

Now for this I have tried substituting $\sin 3x=3\sin x-4\sin^3x$ and then used some log properties to simply it further but it did not turn out to be good.  I also tried to use the facts that $\lim_{x\to 0} \frac {\sin x}{x}=1$ and $\lim_{x\to 0} \dfrac {\ln (1+x)}{x}=1,$ but still had no success.

Any help would be greatly appreciated

I think you can use Taylor expansion of $\sin x$:
$$\sin (\alpha x)=\alpha x\ +\ o(x^2)$$
Then
$$\lim_{x\to 0^+} \frac {\ln (\sin 3x)}{\ln (\sin x)}=\lim_{x\to 0^+} \frac {\ln 3x}{\ln x}=\lim_{x\to 0^+}  \frac{\ln 3}{\ln x}+\frac{\ln x}{\ln x}=1$$


Diagonalizable Operators: An Operational Extension Let $T$ be a diagonalizable operator on a vector space $V$. Prove that the operator
$$a_nT^n + a_{n-1}T^{n-1}+\cdots+a_1T+a_0 Id_V$$
 on $V$ is also diagonalizable for any scalars $a_1, a_1,\dots,a_n$.

First off, what is $Id_V$? I've never seen this before.

Hint:

$$\exists\,P\;\,\text{an invertible operator}\,\;s.t.\;\, P^{-1}TP=D\;\,\text{is diagonal}\implies \forall\,n\in\Bbb N:$$

$$P^{-1}T^nP=\left(P^{-1}TP\right)^n=D^n\;\;\text{is diagonal,  and from here the claim follows}$$

I found this in Michael Artin's Algebra, and I feel it will be of use here:


The number of way to distribute n indistinguishable balls into k indistinguishable boxes arbitrarily Before even attempting at the proof, I need a clarification of a point in the question. I don't understand how the $k$ in $k \geq \lambda_1\geq ...\geq \lambda_r\geq 1$ is relevant. For $n=7$ and $k=3$, for example, isn't $\{1,5,1\}$ a valid multiset of ball distribution, but $(5,1,1)$ is not a valid integer partition of 7, since $\lambda_1 = 5 > 3 = k$?

Also, to prove the statement, I'm thinking of setting up a bijection between the set of ball distributions and the set of integer partitions of n. Would this be a sensible approach?

I don't understand how the $k$ in $k \geq \lambda_1\geq ...\geq \lambda_r\geq 1$ is relevant. For $n=7$ and $k=3$, for example, isn't $\{1,5,1\}$ a valid multiset of ball distribution, but $(5,1,1)$ is not a valid integer partition of 7, since $\lambda_1 = 5 > 3 = k$?

It seems that you are misunderstanding something.

For $n=7,k=3$, the number of the former is $8$ since we have $\{7,0,0\},$$\{6,1,0\},$$\{5,2,0\},$$\{5,1,1\},$$\{4,3,0\},$$\{4,2,1\},$$\{3,3,1\},$$\{3,2,2\}$. The number of the latter is $8$ since we have $(3,3,1),$$(3,2,2),$$(3,2,1,1),$$(3,1,1,1,1),$$(2,2,2,1),$$(2,2,1,1,1),$$(2,1,1,1,1,1),$$(1,1,1,1,1,1,1)$. So, the number of the former is equal to the number of the latter as claimed.

To see why they have the same number, let us take again the case where $n=7,k=3$.
$\quad\qquad$
The above shows :

$\{7,0,0\}$ in the former corresponds to $(1,1,1,1,1,1,1)$ in the latter.
$\{6,1,0\}$ in the former corresponds to $(2,1,1,1,1,1)$ in the latter.
$\{5,2,0\}$ in the former corresponds to $(2,2,1,1,1)$ in the latter.
$\{5,1,1\}$ in the former corresponds to $(3,1,1,1,1)$ in the latter.
$\{4,3,0\}$ in the former corresponds to $(2,2,2,1)$ in the latter.
$\{4,2,1\}$ in the former corresponds to $(3,2,1,1)$ in the latter.
$\{3,3,1\}$ in the former corresponds to $(3,2,2)$ in the latter.
$\{3,2,2\}$ in the former corresponds to $(3,3,1)$ in the latter.

In the former, we count the number of balls "vertically". In the latter, we count the number of balls "horizontally".

This explains why they have the same number.

$n=\lambda_1+\lambda_2+ \ldots+\lambda_r$ corresponds to a partition of $n$ into $r$ parts. The condition $k \ge \lambda_i$ indicates that each part has size at most $k$.

Note that the question refers to number of partitions of $n$ into parts of size at most $k$. So $\{5,1,1\}$ is not the partition we're looking for with $n=7,k=3$ since there is one part (i.e. $5$) that has size greater than $3$. For example, with $n=7,k=3$, $\{3,2,1,1\}$ is a valid partition.

The condition $\lambda_1 \ge \lambda_2 \ge \ldots \ge \lambda_r$ means that we don't have to care about the order of the parts when counting number of partitions. For example, $\{3,2,1,1\},\{3,1,1,2\}$ and $\{3,1,2,1\}$ are considered to be one partition $\{3,2,1,1\}$ for $n=7,k=3$. Why do we need this condition? Try to look back the situation with indistinguishable boxes. Do we have to care about the order of the boxes?

  Also, to prove the statement, I'm thinking of setting up a bijection between the set of ball distributions and the set of integer partitions of n. Would this be a sensible approach?

Yes, bijection is the way to go. Before finding the main bijection, you should "translate" these two situations into one language. In particular, since the balls and boxes are indistinguishable, the number of distributions of $n$ indistinguishable balls into $k$ indistinguishable boxes is the same as number of partitions of $n$ into $k$ parts.


how to calculate determinant value for $3\times2$ matrix I am learning Rank of Matrix chapter, here I am stuck on finding the submatrices for $3\times3$ matrix and also, in case if $3\times2$ matrix is generated as submatrix for $3\times3$ matrix, I wonder how to calculate the determinant value for the following matrix. Since I am new to the topic, any kind of help will be appreciated.

Determinant is not defined for non-square matrices. Probably, you want to check whether rows of submatrix are linear-independent. To do so there are many different approaches, for example Gaussian elimination.


I don't understand the 'idea' behind the method of characteristics Below is an image of my lecture notes explaining the idea behind the method of characteristics for quasilinear first order PDEs.

However I don't understand how the curve $C_s$ is defined and how it translates into the graph $Gr_u$. How does $C_s$ 'start' on the initial curve if there is no $f(s)$ or $g(s)$ in the equation for $C_s$? What does $x(t;s)$ mean, does that mean (say for the $x$ term) that for fixed $s$ one starts at $x(s-\epsilon_s)$ and finishes at $x(s+\epsilon_s)$? If so what is this 'function' $x$?

Thanks, I would really appreciate someone explaining this to me! :

To clarify what your notes imply, the curve ${\cal C}_s$ contains points $(x(t;s),y(t;s),z(t;s))$ defined by solving the ODE $\frac{d}{dt}(x,y,z)=(a(x,y,z),b(x,y,z),c(x,y,z))$ for $-\varepsilon_s<t<\varepsilon_s$, and condition $(x(0;s),y(0;s),z(0;s))=(f(s),g(s),h(s))$ on $\Gamma$ at $t=0$.


Eigenvectors of $\begin{bmatrix}a&-b\\b&a\end{bmatrix}$ I am currently confused as to what the correct eigenvectors are for $\begin{bmatrix}a&-b\\b&a\end{bmatrix}$.

I confirmed through my own calculations that the eigenvalues are a$\pm$bi. My textbook, Linear Algebra and its Applications, states that the corresponding eigenvectors are $\begin{bmatrix}1\\-i\end{bmatrix}$ and $\begin{bmatrix}1\\i\end{bmatrix}$.

This makes sense when verifying that $Ax = \lambda x$, as $Ax = \begin{bmatrix}a&-b\\b&a\end{bmatrix} \begin{bmatrix}1\\-i\end{bmatrix} = \begin{bmatrix}a+bi\\b-ai\end{bmatrix} = (a+bi) \begin{bmatrix}1\\-i\end{bmatrix}$.

However, upon performing the calculations myself, I repeatedly found the eigenvectors to be $ \begin{bmatrix}-i\\1\end{bmatrix} $ and $ \begin{bmatrix}i\\1\end{bmatrix} $ rather than the given solution. Thinking that I could have made a calculation error, I plugged this into WolframAlpha  and got the same values.

My calculation process was to solve for $ (A-(a+bi)I)x = 0 $, which I reduced down to $ \begin{bmatrix}-i&-1\\1&-i\end{bmatrix} $. After multiplying the top equation by $i$, I got $ \begin{bmatrix}1&-i\\1&-i\end{bmatrix} $ ~ $  \begin{bmatrix}1&-i\\0&0\end{bmatrix} $. Thus, $x_1 = ix_2 $ and $x_2$ is free. So $x = \begin{bmatrix}x_1\\x_2\end{bmatrix} = \begin{bmatrix}i\\1\end{bmatrix} x_2 $, so the eigenvalue corresponding to $\lambda = a+bi$ is $\begin{bmatrix}i\\1\end{bmatrix}$. Similarly, the eigenvalue corresponding to $\lambda = a-bi$ is $\begin{bmatrix}-i\\1\end{bmatrix}$.

Where is the discrepancy between my calculated answer and the one given in the textbook?

Eigenvectors are unique upto some (non-zero) constant. Note that the two eigenvectors that are listed as answers are just $\pm i$ times the eigenvectors you found.

Let $r$ be the module and $\theta$ the argument of complex number $a+bi$, thus with

$$a=r \cos \theta \ \ \ \text{and} \ \ \  b=r \sin \theta$$

No answer mentions the geometrical interpretation of matrix

$$S=\begin{bmatrix}a&-b\\b&a\end{bmatrix}=r\begin{bmatrix}\cos \theta&-\sin \theta\\\sin \theta&\cos \theta\end{bmatrix}$$

that helps to understand why such matrices are bound to have non real eigenvalues.

The interpretation of $S$ as a geometrical transformation is clearly a rotation followed (or preceded) by an homothetic transform, i.e., a similitude. It is clear that (unless $\theta=0$!) no real vector can be transformed by $S$ into a real multiple of it. Therefore, the eigenvalues "have to be" complex.


Distinct Values that can be Created An expression is formed using the numbers 7, 16, 25, and 27 according to the following rules.
$\bullet$ Each of the four numbers is used exactly once.
$\bullet$ The four numbers may be used in any order.
$\bullet$ Exactly three operations are used; each one is either $+$ or $\times$.
$\bullet$ An unlimited number of parentheses may be used.
No two distinct expressions have the same simplified value. The two expressions below are not distinct, and therefore must be counted as only one value. What is the greatest number of distinct values, including the one below, that can be obtained when building expressions following these rules?
$(7+16+27) \times 25 = 25 \times (27+7+16)$

With one operation $\cdot$ you can form recursively one pair $(x\cdot x)$, two triples $((x\cdot x)\cdot x)$, $(x\cdot(x\cdot x))$, and five quadruples
$$(((x\cdot x)\cdot x)\cdot x),\quad ((x\cdot(x\cdot x))\cdot x),\quad((x\cdot x)\cdot(x\cdot x)),\quad(x\cdot((x\cdot x)\cdot x)),\quad(x\cdot(x\cdot(x\cdot x)))\ .$$
In each of these quadruples we can replace the single  $\cdot$ by one of the two $*$ or $+$ in eight ways, and the four ixes by a permutation of $7$, $16$, $25$, $27$ in $24$ ways. It follows that there are in all $5\cdot8\cdot 24=960$ ways to do the computation. At the end there result only $52$ different values, namely
$$75, 164, 218, 230, 314, 326, 380, 434, 464, 476, 539, 569, 589, 602, \
607, 646, 698, 787, 839, 866, 880, 944, 1082, 1114, 1196, 1250, 1296, \
1376, 1394, 2827, 2989, 3049, 3199, 3232, 3424, 3475, 3699, 4741, \
4837, 5125, 5157, 5824, 7525, 7749, 10807, 10912, 10975, 10989, \
13600, 13824, 15525, 75600\ .$$

This question may be very interesting. Firstly, we can use a method similar to stars and bars to solve the problem. There are 7 places where 4 numbers and 3 plus or multiply may be used. One case is that all 4 are added. First, when 3 are added and the last is multiplied to all. It contains 4 cases. Then, when two are added and rest are multiplied to it. It contains 6 cases.Then, when 2 are added and multiplied by third and last is added. It contains 6*2=12 cases. Then, 2are added multiply by 2 added. It contains 12 cases. Now all parenthesis are used. Left are without brackets. Then, when two are added and two are multiplied. It contains 6 cases. Then, 2 are multiplied added by 2 multiplied. It contains 6 cases. Then 3 are multiplied and left is added. This contains 4 cases. Last case is when all are multiplied.
So, total number of cases are 1+4+6+6+12+12+6+4+1= 52 cases
Hence the answer is 52.
Hope it was easy for you, too.


area of figure in sector of intersecting circles I need to find an area of blue part of figure APBC. I draw line segments between B and C, between C and A, and got equilateral triangle. I'm stuck here. Please help. Thanks.

|AB| = a, P is midpoint of segment AB

As achille hui says in the coments, $$(1-r)^2  = |AF|^2 =  \frac{1}{2^2} + |PF|^2 =  \frac{1}{2^2} + |DF|^2 - \frac{1}{4^2}  = \frac{1}{2^2} + \left(r + \frac14\right)^2 - \frac{1}{4^2}.$$ Solve for $r$ and you get $r=\dfrac{3}{10}.$

Since area of intersection between the two larger circles is $\dfrac{2\pi}{3}-\dfrac{\sqrt{3}}{2}$ (see here), shaded area is

\begin{align}
&\dfrac{1}{2}\left(\dfrac{2\pi}{3}-\dfrac{\sqrt{3}}{2}\right)-\pi\left(\dfrac{1}{4}\right)^2-\pi\left(\dfrac{3}{10}\right)^2\\
\\
& = \dfrac{217 \pi }{1200}-\dfrac{\sqrt{3}}{4}\\
\end{align}

If you know the radius of the circles it would be really simple.

You could just find the area of the intersection and subtract the area of the circle with diameter AP and the are of the circle between.


Tensor power- Notation question I have a notational question, which is written usually in papers, but I can not figure it out what could be. Let $M$ be an $A$-module. I have seen this notation
$$M^{\otimes -n}$$

I think this would mean $Hom_A(M^{\otimes n},A)$, but in other way this usually is denoted by $(M^{\otimes n})^*$. Am I right?

Perhaps the other answer was describing the Picard group of a commutative ring A.  If M is a finitely generated projective A-module of rank 1, then the tensor product M ⊗A HomA(M, A) is isomorphic to A.  In particular, the set of isomorphism classes of finitely generated projective modules of rank 1 forms a group under tensor product.

If A is an integral domain, then the M we consider are just the isomorphism classes of projective modules that are isomorphic to ideals of A.  This forms the ideal class group.

In particular for principal ideal domains or fields, this is not a useful concept.

Also, none of this works for general modules, not even finitely generated projective modules of rank larger than 1, so I would be very cautious of assuming this is true for a negative tensor power of some random module.

Yes, you are right. If you're curious, the reason for that notation is that you can view the set of isomorphism classes of A-modules as a group, with the operation of tensor product. Since $M \otimes M^*$ is isomorphic to $A$, we write $M^{-1} = M^*$, and similarly for products.


Some examples of applications of Game Theory I'm approaching my junior year of HS now, and I'm looking for a good science fair project to do. I love mathematics, so I decided to a category of mathematics that can help base logical conclusions to real-life events. That's when I came upon the topics of game theory.

I was very interested in the subject, so I decided to read more about it (started looking here).

After reading more about it, I knew that I really wanted to a project on game theory. However, because I've quite new to it, I don't know any good applications that can be solved using game theory. Does anyone know of any? I don't mind the difficulty of the project, I just want to find something to do using my knowledge of game theory that I've read so far.

Any help would be greatly appreciated.

I find one of the very interesting applications is in nature itself, how living things naturally find optimal strategies in their interactions. For instance some occurences of the game "rock-paper-scissors" can be found in nature. The wikipedia page on Evolutionary Game Theory already contains nice material.

The "most classical" application of game theory is economics, but in my opinion it is a slippery slope, because of the political and ideological implications it can lead to. So my advice would be to avoid it, both for staying away of ideological fights and for the sake of originality !

It is also interesting to note that games play important roles in the most unexpected areas of mathematics: set theory, computer science, topology,... For instance the issue of determinacy in two-player games (does one of the player have a winning strategy ?) is very rich and could be the theme of a full project by itself.

My Answer follow by question: "whether, when and how the fight against the scourge called cyber crime and corruption, its contribution can provide game theory methods and, generally, techniques of operations research and Invention of operational researchers? ".


How to solve $\lim_{n\to\infty}$ $(\frac{\sqrt{n^4-2n^3} - n^2}{10n+2})^2$ I'm trying to solve:

$\hspace{40mm}$ $\lim_{n\to\infty}$ $(\frac{\sqrt{n^4-2n^3} - n^2}{10n+2})^2$

This is what I did to try to solve it, but my answer doesn't match the possible answers so I know it's wrong.

$\lim_{n\to\infty}$ $(\frac{\sqrt{n^4-2n^3} - n^2}{10n+2})^2$

Highest power of denominator is n, so pull out $n^2$ from the square root to give an n

$\lim_{n\to\infty}$ $(\frac{n(\sqrt{n^2-2n} )- n^2}{10n+2})^2$

Factor out n from numerator and denominator

$\lim_{n\to\infty}$ $(\frac{n((\sqrt{n^2-2n})- n)}{n(10+\frac{2}{n})})^2$

Cancel n from numerator and denominator

$\lim_{n\to\infty}$ $(\frac{\sqrt{n^2-2n} - n}{10+\frac{2}{n}})^2$

Take $n^2$ out from the square root

$\lim_{n\to\infty}$ $(\frac{n\sqrt{1-\frac{2}{n}} - n}{10+\frac{2}{n}})^2$

Factor out n from numerator

$\lim_{n\to\infty}$ $(\frac{n((\sqrt{1-\frac{2}{n}}) - 1)}{10+\frac{2}{n}})^2$

At this point I would take the limit to be 0, but the possible answers are 0.05, 0.01, 0.001, or 0.1. Where did I go wrong?

You have got everything right till  the end. Use the fact that $(1-\frac 2 n)^{1/2} =1-\frac 1 n +o(\frac 1 n)$ to see that the limit is $(0.1)^{2}=0.01$.


Function grows slower than $\ln(x)$ What function grows slower than $\ln(x)$ as $x \rightarrow\infty$? How am I supposed to find it besides just trying finding limits of all known functions?

You could compare the derivatives of two functions, and prove that the derivative of ln(x) is always larger than the derivative of your other function

$f(x) = ln(x)$

$f'(x) = \frac{1}{x}$

$g(x) = arctan(x)$

$g'(x) = \frac{1}{x^2 + 1}$

For all $x$

$x^2 + 1 > x \Rightarrow \frac{1}{x^2 + 1}< \frac{1}{x}$

What this means is that the slope of $f(x)$ is always greater than the slope of $g(x)$ for any given $x$. Which means $g(x)$ is growing slower than $f(x)$.


What if a conjecture were provably unprovable? Suppose we found a proof that "The Twin Prime Conjecture cannot be proven", without any conclusion as to the conjecture itself being true or false.

Is it then possible for the conjecture to be true?  or, must the conjecture then be false for some reason?  or, must this proof of unprovability be flawed?

(I'm using this specific example to help understand if any similar simple conjectures could be provably unprovable.  I have my doubts on whether a simple conjecture can be provably unprovable since, instead of making conclusions for an infinite number of inputs like the Halting Theorem does, we are here talking about just one true/false conclusion.)

There is an important issue with statements of the form "$X$ is unprovable": in what system is $X$ unprovable? We assume here that $X$ is some consistent statement written in some formal language.

Statements provable with no non-logical axioms are known as logical validities. Few interesting mathematical statements are phrased as logical validities. So the fact that $X$ is unprovable without non-logical axioms tells very little about $X$. On the other hand, if $T$ is the conjunction of some finite set of axioms sufficient to prove $X$, then $T \to X$ is a logical validity.

What if $X$ is unprovable from some larger set of non-logical axioms? That mainly tells us that those axioms are not strong enough to prove $X$. In other words, the fact that $X$ is unprovable from a set of axioms often tells us both about $X$ and about that system of axioms.  As long as $X$ is consistent, we could always choose a stronger set of consistent axioms that is able to prove $X$ (e.g. we could take $X$ itself as an axiom).

In many concrete examples, the unprovable of $X$ from some system of axioms tells us more about the system of axioms than it tells us about $X$. For example, the combinatorial principle in Goodstein's theorem is unprovable in Peano arithmetic. The main interest in this is that, when we really look at the result, it shows us something new about provability in Peano arithmetic. It also helps us understand the combinatorial principle more deeply, but in this case we already have a very good understanding of the combinatorial principle which comes from its proof in ZFC.

This is a common pattern in mathematical logic. When we examine the proof that a principle is unprovable in a particular system, it often tells us as much or more about the system as it tells us about the principle - particularly when we scrutinize the unprovability proof intensively to try to extract the general method.

If a conjecture X is unprovable that can be translated to the following:

if X is true no contradiction can be derived

if X is false no contradiction can be derived

Both of these must be true because otherwise X would not be unprovable (we could merely choose one of the appropriate routes and eventually find a contradiction)

That being said it becomes evident then that the truth/false value of X is independent of the system of proving being used.

So once something has become unprovable in a fixed system of mathematics. It becomes meaningless to ask if the statement is true or false in that same system.

Hope that clarifies a thing or two :)


Volume of revolution generated by revolving region How can I find the volume of the solid when revolving the region bounded by $y=1-\frac{1}{2}x$, $y=0$, and $x=0$ about the line $ x=-2$? How could I set it up?

Would it be $x=2-2y$
so radius $r(y) = 2-2y -(-2) $ => $r(y)= 4-2y$

$π\int (4-2y)^2 dy$ ?

What would be my limits of integration? Would it be from 0 to 2?

$\textbf{1)}$ Using the disc method, you have $R(y)=(2-2y)-(-2)=4-2y$ and $r(y)=0-(-2)=2$,

so $V\displaystyle=\int_0^1\pi\big((4-2y)^2-2^2\big)dy$.

$\textbf{2)}$ Using the shell method, you have $r(x)=x-(-2)$ and $h(x)=(1-\frac{1}{2}x)-0$,

so $V\displaystyle=\int_0^22\pi(x+2)\big(1-\frac{1}{2}x\big) dx$.

First thing you should do is draw a picture of the volume. It looks like a volcano. Then you should decide the shape of the slice you want to use. Either a shell shape or washer shape would work, but not a disk.

For a shell shape, you are integrating dx, and for a washer shape, you are integrating dy. Decide which one you want to use. For a shell, you have to find height and radius. They should be functions of x, no y's in there. For a washer, you have to find inner radius and outer radius. They should be functions of y, no x's.

After you have the dimensions right, THEN you can use your volume formula and integrate.


How to find out the partial fraction decomposition form? How to find out the partial fraction decomposition form (or pattern ) of a rational function $\dfrac{P(X)}{Q(x)}$ , for example we know that the partial fraction decomposition form of $$\frac{x^{2}+3x+1}{(x-1)^{2}(x+3)},$$ would be something like this :  $$\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{x+3}, \qquad A,B,C \in \mathbb{R}$$

but how did we "guess" that pattern ? and how to guess it in general case ?

We are just doing polynomial long division on the numberator, and collecting the remainders.

$\frac {p(x)}{(x+3)(x-1)^2} = \frac {q(x) (x+3)  + r}{(x+3)(x-1)^2} = \frac {r}{(x+3)} + \frac {q(x)+r}{(x-1)^2}$

$r$ will always have degree less than its denominator.  $q(x)$ will have degree less than $p(x).$

Next we do polynomial division to $q(x)+r$ breaking into $\frac {r_2(x-1) + r_3}{(x-1)^2} = \frac {r_2}{x-1} + \frac {r_3}{(x-1)^2}$

Resolve the denominator into factors, break them,and in numerators of each write an arbitrary polynomial expression with degree one less than the denominator. Find it by comparison.


The median minimizes the sum of absolute deviations (the $ {\ell}_{1} $ norm) Suppose we have a set $S$ of real numbers. Show that
  $$\sum_{s\in S}|s-x| $$
  is minimal if $x$ is equal to the median.

This is a sample exam question of one of the exams that I need to take and I don't know how to proceed.

Introduction: The solution below is essentially the same as the solution given by Brian M. Scott, but it will take a lot longer. You are expected  to assume that $S$ is a finite set. with say $k$ elements. Line them up in order,  as $s_1<s_2<\cdots <s_k$.

The situation is a little different when $k$ is odd than when $k$ is even. In particular, if $k$ is even there are (depending on the exact definition of median) many medians. We tell the story first for $k$ odd.
Recall that $|x-s_i|$ is the distance between $x$ and $s_i$, so we are trying to minimize the sum of the distances. For example, we have $k$ people who live at various points on the $x$-axis. We want to find the point(s) $x$ such that the sum of the travel distances of the $k$ people to $x$ is a minimum.

The story: Imagine that the $s_i$ are points on the $x$-axis. For clarity, take $k=7$. Start from well to the left of all the $s_i$, and take a tiny step, say of length $\epsilon$, to the right. Then you have gotten $\epsilon$ closer to every one of the $s_i$, so the sum of the distances has decreased by $7\epsilon$.

Keep taking tiny steps to the right, each time getting a decrease of $7\epsilon$. This continues until you hit $s_1$. If you now take a tiny step to the right, then your distance from $s_1$  increases by $\epsilon$, and your distance from each of the remaining $s_i$  decreases by $\epsilon$. What has happened to the sum of the distances? There is a decrease of $6\epsilon$, and an increase of $\epsilon$, for a net decrease of $5\epsilon$ in the sum.

This continues until you hit $s_2$. Now, when you take a tiny step to the right, your distance from each of $s_1$ and $s_2$ increases by $\epsilon$, and your distance from each of the five others decreases by $\epsilon$, for a
net decrease of $3\epsilon$.

This continues until you hit $s_3$. The next tiny step gives an increase of $3\epsilon$, and a decrease of $4\epsilon$, for a net decrease of $\epsilon$.

This continues until you hit $s_4$. The next little step brings a total increase of $4\epsilon$, and a total decrease of $3\epsilon$, for an increase of $\epsilon$. Things get even worse when you travel further to the right. So the minimum sum of distances is reached at $s_4$, the median.

The situation is quite similar if $k$ is even, say $k=6$. As you travel to the right, there is a net decrease at every step, until you hit $s_3$. When you are between $s_3$ and $s_4$, a tiny step of $\epsilon$ increases your distance from each of $s_1$, $s_2$, and $s_3$ by $\epsilon$. But it decreases your distance from each of the three others, for no net gain. Thus any $x$ in the interval from  $s_3$ to $s_4$, including the endpoints, minimizes the sum of the distances. In the even case, I prefer to say that any point between the two "middle" points is a median. So the conclusion is that the points that minimize the sum are the medians. But some people prefer to define the median in the even case to be the average of the two "middle" points. Then the median does minimize the sum of the distances, but some other points also do.

As a start, i will define the median of a set with an even cardinality to be one of the two elements in the mid, for example {1,2,3,4} the median is either 2 or 3. and for a set with odd cardinality the median is the middle element.
Suppose that the set  has  elements,  and 1<2<⋯<, we will start by showing that the median gets min sum for sets of cardinality (size) 1,2, and that any problem can be reduced to a set of cardinality 1 or 2.
For set {}, the median is , and the sum is zero.
For set {1,2}, the median is either 1,2, and the sum is |1 - 2| always.
Obvioulsy its easy to see and prove for sets of sizes 1,2 or any other size, that if x is not one of the set elements then the sum bigger than than if x was an element in the set.
So we have proved that the median works for sets of sizes 1,2. Now lets consider set of size 3 where its sorted: {1,2,3}, to get the minimum its easy to see and prove, that we have to pick x such that x is between s1,s3. but for such an x, |s1 - x| + |s3 - x| is always the same and equals |1 - 3| and so x that acheives min for set {1,2,3} is the same for set {s1} which is s1, which is the median.
Its also not hard to show that the same logic applies for set of four elements, the element that achieves min for set of 4 elements is the same element that achieves min for a set of 2 elements which is the median. And using the same logic reduce the problem from a set of 7 elements to set of 5 elements, and from a set of 6 elements to a set of 4 elements,and so on.


Differential equation involving matrices I want to show that

$ Y(t)=\cos(At)$ and $Y(t)=\sin(At)$ satisfy the equation

$$Y''+A^2Y=0 $$

subjected to the initial conditions $Y(0)=I, Y'(0)=0$ and $Y(0)=0, Y'(0)=I$ respictively

where $A$ is an constant $n \times n$ matrix..

I found the shifting of indices to be highly confusing so I propose a method that does not rely on that.

We will first show that $Y(t)=\cos (At)$ solves the initial value problem $Y''+A^2 Y=0$ where $Y(0)=I$ and $Y'(0)=0$. Because $\cos At$ is defined as the infinite sum $\sum\limits_{k=0}^{\infty} (-1)^{k}\frac{A^{2k}t^{2k}}{(2k)!}$ and differentiation is linear we can differentiate term by term to get:

\begin{eqnarray*}
Y''(t)&=&\frac{d^2}{dt^2}\left[    I -\frac{A^2t^2}{2!}+\frac{A^4t^4}{4!}-\frac{A^6t^6}{6!}+\ldots   \right]\\
&=&\left[  -2\frac{A^2}{2!} +4\cdot 3 \frac{A^4 t^2}{4!}  -6\cdot 5 \frac{A^6 t^4}{6!} +\ldots \right]\\
&=&-A^2\left[    I -\frac{A^2t^2}{2!}+\frac{A^4t^4}{4!}-\frac{A^6t^6}{6!}+\ldots   \right]\\
&=& -A^2 \sum\limits_{k=0}^{\infty} (-1)^k \frac{A^{2k}t^{2k} }{(2k)!}=-A^2Y(t)
\end{eqnarray*}

So $Y(t)$ satisfies $ Y''+A^2Y=0$. Also $Y(0)=\cos(0)=\sum\limits_{k=0}^{\infty} (-1)^{k}\frac{0^{2k}}{(2k)!}=I$ (The $0^{2k}$ in the summation is the zero matrix $[A\cdot t]_{t=0}=0$  and $0^0=I$, all higher powers of $0$ are of course $0$). And for $Y'(0)$ we find

\begin{eqnarray*}
Y'(0)&=&\frac{d}{dt}\left[  I -\frac{A^2t^2}{2!}+\frac{A^4t^4}{4!}-\frac{A^6t^6}{6!}+\ldots   \right] _{t=0}\\
&=& \left[   -2\frac{A^20}{2!}+4\frac{A^40^3}{4!}-6\frac{A^60^5}{6!}+\ldots   \right] =0
\end{eqnarray*}
(here the $0^k$ is just the scalar $t=0$ )

For $Y(t)=\sin(At)$ with $\sin(At):=\sum\limits_{k=0}^{\infty} (-1)^{k}\frac{A^{2k+1}t^{2k+1}}{(2k+1)!}$ we have:

\begin{eqnarray*}
Y''(t)&=&\frac{d^2}{dt^2}\left[    At -\frac{A^3t^3}{3!}+\frac{A^5t^5}{5!}-\frac{A^7t^7}{7!}+\ldots   \right]\\
&=&\left[  -3\cdot 2\frac{A^3t}{3!} +5\cdot 4 \frac{A^5 t^3}{5!}  -5\cdot 6 \frac{A^7 t^5}{7!} +\ldots \right]\\
&=&-A^2\left[    At -\frac{A^3t^3}{3!}+\frac{A^5t^5}{5!}-\frac{A^7t^7}{7!}+\ldots   \right]\\
&=& -A^2 \sum\limits_{k=0}^{\infty} (-1)^{k}\frac{A^{2k+1}t^{2k+1}}{(2k+1)!}=-A^2Y(t)
\end{eqnarray*}

So $Y(t)=\sin(At)$ also satisfies $Y''(t)+A^2Y(t)=0$. Here however $Y(0)=\sin( 0)=\sum\limits_{k=0}^{\infty} (-1)^{k}\frac{0^{2k+1}}{(2k+1)!}=0$ (again using $0$ as the zero matrix in the summation, and now the lowest power of $0$ is $0^1$ so here we have an infinite sum of zeros). And

\begin{eqnarray}
Y'(0)&=&\frac{d}{dt}\left[    At -\frac{A^3t^3}{3!}+\frac{A^5t^5}{5!}-\frac{A^7t^7}{7!}+\ldots   \right] _{t=0}\\
&=& \left[  A -3\frac{A^30^2}{3!}+5\frac{A^50^4}{5!}-7\frac{A^70^6}{7!}+\ldots   \right] =A
\end{eqnarray}
I am guessing this is what you meant because even if we consider a scalar $A$ then $\left[\frac{d}{dt}\sin(At)\right]_{t=0}=A\cos(0)=A$.


Deriving definition of the complex logarithm Given that:

$$z = Re^{i\theta} = R(a + bi) = R\left( \cos(\theta) + i\sin(\theta) \right)$$

In its polar form.

$$\log(z) = \log(R) + i\theta$$

$$|z| = \sqrt{(Ra)^2 + (Rb)^2} = R\sqrt{a^2 + b^2} = R\sqrt{\cos^2(\theta) + \sin^2(\theta)} = R$$

$$\log(z) = \log(|z|) + i\theta$$

But is taking the absolute value justified?

how do you mean justified? as in is it correct? in either case lets look at
$$
z = R\mathrm{e}^{i\theta}\implies z^* = R\mathrm{e}^{-i\theta}
$$
thus
$$
|z| = \sqrt{zz^*} = \sqrt{ R\mathrm{e}^{i\theta}R\mathrm{e}^{-i\theta}} = \sqrt{R^2} = R
$$
thus
$$
z = |z|\mathrm{e}^{i\theta}
$$
which you take logs and obtain the result you desire..but I do not know if this enlightens any further.


How do I find the derivative of this equation using the fundamental theorem of calculus? Can someone explain how to find the derivative using the fundamental theorem of calculus on the following equation?

$$g(x) = \int_{2}^{x} t^2 \sin(t) dt  $$

Here is one way to think about it. What is your usual approach to calculating a definite integral, like
$$\int_2^x t^2\sin(t)\,dt?$$

Ordinarily, you go through the following steps:

$1$. Find an antiderivative $F(t)$ of $f(t)$, where $f(t)=t^2\sin(t)$. That is, find a function $F(t)$ such that $F'(t)=f(t)=t^2\sin(t)$.

$2$. Then $\int_2^x t^2\sin(t)\,dt=F(x)-F(2)$ (you "plug in").

We are asked for the derivative of $\int_2^x t^2\sin(t)\,dt$ with respect to $x$. Since the integral is $F(x)-F(2)$, you need to differentiate $F(x)-F(2)$. The answer is $F'(x)$. But $F'(x)=f(x)=x^2\sin(x)$, and we are finished.

Note that we did not need to find an explicit formula for $F(t)$. That would be a waste of time, we would be integrating $f(t)$ to find $F(x)-F(2)$, only to turn around and differentiate it!

With a little work (integration by parts), we could in fact find an explicit antiderivative $F(t)$ of $t^2\sin(t)$. But look at the problem of finding the derivative with respect to $x$ of $\int_1^x \sin(t^2)\,dt$. You will not be able to find a "nice" function whose derivative is $\sin(t^2)$. (Neither will I.) But using the same reasoning as above, we can see that the derivative of $\int_1^x\sin(t^2)\,dt$ with respect to $x$ is $\sin(x^2)$.

Here is a fancier example. Find the derivative with respect to $x$ of $\int_1^{x^2}e^{-t^3}\,dt$. Imagine that we were able to find an explicit antiderivative $F(t)$ of $f(t)=e^{-t^3}$. Then our integral would be $F(x^2)-F(1)$. The derivative of this with respect to $x$ is $(2x)F'(x^2)$ (remember the Chain Rule). But $F'(t)=f(t)=e^{-t^3}$, and therefore our derivative is $(2x)e^{-x^6}$.  Note that at no time did we actually find an explicit antiderivative of $e^{-t^3}$.

(2017/8/6) "Amusing" silent downvotes, 5 years after this was posted. Surely for mathematical reasons... :-)

One can go back to the definition of the derivative. To wit, note that, by the first mean value theorem, $g(x+h)-g(x)=hf(z)$ with $f(z)=z^2\sin(z)$, for some $z$ depending on $x$ and $h$, such that $|z-x|\leqslant h$. Now, the function $f$ is continuous at $x$ hence, when $h\to0$, $f(z)\ldots$


Geometric interpretation of $\|a+b\|^2 +\|a-b\|^2 = 2 \|a\|^2 +2\|b\|^2$ For vectors $a,b$ in an inner product space, I have just proven the identity:
$$\|a + b\|^2 + \|a - b \|^2 = 2\|a\|^2 + 2\|b\|^2$$
Now I am trying to visualise this in $\mathbb{R}^2$, but I'm having a hard time drawing out what this means. Isn't
it basically saying that, in a parallelogram the squared lengths of the diagonals add up to the squared lengths of
the sides of the parallelogram?

We can see that $A+B$ and $A-B$ represents the diagonals while vectors $A$ and $B$ are the sides of the parallogram.

No. It means that the sum of the squares of the diagonals is equal to the sum of the squares of all sides.


For all positive real numbers $x$, if $x$ is irrational, at least one of the numbers $x+\sqrt{2} $ and $x^2 - 2$ is irrational. How to prove this theorem?
If I use contrapositive, the theorem becomes "If at least one of the numbers $x+\sqrt{2}$ and $x^2 - 2$ is rational, $x$ is rational". I have no idea how to prove the two numbers are rational.

If both $x+\sqrt{2}$ and $x^2-2$ are rational, then the quotient
$$\frac{x^2-2}{x+\sqrt{2}}=x-\sqrt{2}$$
is rational, and hence
$$x=\frac{1}{2}((x+\sqrt{2})+(x-\sqrt{2}))$$
is rational. Contradiction.

Well, $x+\sqrt{2}$ and $x-\sqrt{2}$ differ by $2\sqrt{2}$ so at most one of them can be rational.  If only one of them is rational, their product $x^2-2$ is irrational.  If $x^2-2$ is rational, both $x-\sqrt{2}$ and $x+\sqrt{2}$ must be irrational since they cannot both be rational.


Prove that there are no composite integers $n=am+1$ such that $m \ | \ \phi(n)$ Let $n=am+1$ where $a $ and $m>1$ are positive integers and let $p$ be the least prime divisor of $m$. Prove that if $a<p$ and $ m \ | \ \phi(n)$ then $n$ is prime.
This question is a generalisation of the question at
Let $n=apq+1$. Prove that if $pq \ | \ \phi(n)$ then $n$ is prime..
Here the special case when $m$ is a product of two distinct odd primes has been proven. The case when $m$ is a prime power has also been proven here https://arxiv.org/abs/2005.02327.
How do we prove that the proposition holds for an arbitrary positive integer integer $m>1 $? ( I have not found any counter - examples).
Note that if $n=am+1$ is prime, we have $\phi(n)= n-1=am$. We see that $m  \ | \ \phi(n) $. Its the converse of this statement that we want to prove i.e. If $m  \ | \ \phi(n) $ then $n$ is prime.
If this conjecture is true, then we have the following theorem which is a generalisation  ( an extension) of Lucas's converse of Fermat's little theorem.
$\textbf {Theorem} \ \  1.$$ \ \ \ $   Let $n=am+1$, where $a$ and $m>1$ are positive integers and let $p$ be the least prime divisor of $m$ with $a<p$. If for each prime $q_i$ dividing $m$, there exists an integer $b_i$ such that ${b_i}^{n-1}\equiv 1\ (\mathrm{mod}\ n)$ and ${b_i}^{(n-1)/q_i} \not \equiv 1(\mathrm{mod}\ n)$ then $n$ is prime.
Proof. $ \ \ \ $  We begin by noting that ${\mathrm{ord}}_nb_i\ |\ n-1$. Let $m={q_1}^{a_1}{q_2}^{a_2}\dots {q_k}^{a_k}$ be the prime power factorization of $m$. The combination of ${\mathrm{ord}}_nb_i\ |\ n-1$ and ${\mathrm{ord}}_nb_i\ \nmid (n-1)/q_i$ implies ${q_i}^{a_i}\ |\ {\mathrm{ord}}_nb_i$. $ \ \ $${\mathrm{ord}}_nb_i\ |\ \phi (n)$  therefore for each $i$, ${q_i}^{a_i}\ |\ \phi (n)$ hence $m\ |\ \phi (n)$. Assuming the above  conjecture is true, we conclude that $n$ is prime.
Taking $a=1$, $m=n-1$ and $p=2$, we obtain Lucas's converse of Fermat's little theorem. Theorem 1 is thus  a generalisation (an extension) of Lucas's converse of Fermat's little theorem.
On recommendation by the users, this question has been asked on the MathOverflow site,
https://mathoverflow.net/questions/373497/prove-that-there-are-no-composite-integers-n-am1-such-that-m-phin

Partial answer:
Lemma: Let $n=am+1$ where $a\ge1$ and $m\ge2$ are integers. Suppose that  $m\mid\phi(n)$ and $a<p$ where $p=\min\{p^*\in\Bbb P:p^*\mid m\}$. If $n$ is not prime then either

$n$ is of the form $\prod p_i$ where $p_i$ are primes, or

$n$ is of the form $2^kr$ where $k,r$ are positive integers.

Proof: Suppose that $n$ is composite. First, note that $m$ must be odd as otherwise, $a=1$ which yields $n-1=m$. The condition $m\mid\phi(n)$ forces $n$ to be prime which is a contradiction.
Next, write $n=q^kr$ where $k,r$ are positive integers and $q$ is a prime such that $(q,r)=1$. As $\phi(n)=q^{k-1}(q-1)\phi(r)$ the condition $m\mid\phi(n)$ yields $$q^{k-1}(q-1)\phi(r)=mt\implies aq^{k-1}(q-1)\phi(r)=t(q^kr-1)$$ for some positive integer $t$. It follows that either $k=1$ or $t=q^{k-1}v$ for some integer $v\ne t$. In the latter case, we obtain $$\frac{q^kr-1}{q^{k-1}(q-1)\phi(r)}=\frac{aps}{mt}=\frac at\implies p>\frac{t(q^kr-1)}{q^{k-1}(q-1)\phi(r)}.$$ Combining this with the trivial result $p<q^{k-1}(q-1)\phi(r)/t$ yields $$t<\frac{q^{k-1}(q-1)\phi(r)}{\sqrt{q^kr-1}}\implies v<\frac{(q-1)\phi(r)}{\sqrt{q^kr-1}}.$$ Substituting back into $n=am+1$ gives $$q^kr-1=\frac av(q-1)\phi(r)\implies aq\phi(r)-vq^kr=a\phi(r)-v>\phi(r)\left(a-\frac{q-1}{\sqrt{q^kr-1}}\right)$$ which is positive since $k\ge2$. This yields $a>vq^{k-1}\ge vq$. Since $p$ is the least prime divisor of $m$, we have $p\le q-1$, unless $q=2$ or $q-1=v$.
Evidently, the first case contradicts $a<p$, so $k=1$. This means that $n$ must be of the form $\prod p_i$ where $p_i$ are primes. The condition $m\mid\phi(n)$ gives $\prod(p_i-1)=bm$ for some positive integer $b$, and substituting this into $n=am+1$ yields $$a=b\frac{\prod p_i-1}{\prod(p_i-1)}.$$ When $m$ is even, we have $a<p\implies a<2$ which implies that $m=\prod p_i-1$. Further, $$b<\frac{2\prod(p_i-1)}{\prod p_i-1}<2\implies m=\prod(p_i-1).$$ The only way that $\prod p_i-1=\prod(p_i-1)$ is when $\prod p_i$ is prime, which solves the problem. Finally, notice that $m$ is odd only when $b=2^{\nu_2(\prod(p_i-1))}d$ for some positive integer $d$, so the condition $a<p$ yields $$2^{\nu_2(\prod(p_i-1))}d\frac{\prod p_i-1}{\prod(p_i-1)}<\frac{p_j-1}{2^{\nu_2(p_j-1)}}$$ for some prime $p_j\mid\prod p_i$.
The second case $q=2$ implies that $n=2^kr=am+1$ where $m\mid\phi(r)$; that is, for some positive integer $g$ we have $g(2^kr-1)=a\phi(r)$.
The third case $q-1=v$ forces $m=\phi(r)$, so $m=1$. This is a contradiction as there is no prime $p$ that can divide $m$.

As mentioned if $2 \mid m$ then $p=2$ and $a=1$.
So we consider the case $m$ odd and $p \ge 3$
$m= \prod_\limits{i=1}^k q_i^{a_i}$  with $q_i$ prime and  $p=q_1<q_2< \dots <q_k$
$n=1+am=\prod_\limits{j=1}^r p_j^{k_j}$ with $p_j$ prime
we take $p_1$  so that $q_1 \mid  (p_1-1)$
$\phi (n)=\prod_\limits{j=1}^r p_j^{k_j-1}(p_j-1)=bm$
$m=\frac{\prod_\limits{j=1}^r p_j^{k_j-1}(p_j-1)}{b}$
if $p_j \mid m$ then $n=1 \bmod p_j$ therefore it must be  $p_j \not \mid   m$
then $b=c\prod_\limits{j=1}^r p_j^{k_j-1}$ ,  $c \ge 1$
$n=1+am=1+a \frac{\prod_\limits{j=1}^r p_j^{k_j-1}(p_j-1)}{b}=\prod_\limits{j=1}^r p_j^{k_j}$
therefore
$ a=b \frac{\prod p_j^{k_j}-\displaystyle 1}{\prod p_j^{k_j-1}(p_j-1)}=c\prod p_j^{k_j-1} \frac{\prod p_j^{k_j}-\displaystyle 1}{\prod p_j^{k_j-1}(p_j-1)}$
but $  \frac{\prod p_j^{k_j}-\displaystyle 1}{\prod p_j^{k_j-1}(p_j-1)}>1$
then if exists $k_j \geq 2 $ for some $j$ with $q_i \mid (p_j-1)$
$a>p_j>(p_j-1)>q_i \geq q_1=p$
in this case $n$ composite number  and $a>p$
therefore we have to analyze the cases in which $k_j =1$ for each $j$ with $q_i \mid (p_j-1)$
therefore
$$n=n_1 \prod_\limits{j=1}^t p_j \text{  where some } q_i \mid (p_j-1) \text{  and } n_1 \ge 1 \text{  with } q_i \not\mid \phi (n_1)$$
Case 1
$m \mid \phi(p_1)$ then $m \mid  (p_1-1)$
$p_1=1+fm$   ,  $f \ge 2$
as demonstrated in the case $n=1+apq$
$n=1+am=ep_1$   with  $e \ge 1$
$e(fm + 1)  = am + 1 $
$(ef)m + e  = am + 1 $
$e - 1  = (a - ef)m$
then $m \mid e - 1$, but $m \gt a \ge ef$ so $e \lt m$, then $e = 1$
therefore in this case it must be $n = p_1$ prime  and $am = \phi(p_1)=fm$ .
Case 2
$m \not\mid  (p_1-1)$
$n=p_1p_2$
$m=q_1d_1d_2$
$q_1d_1 \mid  (p_1-1)$ , $d_1 \ge 1$
$d_2 \mid  \phi(p_2)$ , $d_2 > q_1$
$d_2 \not\mid  (p_1-1)$
$p_1=1+fq_1d_1$ , $f \ge 2$
$n=1+am=1+aq_1d_1d_2=p_2(1+fq_1d_1)$
$aq_1d_1d_2=p_2-1+p_2fq_1d_1$
$q_1d_1(ad_2-p_2f)=p_2-1$
then $q_1d_1 \mid  (p_2-1)$
$p_2=1+gq_1d_1$ , $g \ge 2$
$n=1+aq_1d_1d_2=(1+gq_1d_1)(1+fq_1d_1)=1+fq_1d_1+gq_1d_1+fgq_1^2d_1^2$
$a= \frac{f+g+fgq_1d_1}{d_2}$
but $\phi (p_2)=gd_1q_1$ and $d_2 \mid  gd_1q_1$
but $d_2 \not\mid  (p_1-1)$ then $d_2 \not\mid  q_1d_1$   and $d_2 \mid  g$
then $d_2 <  g$  and $a= \frac{f+g+fgq_1d_1}{d_2}>\frac{fgq_1d_1}{d_2}>\frac{fgq_1d_1}{g}>fq_1d_1>q_1$
in this case $n$ composite number  and $a>p$
Case 3
$m \not\mid  (p_1-1)$
$n=1+am=n_1 \prod_\limits{j=1}^t p_j \text{  where for each p_j there is some } q_i \text{  with }  q_i\mid (p_j-1)$
$q_i \not\mid \phi (n_1)$
$p_j=1+b_jq_{i_1}\dots q_{i_{m_j}}$  with $b_j\ge 2$
$\phi(n_1)=c$
$\phi(n)=\phi(n_1) \prod(p_j-1)=\phi(n_1) \prod (b_jq_{i_1}\dots q_{i_{m_j}})=cbm$
but $\frac{n-1}{\phi(n)}=\frac{am}{cbm}>1$
then $cb= \phi(n_1) \prod b_j <a$ with $b \ge 2^{a_1+\dots +a_k}$
but $n=n_1 \prod_\limits{j=1}^t(1+b_jq_{i_1}\dots q_{i_{m_j}})=1+am$
and $\prod_\limits{j=1}^t(1+b_jq_{i_1}\dots q_{i_{m_j}})=(1+b_1q_{i_1}\dots q_{i_{m_1}}+\dots+\prod_\limits{j=2}^t(b_jq_{i_1}\dots q_{i_{m_j}})+\prod_\limits{j=1}^t(b_jq_{i_1}\dots q_{i_{m_j}}))$
then $m(a-n_1b)=-1+n_1(1+b_1q_{i_1}\dots q_{i_{m_1}}+\dots+\prod_\limits{j=2}^t(b_jq_{i_1}\dots q_{i_{m_j}}))$
from which $(a-n_1b)> 0$
then $a>n_1b$
$m<-1+n_1(1+b_1q_{i_1}\dots q_{i_{m_1}}+\dots+\prod_\limits{j=2}^t(b_jq_{i_1}\dots q_{i_{m_j}})$
but each addend $\prod(b_jq_{i_1}\dots q_{i_{m_j}})<\frac{bm}{q_{i_k} \prod b_k}<\frac{bm}{q_12^{s_k}}$ with $q_{i_k} \not=q_{i_1}\not=\dots\not= q_{i_{m_j}}$ and $ b_k \not= b_j$
then $m<\frac{n_1bm}{2q_1}(1+\sum\frac{1}{2^{s_k-1}})<\frac{n_1bm}{q_1}<\frac{am}{q_1}$
therefore $a>q_1$ and this contradicts the initial condition  $a<q_1$


Permutation in line and circle with conditions. A party of $10$ consists of $2$ Americans, $2$ British men, $2$ Chinese & $4$ men of other nationalities (all different). Find the number of ways in which they can stand in a row so that no two men of the same nationality are next to one another. Find also the number of ways in which they can sit at a round table.
Here for the first one I try to use the gap method. So arrange $5$ to create gaps to accommodate all but we have only four of different nationalities. So a different method should be used.
All permutations - permutations of people of same country being together
But this has cases as only one pair together, two pairs together and three pairs together. Which overlap. How to handle ?

In a row:  There are $10!$ ways to arrange the people. We now count the bad arrangements, in which two A's are next to each other, or two B's, or two C's.

To do the counting, we use Inclusion/Exclusion. How many arrangements have the two A's together? Put them in a bag. We now have $8$ people and $1$ bag. These can be arranged in $9!$ ways. Then when we let the A's out of the bag, they can arrange themselves in $2$ ways, for a total of $2\cdot 9!$.

Do the same for the B's, the C's, and add up. We get $3\cdot 2\cdot 9!$.

However, we have double-counted the bad arrangements in which, for example, the A's and the B's are together. The same bag argument shows there are $2^2\cdot 8!$ such arrangements, for a total of $3\cdot 2^2\cdot 8!$.

Thus our next estimate for the number of bads is $3\cdot 2\cdot 9!-3\cdot 2^2\cdot 8!$.

However, we have subtracted one too many times the arrangements in which the A's are together, and the B's, and the C's. The now familiar technique  shows there are $2^3\cdot 7!$ of these. So add back $2^3\cdot 7!$.

We end up with $7!(6\cdot 9\cdot 8-12\cdot 8+8)$, which is $8!\cdot 43$.

This is the number of bad arrangements. The number of good arrangements is $10!-8!\cdot 43$, which is $8!\cdot 47$.

Circular table: We use the convention that two orders are to be considered the same if they differ by a rotation. Assume that one of the "others" is a Canadian. He/she is probably too polite to complain, so one can put the Canadian in the worst chair.

The remaining chairs can now be thought of as a line. We have $9$ people, including $3$ "others" to put in a line, with the condition that the two A's, two B's, and two C's are separated. Same Inclusion/Exclusion technique.

For the first one, Total no of combinations is, $$ s = \frac{ 10!}{2!.2!.2!} $$ Now let A represent the group in which two Americans are together, B represent 2 British together and C represent group in which 2 Chinese are together, calculate $  A \cup B \cup C$ and subtract it from s.

So the final answer is $$ \frac{ 10!}{2!.2!.2!} - 3.\frac{9!}{2!.2!} + 3.\frac{8!}{2!} - 7! $$


Is the multiplication result of two non-equal unit-length quaternions perpendicular to both operands? The question:
Is the multiplication result of two unit-length quaternions "perpendicular" to both operands? "Perpendicular" means that dot product between either operand and multiplication result will be zero (as with vector3 cross-product). Operands (say q1 and q2) are not equal to each other.

Reason for asking:
I'm trying to understand quaternion rotations from purely geometric point of view (even if it is 4d). For example, 3x3 affine transform matrix in 3d space can be represented as a new coordinate system into which a point should be placed, and each row(or column - depending on notation) of such matrix will represent a vector of a coordinate axis of the "new" coordinate system. So to transform a point with matrix, you simply take every vector that represents axis of a "new" coordinate system, scale it using numeric value of corresponding x/y/z of the point to transform, and add them all together to get new point. Such representation is very easy to understand. I'm looking for similar explanation of quaternions, and If I get confirmation that result of quaternion multiplication is "perpendicular" to both operands that would be at least something to get started.

Regards, SigTerm.

let $x=ai+bj+ck, y=di+ej+fk$ (pure quaternions).  then $xy=-(ad+be+cf)+(bf-ce)i+(af-cd)j+(ae-bd)k$.
if $u=(a,b,c), v=(d,e,f)$ then the real part of $xy$ is $-u\cdot v$ and the "imaginary" part of $xy$ is $u\times v$.

"I'm trying to understand quaternion rotations from purely geometric point of view (even if it is 4d)."

Arbitrarily pick a quaternion A as the operator, and the other B as the data.

The operator A does two separate complex rotations. Take the vector part of A. That is the axis for one complex rotation. It rotates B around that axis.

At the same time, B has a separate plane, defined by A's vector and the scalar. A also rotates B around that axis, by the same amount.

A's scalar (divided by A's euclidean magnitude) is the cosine of the angle for both rotations.

If you apply A on the right instead of on the left, BA instead of AB, everything is the same except one term, and the result is that the complex rotation around the vector part of A goes in the opposite direction.

You don't get gymbal lock because you do both rotations at once.


Why are alternating matrices only used when $n$ is even? Consider $\textrm{GL}_n(k)$, where $k$ is a field.  If $Q \in \textrm{GL}_n(k)$, then $Q$ induces a $k$-bilinear form $B_Q: k^n \times k^n \rightarrow k$ by the formula $$B_Q(v,w) = v^t Q w$$  $Q$ is called alternating if $B_Q(v,v) = 0$ for all $v$.  If $Q = (a_{ij})$ is alternating, then one can see that $a_{ij} + a_{ji} = 0$ for all $i,j$, or in other words $Q = -Q^t$.  Is it true that if $Q$ is invertible and alternating, then $n$ must be even?  It seems like many references on algebraic groups are assuming this.

I can see how if the characteristic of $k$ is not $2$, then $n$ must be even: if $n$ isn't even, then $\textrm{Det } Q = (-1)^n \textrm{Det } Q^t = -\textrm{Det } Q$, which implies $\textrm{Det } Q = 0$, absurd.  But what about when $k$ does have characteristic $2$?

When the field has characteristic two, an alternating matrix is symmetric with zero diagonal. Suppose $A$ is such a matrix and $x^TAy=1$. Consider the matrix
\[
  B = A - Ayx^TA - Axy^TA.
\]
Then $B$ is symmetric with zero diagonal and, since $\mathrm{ker}(B)$ contains $x$ and $y$, it follows that $\mathrm{rk}(B)=\mathrm{rk}(A)-2$.
(I'm using the fact that $z^TAz=0$ for all $z$.) By induction it follows that the rank of $A$ is even.

For a field $k$ of arbitrary characteristic, a bilinear form $B:V \otimes V \to k$ on a vector space $V/k$ is considered alternating if $B(v, v) = 0$ for all $v\in V$. Away from characteristic $2$, this definition reduces to the one you mentioned:
\begin{align*}
B(v, w) + B(w, v) = B(v + w, v + w).
\end{align*}
In a vague and fuzzy way, alternating forms are annoying over characteristic $2$. For example, we can usually decompose an arbitrary form $B$ into symmetric and antisymmetric forms:
\begin{align*}
B^+(x, y) &= \frac{1}{2}(B(x, y) + B(y, x)) &
B^-(x, y) &= \frac{1}{2}(B(x, y) - B(y, x)).
\end{align*}
(These are the projections of $B$ onto the $+1$ and $-1$ eigenspaces of the operator $(TB)(x, y) = B(y, x)$ on the space of bilinear forms.) In characteristic $2$, this doesn't quite work.


inverse of the function whose exponential value is different $f(x)=x^2+2 $

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

$g(x)=x^7+x^3+2$

$g^{-1}(x)= ?$

and for more $x^y$?

for example:

$f(x)=x^{11}+x^8+x^7+x^4+x^3+x+7$


The proof of Raabe's Test for absolute convergence In Introduction to Real Analysis second edition by Bartle & Sherbert's, there is a proof of Raabe's Test for absolute convergence. The problem is that I don't understand why some part of the proof is necessary. I will show you first the proof as it is in the book, and then explain what I don't understand. Part (a) of the test is as follows:

Raabe's Test: Let $X:=(x_n)$ be a sequence of nonzero real numbers. If there exists numbers $a>1$ and $K\in\mathbb{N}$ such that
$$\left|\frac{x_{n+1}}{x_n}\right|\leq 1-\frac{a}{n}\quad\text{for}\quad n\geq K,$$
then $\sum x_n$ is absolutely convergent.

Proof: If the inequality holds, then we have
$$k|x_{k+1}|\leq(k-1)|x_k|-(a-1)|x_k|\quad\text{for}\quad k\geq K$$
On reorganizing the inequality, we have
$$(k-1)|x_k|-k|x_{k+1}|\geq(a-1)|x_k|>0\quad\text{for}\quad k\geq K$$
from which we deduce that the sequence $(k|x_{k+1}|)$ is decreasing for $k\geq K$. If we add the last inequality for $k=K,\ldots,n$ and note that the left side telescopes, we get
$$(K-1)|x_K|-n|x_{n+1}|\geq(a-1)(|x_K|+\cdots+|x_n|).$$
This shows (why?) that the partial sums of $\sum|x_n|$ are bounded and establishes the absolute convergence of the series. Q.E.D.

Now, I don't see why it is important to show that the sequence $(k|x_{k+1}|)$ is decreasing. From the inequality
$$(K-1)|x_K|-n|x_{n+1}|\geq(a-1)(|x_K|+\cdots+|x_n|).$$
we have
$$(a-1)(|x_K|+\cdots+|x_n|)\leq (K-1)|x_K|\quad\text{for}\quad k\geq K$$
independently if $(k|x_{k+1}|)$ is decreasing or not since $n|x_{n+1}|>0$. So the partial sums of $\sum|x_n|$ are bounded anyway. Can someone explain me what I am missing?

Suppose $(k|x_{k+1}|)$ is not decreasing, which means there is a possibility that the terms in the sequence is getting larger and larger. In this case, you won't have a upper bound of the partial sum because $ (K-1)|x_K|-n|x_{n+1}|$ may be a negative number.

In this proof the author desires to prove that the sequence of partial sums converges. He is trying to use the fact that a monotonic sequence in bound has a finite limit. Here the sequence of partial sums is decreasing monotonically and is bounded as shown in the inequality above. The sequence is also bounded by zero below since all terms are non-negative. Thus a finite limit exists and the series absolutely converges.


How many sides does a circle have? My son is in 2nd grade.  His math teacher gave the class a quiz, and one question was this:

  If a triangle has 3 sides, and a rectangle has 4 sides,
  how many sides does a circle have?

My first reaction was "0" or "undefined".  But my son wrote "$\infty$" which I think is a reasonable answer.  However, it was marked wrong with the comment, "the answer is 1".

Is there an accepted correct answer in geometry?

edit: I ran into this teacher recently and mentioned this quiz problem.  She said she thought my son had written "8." She didn't know that a sideways "8" means infinity.

The answer depends on the definition of the word "side." I think this is a terrible question (edit: to put on a quiz) and is the kind of thing that will make children hate math. "Side" is a term that should really be reserved for polygons.

One way to understand the question and demonstrate applicability is to consider the problem and efficiency of finding the area of the union of 2 overlapping circles versus 2 overlapping rectangles or squares.

Let us now constrain the union area to be constant.

In the case of circles it will always be the same shape, no matter which way the circles are positioned in relation to one another.

In the case of squares, it would be the same shape too.

In case of rectangles, the shape would vary.

We could here argue that the circle and square both have 1 side, because they are defined by a single length (radius or diagonal), or in other words, they have no apparent "orientation".


Distance of the Focus of an Hyperbola to the X-Axis For a National Board Exam Review:

  How far from the $x$-axis is the focus of the hyperbola $x^2 -2y^2 + 4x + 4y + 4$?

Answer is $2.73$

Simplify into Standard Form:

$$ \frac{ (y-1)^2 }{} - \frac{ (x+2)^2 }{-2} = 1$$

$$ a^2 = 1 $$
$$ b^2 = 2 $$
$$ c^2 = 5 $$

Hyperbola is Vertical:

$$ C(-2,1) ; y = 1 $$

$$ y = 1 + \sqrt5 = 3.24 $$
$$ y = 1 - \sqrt5 = 1.24 $$

Both answers don't match; What am I doing wrong?