Probability with a chance to try again If I play a game where the chance to win is 40%, and every time I lose I get a 25% chance to try again, what's my actual chance to win?

Every time you try you have an $0.4$ chance of succeeding and an $0.6$ chance of failing. At each failure you have an $0.25$ chance of trying again.

The chance you win after one try is $p_1 = 0.4$. The chance you win at your second try is the chance of first failing, then being allowed to try again, and then actually succeeding, so $p_2 = 0.6 \cdot 0.25 \cdot 0.4$.
Now in general in order to try an $n-th$ time, you first have to fail $n-1$ times and have been allowed to retry $n-1$ times. Since these events are independent, we have that $p_n = 0.4(0.6 \cdot 0.25)^n$, and the chance that you succeed at any time is simply the sum of all these $p_n$'s.

Or, $$p_{win} = \sum_{n=0}^{\infty}p_n=\sum_{n=0}^{\infty}0.4(0.6 \cdot 0.25)^n=0.4\sum_{n=0}^{\infty}(0.6 \cdot 0.25)^n = \frac{0.4}{1-0.6\cdot 0.25}\approx0.47.$$ Where in the second to last step I have used the formula for a geometric sequence.


How to prove, that $f\left(\bigcap\limits_{n=1}^\infty K_n\right)=\bigcap\limits_{n=1}^\infty f(K_n)$? Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function and $\{K_n\}_{n=1}^\infty$ be a decreasing sequence of compact subsets of $\mathbb{R}$. Show that $f\left(\bigcap\limits_{n=1}^\infty K_n\right)=\bigcap\limits_{n=1}^\infty f(K_n)$

I have a problem with this specific task. I would be glad if someone can give a hint in order that I can solve it alone.

Thanks in advance.

If $z\in \bigcap K_n$ then $f(z)\in f(K_n)$ for all integers $n$, so $f(z)\in \bigcap f(K_n)$.

The other inclusion requires more work: if $w\in f(K_n)$ for each integer $n$, there is a $z_n\in K_n$ such that $f(z_n)=w.$ Using compactness , we find a subsequence, which we still call $(z_n)$ for convenience, such that $z_n\to z\in K_1.$ But since $z_n\in K_n$, and the $K_n$ are decreasing, we must have, in fact,  $z\in \bigcap K_n:$

for if not, there is an integer $N$ such that $z\notin K_N$, which implies that $z\notin K_{n}$ if $n\ge N$. But then, because the $K_n$ are closed, there is a positive number $r$ such that $d(z,z_n)>r,$ which is a contradiction since $z_n\to z$.

So we have $f(z)\in f(\bigcap K_n).$

To finish, note that continuity now implies that $w=f(z_n)\to f(z)\in f(\bigcap K_n).$

Let $K = \bigcap_{n=1}^{\infty}K_n$. Since $f(K)\subset f(K_n)$ for all $n$ then $f(K)\subset \cap f(K_n)$. Now let $y\in f(K_n)$. Then $f^{-1}(\{y\})\cap K_n$ is non-empty and compact. Also for any $n$

$$f^{-1}(\{y\})\cap K_{n+1}\subset f^{-1}(\{y\})\cap K_n$$

we have the set

$$\bigcap_{k=1}^{\infty}\left(f^{-1}(\{y\}\cap K_n\right) = f^{-1}(\{y\})\cap K$$

is non-empty, so $y\in f(K)$.

Hope that helps, I may have messed up some details...


Solve $\int\left(\ln5x+2e^{2x}\right)\,dx$ $$\int\left(\ln(5x)+2e^{2x}\right)\,dx$$
I am not getting how to integrate this function. Please help.

You have
$$\int (\ln 5x + e^{2x})~dx = \int \ln 5x~dx + \int 2e^{2x}~dx = I_1 + I_2.$$
For $I_2$ try by subtstitution and for $I_1$ try by parts with $u = \ln 5x$ and $v' = 1$.

Doing it by parts and splitting, We get $\int\left(\ln(5x)+2e^{2x}\right)\,dx=\boxed{x\ln \left(5x\right)-x+e^{2x}+C}$


Mathematical explanation for the Repertoire Method There are a few questions already about this method, which has stumped me for a long while. The process is explained, for instance, here: Repertoire Method Clarification Required ( Concrete Mathematics )

What I absolutely don't get is: What's the meaning of plugging in simple functions (which are not solutions) in the recurrence and how would that help finding what the right solution really is?

As I understand it, we have a function determined by a recurrence relation and we want to find a closed formula for it. The recurrence being linear, we assume this closed form is a linear combination of 3 other functions, with coefficients $\alpha, \beta$ and $ \gamma$.

The first method suggested in the book is clear to me: set simple values for the constants, since the function will be the same for all of them, and try to guess A, B and C and prove it by induction. What I don't get is the "dual" repertoire method.

$f$ could be anything, but it's something fixed, so I don't see any meaning in this process of plugging in $f(n) = 1$. $f$ clearly is not 1. What is going on?

I expect everything coming down to seeing a vector space / module in two different ways and then somehow picking some basis vectors from each. However, I can't work out the right abstraction.

I'd like to ask for a very explicit top down explanation of this repertoire method, if possible with linear algebra vocabulary.

Note: At first I'll try to describe some key ideas of the repertoire method using a (very) simple example. This should give you some basic information about this method. But to get a good impression what's really going on, we will also look at a considerably more complex example.

  Repertoire method (some basics)

  This method is based upon two essential ingredients. The first one is the possibility to build linear combinations of already known recurrences.

  Let's assume we have a recurrence for $x_n$
  \begin{align*}
x_0&=a_0\\
x_n&=a_n+x_{n-1},\quad n>0
\end{align*}
  and we have another recurrence for $y_n$
  \begin{align*}
y_0&=b_0\\
y_n&=b_n+y_{n-1},\quad n>0
\end{align*}
  Then provided these recurrences are known, we also know by linearity that an equation with additive term
  $$\alpha a_n+\beta b_n$$
  will have the solution
  $$\alpha x_n + \beta y_n$$

To make it more concrete, let $a_n=3$ and $b_n=5n^2+1$. Assuming we know the solutions $x_n$ and $y_n$ of the recurrences
\begin{align*}
x_0&=3&y_0&=1\\
x_n&=3+x_{n-1},\quad n>0&y_n&=5n^2+1+y_{n-1},\quad n>0
\end{align*}

then we also know by linearity that the solution of the recurrence
\begin{align*}
z_0&=7\\
z_n&=2n^2+7+z_{n-1}
\end{align*}
is

\begin{align*}
z_n=\frac{11}{5}x_n+\frac{2}{5}y_n
\end{align*}

  We observe that in this case we have a repertoire of two solutions $x_n$ and $y_n$ which we can linearly combine in order to find the wanted solution $z_n$.

BUT: We typically start with a recurrence $z_n$ without having a repertoire of proper candidates. And this will be the second ingredient. We have to build a repertoire.

  The second ingredient is building a repertoire which can be used for linear combinations. Let's assume we start with a recurrence
  \begin{align*}
z_0&=7\\
z_n&=2n^2+7+z_{n-1},\quad n>0\tag{1}
\end{align*}
  So, if we try to solve this recurrence by the method of repertoire, we first generalise the recurrence and consider instead
  \begin{align*}
\mathcal{Z}_0&=a_0\\
\mathcal{Z}_n&=a_n+\mathcal{Z}_{n-1},\quad n>0
\end{align*}
  Now it's time for some creative ideas which is typically the most challenging phase when using this method. We want to find a repertoire consisting of two members. One of them with $a_n=const.$ and the other with $a_n=$ square of $n$. In order to do so we have to guess some proper candidates for $\mathcal{Z}_n$ which provides us with the wanted $a_n$.

So let's guess a first candidate. This is not too hard (in this case) since setting $\mathcal{Z}_n=n$ gives
\begin{align*}
a_0&=0\\
n&=a_n+n-1\quad n>0
\end{align*}
and we find: $a_0=0$ and $a_n=1,n>0$.

  We get the first candidate, let's say $x_n$ with

  \begin{align*}
x_0&=0\\
x_n&=1+x_{n-1},\quad n>0\tag{2}
\end{align*}

We can similarly find a proper second candidate which provides us with $a_n=$ square of $n$ by observing that typically the sum of $k-th$ powers of natural numbers is something with a $(k+1)$-th power. So we guess $\mathcal{Z}_n=n^3$  which gives
\begin{align*}
a_0&=0\\
n^3&=a_n+(n-1)^3\quad n>0
\end{align*}
and we find $a_0=0$ and $a_n=3n^2-3n+1,n>0$.

  We get the second candiate, let's say $y_n$ with

  \begin{align*}
y_0&=0\\
y_n&=3n^2-3n+1+y_{n-1},\quad n>0\tag{3}
\end{align*}

We observe that $y_n$ also contains a linear term in $n$ which is not wanted, since we need according to (1) $a_n=n^2+7$. So we extend our repertoire by introducing a third member which provides us with $a_n=$ linear term in $n$. Then we should be able to eliminate the linear term in $n$ by a proper linear combination of the three members. We guess $\mathcal{Z}_n=n^2$  which gives
\begin{align*}
a_0&=0\\
n^2&=a_n+(n-1)^2\quad n>0
\end{align*}
and we find $a_0=0$ and $a_n=2n-1,n>0$.

  So we get the third candiate, let's say $u_n$ with

  \begin{align*}
u_0&=0\\
u_n&=2n-1+u_{n-1},\quad n>0\tag{4}
\end{align*}

Let's have a look at the repertoire:

  Overview of the three candidates

  \begin{array}{rlcr}
\mathcal{Z}_n&&a_n&\\
\hline\\
x_n=&n\qquad&1&\qquad\qquad\text{acc. to }(2)\\
y_n=&n^3\qquad&3n^2-3n+1&\qquad\qquad\text{acc. to }(3)\\
u_n=&n^2\qquad&2n-1&\qquad\qquad\text{acc. to }(4)\\
\end{array}

We observe when using an appropriate linear combination
\begin{align*}
a_n&=2n^2+7\\
&=\frac{2}{3}\left(3n^2-3n+1\right)+\left(2n-1\right)+\frac{22}{3}
\end{align*}
and we conclude
\begin{align*}
z_n&=\frac{22}{3}x_n+\frac{2}{3}y_n+u_n+c_0\\
&=\frac{1}{3}n\left(2n^2+3n+22\right)+c_0
\end{align*}

  Observe, that we have to determine a constant $c_0$ since we also have to properly respect the initial condition $z_0=7$. We do so by setting $c_0=7$ and so we finally get:
  $$z_n=\frac{1}{3}n\left(2n^2+3n+22\right)+7,\quad n\geq 0$$

Let's summarise

  Summary of the method by Repertoire

  In order to solve a recurrence of the form
  \begin{align*}
x_n=f(n)+g(x_{n-1},x_{n-2},\ldots,x_0)\tag{5}
\end{align*}

  we identify building blocks $f_1(n),\ldots,f_k(n)$ of $f(n)$, so that they can be linearly combined in order to form $f(n)$

  \begin{align*}
f(n)=\lambda_1 f_1(n)+\ldots+\lambda_k f_k(n)
\end{align*}

  then we consider the generalised representation of (5) by substituting $f(n)$ with $a_n$.
  \begin{align*}
\mathcal{Z}_n=a_n+g(\mathcal{Z}_{n-1},\mathcal{Z}_{n-2},\ldots,\mathcal{Z}_0)
\end{align*}
  and solve the simpler recurrences
  \begin{align*}
x_n^{(l)}=f_l(n)+g(x_{n-1},x_{n-2},\ldots,x_0)\quad 1 \leq l \leq k
\end{align*}
  by proper guessing of $x_n^{(l)}$ in order to get $f_l(n)$.
  The solutions $x_n^{(l)}$ with $1\leq l \leq k$ form the repertoire of the method in order to solve $x_n$.
  Determine the linear combination
  $$f(n)=\lambda_1 f_1(n)+\ldots+\lambda_k f_k(n)$$
  and deduce the solution

  $$x_n=\lambda_1 x_n^{(1)}+\ldots+\lambda_k x_n^{(k)}+c_0$$

  Finally determine $c_0$ according to initial conditions

Note: There may be more than one initial condition which are to determine

Note: It may be necessary to extend the repertoire in order to remove unwanted terms which additionally occur during the calculation of the $x_n^{(l)}$.

What follows is no longer a core part of the answer. But if you are curious ... :-)

  Note: You may object that the example above uses quite a heavy machinery for a simple recurrence. But this was only for demonstration.

  A real life application is presented by D.E.Knuth and D.H.Greene in their Mathematics for the Analysis of Algorithms section 2.1.2.2. Albeit this example is somewhat lengthy, I can't resist to present at least the essential steps, since it's instructive to see the method in action.

  Real life application of method by Repertoire:  from D.E.Knuth and D.H.Greene

  The sorting algorithm median-of-three quicksort leads to the following recurrence.

  \begin{align*}
x_n=n+1+\frac{1}{\binom{n}{3}}\sum_{1\leq k \leq n}(k-1)(n-k)\left(x_{k-1}+x_{n-k}\right)
\end{align*}

At first we observe that the sum is symmetric and we replace the additive term $n+1$ by $a_n$ in preparation for the repertoire approach:

\begin{align*}
x_n=a_n+\frac{2}{\binom{n}{3}}\sum_{1< k <n}(k-1)(n-k)x_{k-1}
\end{align*}

  First step: Create a Repertoire

  We choose $x_n$ equal to the falling factorial $(n-1)^{\underline{s}}=(n-1)(n-2)\cdot\ldots\cdot(n-s)$

  $$x_n=(n-1)^{\underline{s}}$$

and get

\begin{align*}
(n-1)^{\underline{s}}&=a_n+\frac{12}{n^{\underline{3}}}\sum_{1< k <n}(n-k)(k-1)^{\underline{s+1}}\\
&=\ldots\\
&=a_n+\frac{12(s+1)!}{n^{\underline{3}}}\binom{n}{s+3}
\end{align*}

  This yields
  \begin{align*}
a_n=(n-1)^{\underline{s}}-\frac{12}{(s+2)(s+3)}(n-3)^{\underline{s}}
\end{align*}
  which is a family of solutions for the repertoire parametrised with $s$.

  We observe:

  \begin{array}{ccc}
&x_n&a_n\\
\hline\\
s=0\qquad&1&\qquad-1\\
s=1\qquad&(n-1)&\qquad2\\
s=2\qquad&(n-1)(n-2)&\qquad\frac{1}{5}\left(2n^2+6n-26\right)
\end{array}

Attention: Regrettably this family is inadequate; it lacks a member with linear $a_n$. You might want to read section 2.1.2.2 which clarifies why this lack is not surprising (:-))! But we have at least with $s=0$ a candidate for a constant amount.

  Second step: Extend the repertoire

  Since these types of algorithms are of order $O(n \log n)$ we introduce the Harmonic numbers $H_n=\sum_{k=1}^n\frac{1}{k}$ which contributes $O(\log n)$ and consider a new family
  $$x_n=(n-1)^{\underline{t}}H_n$$

Solving for $a_n$ yields

\begin{align*}
(n-1)^{\underline{t}}H_n&=a_n+\frac{12}{n^{\underline{3}}}\sum_{1< k <n}(n-k)(k-1)^{\underline{t+1}}H_{k-1}\\
&=\ldots\\
&=a_n+\frac{12}{n^{\underline{3}}}\left(\frac{n^{\underline{t+3}}}{t+2}\left(H_{n-1}-\frac{1}{t+2}\right)\right.\\
&\qquad\qquad-\left.\frac{n^{\underline{t+3}}}{t+3}\left(H_n-\frac{1}{t+3}\right)+\frac{(n-1)^{\underline{t+2}}}{t+2}\right)
\end{align*}

The calculation above were done using the identity:

$$\sum_{k=1}^{n}\binom{k}{m}H_k=\binom{n+1}{m+1}\left(H_{n+1}-\frac{1}{m+1}\right)$$

  Some simplifications lead to
  \begin{align*}
a_n=H_n\left((n-1)^{\underline{t}}-\frac{12}{(t+2)(t+3)}(n-3)^{\underline{t}}\right)+\frac{12(2t+5)}{(t+2)^2(t+3)^2}(n-3)^{\underline{t}}
\end{align*}

(... The following text is nearly verbatim from the book ...)

  Now when examining the small members of the family of solutions we discover a fortunate alignment:

  \begin{array}{ccc}
&x_n&a_n\\
\hline\\
t=0\qquad&H_n&\qquad-H_n+\frac{5}{3}\\
t=1\qquad&(n-1)H_n&\qquad2H_n+\frac{7}{12}(n-3)\\
\end{array}

The smallest two solutions for $a_n$ both have leading term $H_n$. By an appropriate linear combination we can eliminate $H_n$ and obtain an $a_n$ that grows as order $n$:

$$x_n=(n+1)H_n\qquad\longleftrightarrow\qquad a_n=\frac{7n+19}{12}$$

The $s=0$ solution from the first family is used to adjust the constant term, enabling us to reconstruct the $a_n$ given in the original problem:

\begin{align*}
x_n=\frac{12}{7}\left((n+1)H_n+1\right)\qquad\longleftrightarrow\qquad a_n=n+1\tag{6}
\end{align*}

  Third step: Adjustment for initial values

  This solution for $x_n$ may not agree with the inital values $x_1$ and $x_2$. To accommodate arbitrary initial values we need to discover two extra degrees of freedom in the solution.

  One degree of freedom can be observed in the first family of solutions. Combining $s=0$ with $s=1$ yields
  $$x_n=n+1\qquad\longleftrightarrow\qquad a_n=0$$
  So any multiple of $n+1$ can be added to the solution in (6).

  The second degree of freedom is not quite so obvious. Since $a_n=0$ we have a simplified recurrence for $x_n$,

  \begin{align*}
n(n-1)(n-2)x_n=12\sum_{1<k<n}(n-k)(k-1)x_{k-1}\tag{7}
\end{align*}

Using a generating function, $G(z)$, for the sequence $x_n$, the convolution on the right of (7) is represented by the product $\frac{1}{(1-z)^2}$ corresponding to $(n-k)$ and $G^{\prime}(z)$ corresponding to $(k-1)x_{k-1}$. We obtain the differential equation
$$G^{\prime\prime\prime}(z)=\frac{12}{(1-z)^2}G^{\prime}(z)$$

The nature of the equation suggests a solution of the form $G(z)=(1-z)^\alpha$, and testing this solution yields $\alpha=-2$ or $5$. The case $\alpha=-2$ corresponds to the previous observation that multiples of $n+1$ do not affect the solution. But for $\alpha =5$ we obtain an unusual  solution that is zero after its first five values:

$$x_1=-5,\quad x_2=10,\quad x_3=-10,\quad x_4=5,\quad x_5=-1$$

  This provides a second degree of freedom and yields the final solution

  $$x_n=\frac{12}{7}\left((n+1)H_n+1\right)+c_1(n+1)+c_2(-1)^n\binom{5}{n}$$

where $c_1$ and $c_2$ are determined by the initial conditions.

  Conclusion: It's instructive to see which candidates were selected for the repertoire in the first step. And it's very instructive to see how to cope with the inadequacy of the first family and select $x_n=(n-1)^{\underline{t}}H_n$ to extend the repertoire. But the last step indicates (at least for me) that finding a solution is also an art based upon a high level of creativity and a high level of experience.

Lets take a recurrence relation for example
$f(0) = 0$
$f(n) = 1 + f(n - 1), n > 0$
for this recurrence the closed form is
$f(n) = n$
lets take another recurrence relation for example
$g(0) = 0$
$g(n) = n + g(n - 1), n > 0$
lets first construct a table of values for $g(n)$
$g(0)$ -> $0$
$g(1)$ -> $1$
$g(2)$ -> $3$
$g(3)$ -> $6$
$g(4)$ -> $10$
By sheer observation, we get the closed form for $g(n)$
$g(n) = (n * (n + 1)) / 2$
Now lets take a third recurrence relation
$h(0) = 0$
$h(n) = n + 1 + h(n - 1), n > 0$
Now lets construct a table of values for this recurrence $h(n)$
$h(0)$ -> $0$
$h(1)$ -> $2$
$h(2)$ -> $5$
$h(3)$ -> $9$
$h(4)$ -> $14$
Now we would like to find a closed form for this recurrence relation h(n).
But it is not very clear to find the closed form by sheer observation.
But first let's try out a formula.
$h(n) = ((n * (n + 1)) / 2) + n$
Considering the above formula is a closed form for $h(n)$.
lets create another table using the above formula
$h(0)$ -> $0$
$h(1)$ -> $2$
$h(2)$ -> $5$
$h(3)$ -> $9$
$h(4)$ -> $14$
and viola the values match. If we properly observe the closed form of $h(n)$,
we can see that it consists of two closed form formulas.
first one being closed form of $g(n)$
$g(n) = (n * (n + 1)) / 2$
second one being closed form of $f(n)$
$f(n) = n$
It turns out that $h(n)$ is a linear combination of the
two recurrences $f(n)$ and $g(n)$.
We can write $h(n)$ in closed form as follows
$h(n) = ((n * (n + 1)) / 2) + n$
$h(n) = g(n) + f(n)$, here $f(n)$ and $g(n)$ are in closed form
or we can also write $h(n)$ in recurrence form as
$h(0) = 0$
$h(n) = (n + g(n - 1)) + (1 + f(n - 1))$, $n > 0$ here $f(n)$ and $g(n)$ are in recurrence form
now lets compare the above recurrence with the original recurrence of $h(n)$
$h(0) = 0$
$h(n) = n + 1 + h(n - 1), n > 0$
Now by sheer observation we can see that, in the original recurrence of
$h(n)$ we only see the variable "$n$" and the constant "$1$", but in the modified recurrence of
$h(n)$ we have associated the variable "$n$" with another recurrence called $g(n)$ and
similarly we have associated the constant "$1$" with another recurrence called $f(n)$.
$n$ -> $(n + g(n - 1))$
$1$ -> $(1 + f(n - 1))$
From this we can say that if we are unable to find the closed form of some recurrence relation by
just sheer observation. Then, we can actually look into the constants and the variables and try to
make small recurrence relations for those individual terms. Once, we figure out these individual
recurrence relation of every term and also find their corresponding closed form. We can
then combine the closed forms of those bits of recurrence relation to get the closed form of the
orginal recurrence relation.


Simplify $\tan^{-1}[(\cos x - \sin x)/(\cos x + \sin x)]$ Write the following functions in simplest form:
    $$\tan^{-1}\left(\frac{\cos(x)-\sin(x)}{\cos(x)+\sin(x)}\right), \quad 0<x<\pi$$

Please help me to solve this problem. I have been trying to solve this from last 3 hours. I can solve simple inverse trigonometric functions

Let's multiply the numerator and denominator by $\ \cos\bigl(\frac {\pi}4\bigr)=\sin\bigl(\frac {\pi}4\bigr)\ $ :
$$\tan^{-1}\left(\frac{\sin\bigl(\frac {\pi}4\bigr)\cos(x)-\cos\bigl(\frac {\pi}4\bigr)\sin(x)}{\cos\bigl(\frac {\pi}4\bigr)\cos(x)+\sin\bigl(\frac {\pi}4\bigr)\sin(x)}\right),\quad 0<x<\pi$$

$$=\tan^{-1}\left(\frac{\sin\left(\frac {\pi}4-x\right)}{\cos\left(\frac {\pi}4-x\right)}\right),\quad 0<x<\pi$$
$$=\begin{cases}
&\frac {\pi}4-x&\quad\text{if}\ \ 0 < x < \frac{3\pi}4\\
&\text{not defined}&\quad\text{if}\quad x = \frac{3\pi}4\\
&\frac {5\pi}4-x&\quad\text{if}\ \ \frac{3\pi}4< x < \pi\\
\end{cases}
$$

It is  $\tan  ^{-1}\dfrac{\cos  \left(x\right)-\sin  \left(x\right)}{\cos  \left(x\right)+\sin  \left(x\right)}$

We will divide in bracket with cosx to get it in tan form which will be easy for us to simplify

$\implies \tan  ^{-1}\dfrac{\dfrac{\cos  \left(x\right)-\sin  \left(x\right)}{\cos  \left(x\right)}}{\dfrac{\cos  \left(x\right)-\sin  \left(x\right)}{\cos  \left(x\right)}}$

$\implies \tan  ^{-1}\dfrac{1-\tan  \left(x\right)}{1+\cos  \left(x\right)}$

Now we know that $\tan  ^{-1}x + \tan  ^{-1}y= \dfrac{x-y}{1-xy}$ when $xy<1$

and we have the bracket in the same form as $\dfrac{\tan  \left(1\right)-\tan  \left(x\right)}{1-\tan  \left(1\right)\tan  \left(x\right)}$

So we get $\tan  ^{-1} \left(\tan   \left(\dfrac{π}{4}\right) + \tan   \left(X\right)\right)$
i.e $\dfrac{π}{4-x}$.


Is every odd order skew-symmetric matrix singular? We call a square matrix $A$ a skew-symmetric matrix if $A=-A^T$. A matrix is said to be singular if its determinant is zero. Is every odd order skew-symmetric matrix with complex entries singular?

Yes, that holds, since: $$\det A=\det{(-A^T)}=(-1)^{odd}\det{A^T}=-\det A,$$ from where we get $\det{A}=0$.

Consider the example
$$
\begin{bmatrix}
    0 & i & -3\\
    -i & 0 & 2i\\
    3 & -2i & 0\\
\end{bmatrix}
$$


How to solve this nonstandard system of equations? How to solve this system of equations
$$\begin{cases}
2x^2+y^2=1,\\
x^2 + y \sqrt{1-x^2}=1+(1-y)\sqrt{x}.
\end{cases}$$
I see $(0,1)$ is a root.

Solution.

First way

From the first equation, we have
$$\begin{cases}
2x^2\leqslant 1,\\
y^2 \leqslant 1
\end{cases}
\Leftrightarrow
\begin{cases}
-\dfrac{1}{\sqrt{2}} \leqslant x \leqslant \dfrac{1}{\sqrt{2}},\\
- 1 \leqslant y \leqslant 1.
\end{cases}$$
Then, the conditions of $x$ and $y$ are
$$\begin{cases}
0 \leqslant x \leqslant \dfrac{1}{\sqrt{2}},\\
- 1 \leqslant y \leqslant 1.
\end{cases}$$
We have $x^2 + y^2 = 1-x^2$. Therefore $x^2 + y^2 \leqslant 1$.
Another way,
$$1-x^2 =  y \sqrt{1-x^2} -(1-y)\sqrt{x} \leqslant  y \sqrt{1-x^2}.$$
Because
$$ y \sqrt{1-x^2} \leqslant \dfrac{y^2 + 1 - x^2}{2}.$$
Implies
$$1-x^2  \leqslant \dfrac{y^2 + 1 - x^2}{2}  \Leftrightarrow x^2 + y^2 \geqslant 1 .$$
From $x^2 + y^2 \leqslant 1$ and $x^2 + y^2 \geqslant  1$, we have $x^2 + y^2 = 1.$
Solve
$$\begin{cases}
x^2 + y^2 = 1,\\
2x^2 + y^2 = 1,\\
0 \leqslant  x \leqslant 1,\\
- 1 \leqslant y \leqslant \dfrac{1}{\sqrt{2}}
\end{cases} \Leftrightarrow \begin{cases}
x = 0,\\
y = 1.\end{cases}$$

Second way.
We have $2x^2 + y^2 = 1$, therefore $y=\sqrt{1 - 2x^2}$ or $y=-\sqrt{1 - 2x^2}.$

First case, $y=\sqrt{1 - 2x^2}$, subtitution the second equation, we get
$$x^2 + \sqrt{1 - 2x^2}\cdot\sqrt{1-x^2}=1+(1-\sqrt{1 - 2x^2})\sqrt{x}.$$
equavalent to
$$1 - x^2 - \sqrt{1 - 2x^2}\cdot\sqrt{1-x^2} + (1-\sqrt{1 - 2x^2})\sqrt{x}=0$$
or
$$\sqrt{1-x^2} (\sqrt{1-x^2} - \sqrt{1 - 2x^2}) + (1-\sqrt{1 - 2x^2})\sqrt{x}=0.$$
This is equavalent to
$$\dfrac{\sqrt{1-x^2}\cdot(1-x^2-1+2x^2)}{\sqrt{1-x^2}+\sqrt{1 - 2x^2}}+\dfrac{(1-1+2x^2)\sqrt{x}}{1+\sqrt{1 - 2x^2}} = 0$$
or
$$x^2\left (\dfrac{\sqrt{1-x^2}}{\sqrt{1-x^2}+\sqrt{1 - 2x^2}} + \dfrac{2\sqrt{x}}{1+\sqrt{1 - 2x^2}}\right )=0.$$
It is easy to see that
$$\dfrac{\sqrt{1-x^2}}{\sqrt{1-x^2}+\sqrt{1 - 2x^2}} + \dfrac{2\sqrt{x}}{1+\sqrt{1 - 2x^2}}> 0$$
Thus $x = 0$.

Second cases.  $y=-\sqrt{1 - 2x^2}.$
 We can check that
$$x^2 + y \sqrt{1-x^2} \leqslant \dfrac{1}{2}$$
and
$$1+(1-y)\sqrt{x} >1.$$
In this case, the given system of equations has no solution.

(More a comment.) If we allow the non-principal square root,
$$\begin{cases}
2x^2+y^2=1,\\
x^2 + y \sqrt{1-x^2}=1\color{red}{\pm} (1-y)\sqrt{x}
\end{cases}$$
the $+$ case has one real solution, but the $-$ case has three real solutions: $(x,y)=(0,1)$ and two which surprisingly are roots of $11$-deg equations. Using,
$$x=+\sqrt{\frac{1-y^2}{2}}$$
and with $y$ as two appropriate real roots of,
$$\small49 - 301 y + 327 y^2 - 51 y^3 - 1038 y^4 + 1334 y^5 - 1066 y^6 + 850 y^7 - 259 y^8 + 279 y^9 + 3 y^{10} + y^{11}=0$$
namely $y_1 \approx -0.619107$, and $y_2 \approx 0.939251$.


Find a formula for a linear transformation Find an example of a linear transformation formula $\varphi$ so that:
$$\ker\varphi = \{(x,y,z,t) \in \mathbb{R}^4 : x-y+6z+2t=0\},$$
$$\operatorname{im}\varphi = \operatorname{span}((2,3,1))$$
How to approach such problem? The most standard way to is appreciated.

$\varphi$ is a linear transformation $\mathbb{R}^4\rightarrow\mathbb{R}^3$, so the matrix $A$ representing $\varphi$ (with respect to standard basis) is $3$ by $4$. Now, if
$$\ker\varphi=\{(x,y,z,t)\in\mathbb{R}^4:x-y+6z+2t=0\}$$
then everything in the kernel of $A$ is orthogonal to $(1,-1,6,2)$, so let's set
$$A=\begin{bmatrix}1&-1&6&2\\ ?&?&?&?\\?&?&?&?\end{bmatrix}.$$
We are not done yet, because we haven't specified the remaining entries. But this is not difficult, because we know
$$\text{im}\varphi=\text{span}((2,3,1))$$
which implies that all the column vectors are scalar multiples of $(2,3,1)$. So for example, the first column is just $1/2$ times $(2,3,1)$, which gives
$$A=\begin{bmatrix}1&-1&6&2\\ 3/2&?&?&?\\1/2&?&?&?\end{bmatrix}.$$
Continuing this logic, we can fill out the last three columns similarly, giving us
$$A=\begin{bmatrix}1&-1&6&2\\ 3/2&-3/2&9&3\\1/2&-1/2&3&1\end{bmatrix}.$$
Now we are done.

In standard basis, we  need to find a $4\times 3$ matrix  which has rank $1$, such that all column vectors are collinear to the vector $^{\mathrm t\mkern-1.5mu}(2,3,1)$ and also such that all row vectors (which are collinear) are collinear to the vector $(1,-1,6,2)$.
To make it simple, we're  looking for a matrix of the form
$$\begin{bmatrix}
1&-1&6&2 \\
. & . & . & 3 \\
. & . & . & 1
\end{bmatrix}$$
Therefore, we can start from the first row and multiply it by the relevant coefficients to obtain the last column:
To make it simple, we're  looking for a matrix of the form
$$\begin{bmatrix}
1&-1&6&2 \\
\frac 32 & -\frac32  & 9 & 3 \\
\frac12 & -\frac12  & 3 & 1
\end{bmatrix},\quad\text{or}\quad\begin{bmatrix}
2 & -2 & 12 & 4 \\
 3 & -3  & 18 & 6 \\
 1 & -1  & 6 &  2
\end{bmatrix}\quad\text{if we prefer integer coefficients.}$$


Algebraic Representability of Prime Number Generators Does anyone happen to have at hand a short proof that demonstrates that there do (or do not) exist one or more algebraically representable prime number generating functions?

P×p=24×x+1(p is prime no )(any prime no square when divded by 24 remainder is 1)


Guessing one root of a cubic equation for Hit and Trial Suppose I have a cubic equation as $$15x^3-4x^2-25x+14=0$$
By Hit and Trial method I know that one of the roots is $x=1$.
And hence I can solve the cubic equation wit ease as it will take the form of $$(x-1)(ax^2+bx+c)=0$$

But what if the cubic equation is $$15 x^3-64 x^2-69 x+70 = 0$$
One of the roots is $x=5$

How do I guess that? Like...I have to try for $x=0,1,-1,2,-2,3,4,5$. It takes time and huge calculations...
I am only talking about simple roots like $x \in \mathbb{Z}$ (integers) and maybe $x \in [-5,5]$ or maybe $x=\frac{1}{2}$ or $x=\frac{-1}{2}$

The methods I am aware of is drawing an approx graph of the cubic equation using maxima/minima or using Newton's method with maybe 2 iterations and check for integers near that.

So are there any easy yet reliable methods of guessing one root.

You can use the rational root theorem to guess some roots.

Rational root theorem. All rational roots have the form $\frac{p}{q}$,with $p$ a divisior of the constant term and $q$ a divisior of the first coefficient.

For example, in the integer case, one can take $q=1$, and one only has to test $p = \pm 1$, $p = \pm 2$, $p= \pm 5$, and then you have it. (you of course could test $\pm 7$, $\pm 10$, $\pm 14$, $\pm 35$ and $\pm 70$ as well, but it would be less work to just divide by the $x-5$)
Another technique:

Descartes rule of Signs. The number of positive roots of $P(x)$ is equal to the number of sign changes is the sequence formed by the coefficients of $P(x)$ or it is less by an even number.
The number of negative roots of $P(x)$ is equal to the number of sign changes is the sequence formed by the coefficients of $P(-x)$ or it is less by an even number.
Roots with multiplicity $n$ are counted $n$ times.

In this case it is useless. But if it is a polynomial like $x^3+3x^2+4x+2$, you now know that there are no positive roots.
Last technique:

Estimating roots. If you've tested all eligible roots that are smaller than, say, 10, then you can use this. If the first two coefficients are not extremely small compared to the last two, then its a good idea to look to you eligible roots near to $$-\frac{\mathrm{second} \; \mathrm{coefficient}}{\mathrm{first} \; \mathrm{coefficient}}$$

If the root is larger than 10, the $x^3$ and $x^2$ term are large enough to be able to ignore the smaller terms, thus this is a good estimation. Write $a$ for the first coefficient and $b$ for the second, $c$ for the third and $d$ for the fourth.
Now, if $x$ is large, then $ax^3+bx^2+cx+d \sim ax^3+bx^2$. Therefore the roots will be comparable. But $ax^3+bx^2=0$ gives $x^2=0$ or $ax+b=0$. In the first case $x$ isn't large enough. In the second case we have $ax+b=0$, thus $ax=-b$, thus $x=\frac{-b}{a}=-\frac{\mathrm{second} \; \mathrm{coefficient}}{\mathrm{first} \; \mathrm{coefficient}}$.

Considering the equation to be $ax^3+bx^2+cx+d=0$ check for $d$ is product of roots. so you can take the divisors of $d$ for trial
Example : $15x^3-64x^2-69x+70=0$
Here the product of the roots is $70$ divisors of $70$ are $1,2,5,7,10,14,35,70$ so you can try these numbers instead of $1,2,3,4,5,6,7,8,9,10$.
Make sure you try the negative numbers too.


Is there a function that will give me nth prime number? Possible Duplicate:
  Is there a known mathematical equation to find the nth prime?

Is there a function, like f(n), I insert n  into the function and it outputs nth prime number? I have been trying to find a pattern between the prime numbers, 1st differences, 2nd differences, but I can't find it.

To be clear: I want something like f(x)=x^2+5 or f(x)=2.sin(x)+x^2. You know what I mean...

Well yes, the function $f$ does exist, but that doesn't mean we know what it looks like.


Union of three independent events Let  $ A_1, A_2, A_3 $ be independent events with probabilities $ \frac 12,\frac13,\frac14,$ respectively. How to compute for  $P(A_1\cup A_2 \cup A_3).$

My solution starts from using the probability of their complements, I do not know how to answer this question. Please help.

$P(A_1\cup A_2 \cup A_3) = 1- P(A_1^c\cap A_2^c \cap A_3^c)$

$P(A_1\cup A_2 \cup A_3)= P(A_1\cup A_2) + P(A_3) - P((A_1\cup A_2)\cap A_3 )$
$=P(A_1\cup A_2) + P(A_3) - P(A_1\cup A_2).P(A_3 )$ ..since $(A_1\cup A_2)$ and $(A_3)$  are independent

Similarly,
$P(A_1\cup A_2)=P(A_1)+P(A_2)-P(A_1\cap A_2)=P(A_1)+P(A_2) - P(A_1).P(A_2)$

Now you can just do the calculation. I guess the final answer is 3/4

Use the following fact

  If $A$ and $B$ are independent events, then $A^{c}$ and $B^{c}$ are independent evetns

Proof: We see that
$$P(A^c\cap B^c)=P(A^c)+P(B^c)-P(A^c\cup B^c)=P(A^c)+P(B^c)-P((A\cap B)^c)=$$
$$P(A^c)+P(B^c)-(1-P(A\cap B))$$
which since $A$ and $B$ are independent then we have
$$=P(A^c)+P(B^c)-(1-P(A)P(B))=(1-P(A))+(1-P(B))-(1-P(A)P(B))=$$
$$1-P(A)-P(B)+P(B)P(A)= (1-P(A))(1-P(B))=P(A^c)P(B^c)$$

Thus $A^c$ and $B^c$ are independent


Mathematical Induction for an alternating series It's been a while since I have done a problem like this. I have this problem

$$\sum_{i=0}^n (-1)^{i+1} i^2 = \frac {(-1)^{n+1}n(n+1)}{2}$$

So I have gotten this far:

Base Case:
$$n=1$$

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

Assume:

$$\sum_{i=0}^n (-1)^{i+1} i^2 = \frac {(-1)^{n+1}n(n+1)}{2}$$

Prove:
$n = n+1$

$$\sum_{i=0}^n+1 (-1)^ {i+1} i^2 = \frac {(-1)^{n+2}(n+1)(n+2)}{2} $$

I think that I can just replace the i's with (n+1) too

Which would be this

$$\sum_{i=0}^{n+1} (-1)^ {n+2} (n+1)^2 = \frac {(-1)^{n+2}(n+1)(n+2)}{2} $$

I don't remember where to go from here.

You can most certainly not replace $i$s with $n$s. That's a major mistake. For example

$$\sum_{i=1}^n i = \frac{n(n+1)}{2}$$
while
$$\sum_{i=1}^n n = n^2$$ and those are certainly not the same thing.

What you do have to do is replace $n$ with $n+1$ everywhere, including the sum, so what you have to prove is

$$\sum_{i=0}^{n+1} (-1)^{i+1} i^2 = \frac {(-1)^{n+2}(n+1)(n+2)}{2}$$

while you can assume that

$$\sum_{i=0}^n (-1)^{i+1} i^2 = \frac {(-1)^{n+1}n(n+1)}{2}$$

is true.

To prove the above equation, remember that

$$\sum_{i=0}^{n+1} a_i = \left[\sum_{i=0}^{n}a_i\right] + a_{n+1}$$

and now use this fact on your particular sum, then use the induction assumption that tells you something about the sum in parentheses.

See the summation of the n+1 terms is just the summation of n terms (which is known) + $(-1)^{n+2}(n+1)$. Now you see the minus sign $(-1)^{n+1}$can be taken common the remaining term is :

$ \frac{n(n+1)}{2} -(n+1)^2 = (n+1) .\frac{n-2(n+1)}{2} = -\frac{(n+1) (n+2)}{2} $.

Thus at last the minus sign turns out to e $(-1)^{n+2}$ and hence proved.


Are there infinite possible values of $1^i$ where $i=\sqrt{-1}$ Are there infinite possible values of $$z=1^i$$

Since $$e^{i2k\pi}=1$$ we have

$$1^i=\left(e^{i2k\pi}\right)^i=e^{-2k\pi}$$

so for different values of $k \in \mathbb{Z}$ we get different values. Is this concept correct?

You cannot use power laws of the reals for complex values. They do not hold on that wider space. So $$\left(e^{i2k\pi}\right)^i\not=e^{-2k\pi}$$

Otherwise you can easy "proof" that $-1=1$ by $$-1 = e^{-i\pi} = \left(e^{-2i\pi}\right)^\frac{1}{2} = 1^\frac{1}{2} = 1$$

So you assume a false condition…

I agree with
the OP's statement that
$1^i
=\left(e^{i2k\pi}\right)^i
=e^{-2k\pi}$
for any integer $k$.
For $k=0$,
$1$ is one of the possible values.

To "check",
note that
$(1^i)^{-i}
=(1)^{i\cdot (-i)}
=1^1 = 1
$
and
$(e^{-2k\pi})^{-i}
=e^{(-2k\pi)(-i)}
=e^{2k\pi i}
=1
$.


Proof of Absolute Convergence of $\sum\limits_{n=1}^{\infty} (-1)^{n-1}\tan\left(\frac{1}{n\sqrt{n}}\right)$ $$\sum\limits_{n=1}^{\infty} (-1)^{n-1}\tan\left(\frac{1}{n\sqrt{n}}\right)$$
My Attempt: To prove absolute convergence, we must consider $\sum\limits_{n=1}^{\infty}a_n$ and $\sum\limits_{n=1}^{\infty}|a_n|$.
I know that as $\theta \to0$ we can approximate $\tan\theta \sim \theta$. Hence $\sum\limits_{n=1}^{\infty}|a_n|$ becomes:
$$\sum\limits_{n=1}^{\infty}\left|\tan{\frac{1}{n\sqrt{n}}}\right| \sim_{\infty} \sum\limits_{n=1}^{\infty} \frac{1}{n\sqrt{n}}, 0 \leq a_n$$
Which converges by the $p$ test. As noted, the $\theta \to 0$ which means that $a_n > a_{n+1}$. Furthermore:
$$\lim_{n \to \infty} \tan{\frac{1}{n\sqrt{n}}} = 0$$
By the alternating series test, it converges. Since the sums for $a_n$ and $|a_{n}
|$ are both converging, it is absolute convergence by definition. Is this approach correct?

Almost. Just saying that $\tan\theta\sim\theta$ is a bit vague though. I suggest that you add$$\lim_{n\to\infty}\frac{\tan\left(\frac1{n\sqrt n}\right)}{\frac1{n\sqrt n}}=1$$to your proof, which is something that follows from that fact that $\tan0=0$ and that $\tan'(0)=1$. In other words, use the comparison test.

we have  $$n\geq1 , |\tan(\frac{1}{n\sqrt{n}})|\leq|\frac{3}{n\sqrt{n}}|   \implies \sum_{n=1}^{\infty}|\tan(\frac{1}{n\sqrt{n}})(-1)^{n-1}|\leq \sum_{n=1}^{\infty}|\frac{3}{n\sqrt{n}}|$$
and
$$\sum_{n\geq 1}^{}|\frac{3}{n\sqrt{n}}| $$  is convergent $$\implies \sum_{n\geq 1}^{}|\tan(\frac{1}{n\sqrt{n}})(-1)^{n-1}|$$ is convergent $$\implies \sum_{n\geq}{}(-1)^{n-1}\tan(\frac{1}{n\sqrt{n}})$$ Absolutely converges


How to integrate: $\frac{1}{(x^2 + z^2)^{\frac{-3}{2}}} dx$ I have tried to use u-substitution but for some reason am not doing it right and thus not getting the correct answer. I want to know the most obvious/ intuitive way to solve this integral.

$x=z\tan(k)$

$\mathrm{d}x=z\sec^{2}(k)\mathrm{d}k$

$(a^2 + x^2)^{\frac32} = z^4 (\sec k)^3 (sec(k))^2 \mathrm{d}k
=z^4 (1+(\tan k)(\tan k))^{\frac32}(\sec k)^2\mathrm{d}k$

now put $\tan(k)=t$, $\mathrm{d}k=(\sec k)(\sec k)\mathrm{d}k$ .
And then integrate it very easily


Find power series of $\frac{1}{\cosh z}$ I`m not sure what is the best way to find the power series for $\frac{1}{\cosh z}$ about the point 0.
I would appreciate it if you give me any help.

Thank you

Let $f(z)=\displaystyle{\frac{1}{\cosh z}}$. Since $f(z)$ is analytic at $z=0$, so it has a Taylor series expansion at $0$. We can try by finding out the derivatives $f^{(n)}(z) $  of $f$ and put it in the formula
$f(z)=f(0)+\displaystyle{\frac{f'(0)}{1}z+\frac{f''(0)}{2!}z^2 +\cdots}$.

I have found $f(0)=1$. $f'(z) =\displaystyle{-\frac{\sinh z}{\cosh^2 z}}$, $f''(z) =\displaystyle{\frac{\sinh^ z-1}{\cosh^3 z}}$ and $f'''(z) =\displaystyle{\frac{\sinh z(2-\sinh z) }{\cosh^4 z}}$ and so on.


Help me understand 'equivalence classes' and relations I'm reading up on binary relations and I understand them to be a mapping from one set into another.

However I'm having problems understand 'equivalence classes'. My book only gives a pretty dry explanation.

If you could give a good tangible explanation of equivalence relations and equivalence classes, that would be great!

Let $X$ be a finite set and $R \subseteq X^2$ be some binary relation. Then $R$ is very similar to a directed graph that has elements of $X$ as vertices, or in other words, $R$ can be drawn as dots and arrows. The following example depicts a graph and corresponding relation.
$$
\begin{array}{ccc}
1 &\to& \stackrel{\curvearrowright} 2 \\
\downarrow & & \updownarrow\\
3 & \gets & 4
\end{array}
$$
$$
\{(1,2),(1,3),(2,2),(2,4),(4,2),(4,3)\}
$$

Now, to talk about equivalence classes we need an equivalence relation. How an appropriate graph looks like?

It has to be reflexive, that is, each vertex $v$ has to have a loop $\stackrel{\curvearrowleft}v$ like $2$ in the previous example.
It has to be symmetric, i.e. each edge of the graph has to be bidirectional, like $\{2,4\}$ in the above diagram. If all the edges are bidirectional, then such graph is called undirected.
It has to be transitive. This means that the graphs includes all the possible "shortcuts", i.e. if you can get from vertex $u$ to $v$, then there is an edge $u \to v$, or in case of undirected graphs $u \leftrightarrow v$.

The above conditions together imply that if some two vertices are connected, then they belong to a clique (a graph that has all the possible edges, modulo loops), that is, each connected component forms a clique. These cliques are precisely the equivalence classes. In the following diagram we have two: $\{1,2,3,4\}$ and $\{5,6,7\}$ (loops omitted for clarity).

$$
\begin{array}{ccc}
1 &\leftrightarrow & 2 &  & 5 & \leftrightarrow & 6\\
\updownarrow & \swarrow\hspace{-10pt}\nearrow\hspace{-10pt}\nwarrow\hspace{-10pt}\searrow & \updownarrow & &\updownarrow & \swarrow\hspace{-10pt}\nearrow\\
3 & \leftrightarrow & 4 & & 7
\end{array}
$$

Observe, that if we would add edge $4 \to 5$ then symmetry would require also $5  \to 4$ and then transitivity would add all the other edges. The two clusters would merge into a single equivalence class, i.e. the graph would become the $K_7$ clique.

This intuition also holds for infinite sets, but it's hard to draw infinite graphs.

I hope this helps $\ddot\smile$

Binary relations and equivalence classes are two basic notions that we all learn at the very beginning of school. By this I am saying that you actually very well know what they are and how they work.

It is only due to the bad influence of Bourbaki, the formalism that was important to address some problem in logic (like consistency) that those dry expositions have permeated areas of mathematics in which they are absolutely not needed, and even become detrimental.

I will use pictures to explain what are these two notions. I will use pictures because it is with pictures we (I) learned (in the first year we entered school in our life) what this concepts are.

Binary relation

In kindergarten they used to gives us pictures of a row of bees and a row of flower pots.
We were supposed to draw arrows indicating what flowers each bee likes. There were different requirements about the preferences depending on the exercise and the skill the teacher wanted to teach. But a totally freely chosen set of arrows going from the bees to the flowers is what a relation is.

If we require, for example, that a bee has only one favorite flower then we get a a relation that is what we call a function from bees to flowers.

If we add to this the requirement, for example, that each flower should have no more than one bee that likes it, they we get a relation that is what we call an injective function from the bees to the flowers.

And so on.

Equivalence class

We had this other type of exercise in kindergarten. We were given a figure with many objects: flowers, tools, people, ...

We were given a feature, and we were supposed to enclose in blobs figures that had that feature in common.

For example, we could be told to group figures by color, or by use, or by kingdom, etc.

Something like this.

This breaking into classes is what an equivalence relation is.


Determining a slope and y-intercept from a piecewise function How would I begin solving a problem that is asking me to determine the value of $b$ such that $$f(x)= \begin{cases}3x^2+5& x \le 2\\ mx+b & x > 2\end{cases}$$ is differentiable for all values of $x$?

Not asking for the answer here obviously, just the steps needed to get off the ground.

You must satisfy two requirements for differentiability at $x=2$:

$f(x)$ must be continuous at $x=2$.
The left limit of the derivative must equal the right limit of the derivative at $x = 2$.

That is,

$$\lim_{x\to 2^+}f(x)=\lim_{x\to 2^-}f(x)$$

and

$$\lim_{h\to 0^+}\frac{f(2+h)-f(2)}{h}=\lim_{h\to 0^-}\frac{f(2+h)-f(2)}{h}$$

You need to find values of $m$ and $b$ that satisfy these two equations. Do you think you can do it from here?

Hint: Differentiability at a point implies continuity at that point and that the derivatives are equal at that point.


why integration of $y=\frac{1}{1+x^2}$ is $\arctan(x)+c$? I manually integrated $$\int  \frac {dx}{x^2+1}$$
 using u substitution and I got $$\frac {\ln(1+x^2)}{2x}$$
 where did I go wrong?

u substitution

$x$ depends on $u$ hence you have to express $x$ as a function of $u$.

However, if $g=f^{-1}$ then
$$
g'(y)=\frac{1}{f'\left(g\left(y\right)\right)}
$$
With $f=\tan$ then
$$
\text{arctan}'\left(y\right)=\frac{1}{\tan'\left(\text{arctan}\left(y\right)\right)}
$$
Using that $\tan'\left(x\right)=1+\tan^2(x)$ we have
$$
\text{arctan}'\left(y\right)=\frac{1}{1+\tan\left(\text{arctan}\left(y\right)\right)}=\frac{1}{1+y^2}
$$

You can get the result indirectly through the chain rule, as follows:

Let y = arctanx. Then tany = x. Differentiate w/r to x: you get dy/dx = the square of cosy. But cosy = 1/[sq. root of(1+x-squared]) ....[To see this, draw a right triangle with legs 1 and x as the adjacent sides of angle y (to conform to the condition that tany=x)]. Thus we have dy/dx=1/(1+x-squared), i.e.
d(arctanx)/dx=1/(1+x-squared), hence arctanx = the integral (antiderivative) of
1/(1+x-squared).


Does this series of functions converge I am looking at

$$ g(x) = \sum_{k=1}^\infty {\cos (kx) \over k^2}$$

I know I can't apply the M-test in this case. Is this function differentiable?

Context: Previously I proved that $$ f(x) = \sum_{k=1}^\infty {\sin (kx) \over k^3}$$

is differentiable and that $f'$ is continuous on $\mathbb R$ by using the Weierstrass M-test.

If you look at the partial sums

$$g_n(x) = \sum_{k=1}^n \frac{\cos (kx)}{k^2},$$

it is clear that these are differentiable, with derivative

$$g_n'(x) = - \sum_{k=1}^n \frac{\sin (kx)}{k}.$$

Then you look at the (for the moment only formal) series

$$h(x) = -\sum_{k=1}^\infty \frac{\sin (kx)}{k}.$$

You know that if that series converged uniformly, then $g$ would be differentiable with derivative $h$. But alas, not only can we not apply Weierstraß' $M$-test to the series $h$, the series actually does not converge uniformly.

However, the series converges in $L^2([0,2\pi])$: the $\sin (kx)$ are mutually orthogonal and $\int_0^{2\pi} \sin^2(kx)\,dx = \pi$ for all $k \in \mathbb{Z}^+$, so

$$\lVert g_n' - g_m'\rVert_{L^2([0,2\pi])}^2 = \sum_{k=m+1}^n \frac{1}{k^2} \int_0^{2\pi} \sin^2(kx)\,dx < \frac{\pi}{m}$$

for $1 \leqslant m < n$, which shows that $(g_n')$ is a Cauchy sequence in $L^2([0,2\pi])$. From that and the periodicity it follows that $g$ is the integral of $h$,

$$\begin{align}
g(y) - g(x) &= \lim_{n\to\infty} g_n(y) - g_n(x) \\
&= \lim_{n\to \infty} \int_x^y g_n'(t)\,dt\\
&= \int_x^y \lim_{n\to\infty} g_n'(t)\,dt\\
&= \int_x^y h(t)\,dt,
\end{align}$$

where the legitimacy of the interchange of limit and integral follows from the Cauchy-Bun'akovskij-Schwarz inequality,

$$\int_x^y 1\cdot\lvert u(t) - v(t)\rvert\,dt \leqslant \sqrt{\int_x^y 1^2\,dt}\cdot\sqrt{\int_x^y \lvert u(t)-v(t)\rvert^2\,dt} \leqslant \sqrt{y-x}\cdot \lVert u-v\rVert_{L^2}.$$

Thus $g$ is differentiable with derivative $h(x)$ in all points where $h$ is continuous, but $g$ is not differentiable in points where $h$ has jump discontinuities - in such points, $g$ has left and right derivatives, but they differ. (Other discontinuities of $h$ would not allow such an easy conclusion, but luckily, $h$ is very well-behaved, continuous except for jump discontinuities in multiples of $2\pi$, so the differentiability of $g$ is easy to determine knowing $h$.)

So it remains to see that $h$ indeed has the properties mentioned. In fact,

$$h(x) = \frac{x-\pi}{2}$$

for $0 < x < 2\pi$, $h(0) = 0$, and $h$ is periodic with period $2\pi$ (the latter is immediate from the periodicity of the sine, and $h(0) = 0$ also is easy to see). If you know the theory of Fourier series, it is probably easiest to see that this function has the indicated Fourier series. Otherwise, note that

$$-\frac{\sin (kx)}{k} = \int_x^\pi \cos (kt)\,dt,$$

whence

$$\begin{align}
g_n'(x) &= - \sum_{k=1}^n \frac{\sin (kx)}{k}\\
&= \int_x^\pi \sum_{k=1}^n \cos (kt)\,dt\\
&= \int_x^\pi \frac{\sin \left(n+\frac{1}{2}\right)t}{2\sin\frac{t}{2}} - \frac{1}{2}\,dt\\
&= \frac{x-\pi}{2} + \int_x^\pi \frac{1}{2\sin \frac{t}{2}}\cdot \sin \left(n+\tfrac{1}{2}\right)t\,dt.
\end{align}$$

The function $\varphi(t) = \frac{1}{2\sin (t/2)}$ is continuously differentiable on $(0,2\pi)$, hence for $0 < x < 2\pi$

$$\int_x^\pi \varphi(t)\sin (ct)\,dt = -\left.\frac{\varphi(t)\cos (ct)}{c}\right\rvert_{x}^\pi + \frac{1}{c} \int_x^\pi \varphi'(t)\cos(ct)\,dt,$$

so we have

$$\left\lvert \int_x^\pi \frac{\sin \left(n+\frac{1}{2}\right)t}{2\sin \frac{t}{2}}\,dt\right\rvert \leqslant \frac{K(x)}{n+\frac{1}{2}},$$

where $K(x)$ depends on the values of $\varphi$ on $[x,\pi]$ resp. $[\pi,x]$. This shows that the series

$$-\sum_{k=1}^\infty \frac{\sin (kx)}{k}$$

converges uniformly on every interval $[\delta, 2\pi - \delta]$ for $0 < \delta < \pi$, but since the limit function has discontinuities in multiples of $2\pi$, the convergence is not uniform on $[0,2\pi]$.

Weierstrass M-test gives us that $f(x)$ converges uniformly, so it also converges pointwise, for $M_k=1/k^3$.


Tying Span, Basis, Linear Independence, Dimension Together I'm going through Linear Algebra by Friedberg et al. and I'm having trouble with the buildup to the main ideas regarding span, linear independence, basis, and dimension. I'll detail what I've taken away so far about each concept and where I get confused. (Note - I'm mainly focusing on the finite dimensional vector spaces).

A set spans a vector space $V$ if all linear combinations (LCs) of the set result in the vector space. There are many possible spanning sets of a vector space.

Spanning sets that have fewer elements in them can express other vectors in $V$ more efficiently (computation-wise) since less vectors are used in forming LCs. Finding such sets motivates the notion of linear independence.

Question A: Sometimes the book refers to spanning sets that have less vectors in them as "smaller" spanning sets, and other times "minimal spanning sets". Are these equivalent?

The book then covers linear independence, which makes sense to me.

After this the book relates linear independence and span of a set: adding/removing dependent vectors does not affect the span (since they're already in the span), whereas adding/removing independent vectors does affect span. There's a theorem which states this more formally, using $S \cup {v}$, where $S$ is a linearly independent set and $v$ is a vector not in $S$. Maybe I'm missing something here conceptually from the formal statement in the book.

Question B: At this point can we conclude that the smallest spanning set is a linearly independent set?

A basis is a linearly independent set which spans $V$. A basis can express each vector in $V$ with a unique LC. A finite spanning set can be reduced to a basis for $V$.

Question C: So far not much has been said about the sizes of spanning sets, linearly independent  sets, and bases of $V$, which is why I get confused about the last sentence in 5). We haven't definitively stated how big/small each of the three sets can be, so can make this remark?

Replacement Theorem: my understanding for this is as follows - the size of a linearly independent subset in $V$ always $\leq$ the size of a spanning set of $$V, and the linearly independent subset can be made to span $V$ if elements from the spanning set are added to it.

Some results from theorem are: number of basis elements is unique = dimension, the maximum number of linearly independent vectors in $V = dim(V)$.

Thanks in advance, I'm a current engineering university student! I have taken a couple of classes in linear algebra before: one was computational, the other was definitely more theoretical but probably not at the level required for mathematics majors (it was a class for engineering students). Out of curiosity and to attain a deeper understanding I decided to go through this book.

Question A: Sometimes the book refers to spanning sets that have less vectors in them as "smaller" spanning sets, and other times "minimal spanning sets". Are these equivalent?

Not quite. If you remove some vectors from a spanning set, in such a way that you don't damage the span, you get a smaller spanning set. Such a set is not necessarily minimal; it's possible that you could remove even more vectors without damaging the span. If you keep removing vectors without damaging the span, eventually you'll get a spanning subset with the property that removing any vector from it will cause the subset to no longer be spanning. This is what it means to be minimal: there's guaranteed to be no more reductions possible.
As an example:
This is a spanning set of $\Bbb{R}^2$: $\{(1, 0), (1, 1), (3, 2), (2, -5)\}$.
This is an example of a spanning set that is "smaller" than the above: $\{(1, 0), (1, 1), (2, -5)\}$.
This is an example of a minimal spanning set (contained in the above spanning sets): $\{(1, 0), (1, 1)\}$.

After this the book relates linear independence and span of a set: adding/removing dependent vectors does not affect the span (since they're already in the span), whereas adding/removing independent vectors does affect span. There's a theorem which states this more formally, using $S \cup v$, where $S$ is a linearly independent set and $v$ is a vector not in $S$. Maybe I'm missing something here conceptually from the formal statement in the book.
Question B: At this point can we conclude that the smallest spanning set is a linearly independent set?

I should just correct you there: $v$ is not just vector not in $S$. We actually need $v \notin \operatorname{span} S$. Note that if you cannot add such a vector, then it means that $S$ is spanning (there is no vector in the space that is not already in the span).
Your understanding is correct otherwise: if a vector is dependent on other vectors in the set $S$ (i.e. if $v \in \operatorname{span} S$), then adding $v$ to $S$ will not increase the span, or removing $v$ from $S$ will not decrease the span. Inversely, if $v$ is not a linear combination of other vectors in $S$, then adding/removing $V$ from $S$ will strictly increase/decrease the span.
And yes, using this result, we can conclude that minimal spanning sets are linearly independent. If $S$ were linearly dependent, then one of the vectors in $S$ lies in the span of the other vectors in $S$. Removing this vector does not decrease the span (i.e. the set with this vector removed is still spanning), contradicting minimiality.

Question C: So far not much has been said about the sizes of spanning sets, linearly independent sets, and bases of V, which is why I get confused about the last sentence in 5). We haven't definitively stated how big/small each of the three sets can be, so can make this remark?

We don't need any solid numbers for the sizes of these sets. The point is, if we start with a finite spanning set, and start removing redundant elements (i.e. remove $v$ from $S$ such that $v \in \operatorname{span}(S \setminus \{v\})$), the set is going to get smaller and smaller. Eventually, we need to stop removing elements, since we only have a finite number of them!
If we only remove redundant elements, then we don't damage the span, but at some point we have to stop. At that point, there will be no redundant vectors left, and the resulting subset will be minimal and hence linearly independent. So, using this process of greedily removing redundant vectors, we can pare down any spanning set to a basis.


Sequence formed by points of continuity Let $f:\mathbb{R} \longrightarrow \mathbb{R}$ be a monotone increasing function. It is well known that $f$ has at most a countably infinite number of discontinuities.
I would like to know if, given a discontinuity point $a$, there are sequences formed by points of continuity $(a_n)$ and $(b_n)$ such that $a_n<a$, $a<b_n$ and $a_n \rightarrow a$, $b_n \rightarrow a$ as $n \rightarrow \infty$.

Given the interval $\left(a-\frac{1}{n},a\right),$ there are uncountably many points, and only countably many points where $f$ is discontinuous, so we can find an $a_n\in\left(a-\frac{1}{n},a\right)$ such that $f$ is continuous at $a_n.$ Then $a_n<a$ for all $n$ and $a_n\to a.$

Similarly, choose $b_n\in\left(a,a+\frac{1}{n}\right)$ where $f$ is continuous, and then $b_n>a$ and $b_n\to a.$

If the function has only finitely many discontinuities in total, then the only such sequences are those that eventually (i.e., starting with some index)  constantly equal $a$.


Why is this inequality true? Trigonometry \begin{align}\lvert \cos\theta(\cos\theta -i\sin\theta)\rvert&\leq \lvert \cos\theta -i\sin\theta\rvert\\&=\sqrt{\cos^2\theta+\sin^2\theta}=1
\end{align}
I know $0\leq\lvert \cos\theta \rvert\leq 1$, but why can we just remove $\lvert \cos \theta \rvert$ on the right hand side of the inequality?

Note that since $\vert\cos(x)\vert\le1$ and $\vert x_1x_2\vert=\vert x_1\vert\cdot\vert x_2\vert$ (you should verify this yourself), then $$\left\vert\cos(x)(\cos(x)-i\sin(x))\right\vert=\left\vert\cos(x)\right\vert\cdot\left\vert(\cos(x)-i\sin(x))\right\vert\le1\cdot\left\vert(\cos(x)-i\sin(x))\right\vert=\sqrt{\cos^2(x)+\sin^2(x)}=1$$

It is generally true that $|ab|=|a||b|$, so that in this case:
$|cos(\theta)(cos(\theta)-isin(\theta))|=|cos(\theta)|(cos(\theta)-isin(\theta))|\leq|(cos(\theta)-isin(\theta))| $.


How to formulate a theorem about bijections between several sets I have several sets $A_i$ and bijections between them. (As stated in my theorem) no composition of these bijections produces a permutation of $A_i$ not equal to identity. So every bijection is identified by the pair of sets between which it acts.

It would be to cumbersome to denote every bijection with a special letter (such as $\Phi$). I want to write it in some consise way.

For example I could denote the bijection from $A_i$ to $A_j$ as $\phi_{A_i,A_j}$ but this is not formally right as I would first to prove $A_i\ne A_j$. Then I would denote it $\phi_{i,j}$ but this way I would need explicitly number my sets, but I'd better to use English names or maybe letters to denote the sets not numbers.

The best solution I found insofar is to denote every set with some letter and denote my bijections as $\phi_{\alpha,\beta}$, where $\alpha$ and $\beta$ determine some sets. This solution is not ideal, because it would involve for each considered set two different notations to denote it: say $A_1$ and $\alpha$.

This is insofar the best solution I know. But maybe somebody may suggest me a better language to formulate my theorem?

Each bijection $\phi$ determines its domain and range. Why not phrase your theorem in terms of the indexed set of bijections $\phi_j$ where $j$ ranges over some arbitrary index set $I$? You can define $D_j$ and $R_j$ (say) to be the domain and range of $\phi_j$ respectively. The sets $D_j$ and $R_j$ are your $A_i$ but you don't have to worry about whether they are distinct.

The best idea I got is the following:

Consider only bijections of the set $A_0$ with $A_i$ for every $i\ne 0$. There are no need to explicitly say about bijections of every $A_i$ with $A_j$.

The rest bijections can be easily figured out by a reader of my writing.


Spring Calculation - find mass A spring with an $-kg$ mass and a damping constant $9$ can be held stretched $2.5 \text{ meters}$ beyond its natural length by a force of $7.5 \text{ Newtons}$. If the spring is stretched $5 \text{ meters}$ beyond its natural length and then released with zero velocity, find the mass that would produce critical damping.

My work:

The restoring force is $-kx$.  Then

$$7.5 = -k(2.5) \\
-\frac{7.5}{2.5} = k \\
ma = -\frac{7.5x}{2.5} \\
my’’ + 9y’ + -3y = 0,\quad y(0) = 2.5, y(5) = 0 \\
\frac{-9 \pm \sqrt{81 + 4(m)(3)}}{2m} \\
-\frac{9}{2m} \pm \frac{\sqrt{81+12m}}{2m} \\
y = Ae^{-(9/2)x}\cos\left(\frac{\sqrt{81+12m}}{2m}x\right) + Be^{-(9/2)x}\sin\left(\frac{\sqrt{81+12m}}{2m}x\right) \\
2.5 = A + B\cdot 0 \\
0 = (2.5)e^{-45/2}\cos\left(\sqrt{81+12m}\frac{5}{2m}\right) + Be^{-45/2}\sin\left(\sqrt{81+12m}\frac{5}{2m}\right)$$

Any help would be appreciated

1) find the $k$

2)$\zeta=c/(2\sqrt{k/m})=1 $ for critical damping. So solve for $m$ as you alreaday know $c$ and $k$.

Critical damping occurs when $(\gamma^2)-4mk=0$
Therefore,
$$(9^2)-4m(7.5/2.5)=0$$
$$81=12m$$
$$m=6.75 kg$$
Bob Shannon also got the same answer, but I wanted to present it this way because I thought it might be more clear to some.


A ''corollary'' to the Chebotarev density theorem Milne's notes on class field theory has the following corollary to the Chebotarev density theorem:

  If a polynomial $f(X)\in K[X]$ splits into linear factors modulo $\mathfrak{p}$ for all but finitely many prime ideals $\mathfrak{p}$, then it splits in $K[X]$.

Here $K$ is a number field. Milne's proof is the following:

  Apply the Chebotarev density theorem to the splitting field.

I've managed to prove this by first showing that the polynomial must be separable modulo all but finitely many primes $\mathfrak{p}$. If $\mathfrak{P}\mid \mathfrak{p}$, we may assume $f(X)$ splits and is separable modulo the prime. It follows that $G(\mathfrak{P})=1$ for all but finitely many primes, since $f$ must split in the completion of $K$ at $\mathfrak{p}$. Hence, almost all primes split, so the splitting field equals $K$.

I'm wondering if there's a shorter or even trivial proof of this since, Milne didn't bother writing down anything at all? The problem as I see it is that one can't use Chebotarev directly since the factorization of $f(X)$ does not necessarily give the factorization of a prime $\mathfrak{p}$. This would be because we don't know what the generators of the ring of integers look like.

As you note, the assumption that $f(x)$ splits into linear factors modulo almost all $\mathfrak p$ implies that the decomposition group at $\mathfrak P$ is trivial
for almost all $\mathfrak P$.  (This is standard algebraic number theory,
and presumably Milne expects the reader to realize this without any explanation being necessary.)

With this fact in hand, Milne's claim follows directly from Cebotarev:

Let $G$ be the splitting field of $L$ over $K$.
Cebotarev says that any element of $G$ can be realized as a Frobenius
element (in particular as an element of the decomposition group $G({\mathfrak P})$) for infintely many $\mathfrak P$.  But since $G(\mathfrak P)$ is trivial
for all but finitely many $\mathfrak P$, we see that every element of $G$
must be trivial, hence that $G$ is trivial, i.e. that $L = K$, i.e. that $f(x)$ splits over $K$.

(If this is the argument you had in mind, then I'm not sure what you're asking.)

A side remark:

I'm not sure what you mean when you write that the factorization of $f(X)$ does not necessarily give the factorization of a prime $\mathfrak p$. As I mentioned in my answer to this question of yours, if $f(x)$ is any element of $K[x]$,
then we may write $f(x) \in \mathcal O_K[1/a][x]$, for some $a \in \mathcal O_K\setminus \{0\}.$
Choosing $a$ appropriately, we may furthermore assume that the discrminant of $f$ is invertible in $\mathcal O_K[1/a]$, so that for every prime $\mathfrak p$ of $\mathcal O_K[1/a]$, the reduction of $f(x)$ modulo $\mathfrak p$ is separable.

The factorization of any prime $\mathfrak p$ of $\mathcal O_K[1/a]$ in the splitting field $L$ of $f(x)$ then corresponds precisely to the factorization of $f(x)$ modulo $\mathfrak p$.

Of course, the primes of $\mathcal O_K[1/a]$ are precisely the primes of $\mathcal O_K$ which do not divide $a$, and so consist of all but finitely many
prime of $\mathcal O_K$.  Thus if one is willing to throw away finitely many primes (which is obviously harmless in the case of e.g. Milne's claim), then one
can just work with the ring $\mathcal O_K[1/a]$ and apply Kummer's lemma in this ring.  Everything proceeds just as if $f(x)$ has integral coefficients (and there is no need whatsoever to worry about generators of the ring of integers in either $K$ or $L$; e.g., since the discriminant of $f(x)$ is invertible in $\mathcal O_K[1/a]$, the ring
$\mathcal O_L[1/a]$ is generated by the roots of $f(X)$ over $\mathcal O_K$).

Refer to milne's ANT(p146,version3.03).
$\mathfrak{P}\in Spl(L/K) => \mathfrak{P}\in Spl(K/K)$ for almost many primes.
So, $Spl(L/K) = Spl(K/K)$.


Find the principal solutions of the trigonometric equation $\cos x-\sin x+\sin 2x+3\cos2x+1=0$ I am unable to simplify the expression. If I simplify the double angles, it leaves me with a nasty expression,
$\cos x-\sin x+2\sin x\cos x+6\cos^2 x-2=0$. What do I do next. Some hints, please. Also, is there some elegant solution? Thanks!

Hint: Let $a = \cos x$ and $b = \sin x$ so that you get $$a - b + 2ab + 6a^2 - 2 = 0 \iff (2a-1)(3a + b + 2) = 0$$

One solution is already apparent, $\cos x = \frac{1}{2}$, can you work out the others?

You have $\cos x-\sin x+\sin 2x+3\cos 2x+1=0$. Or $\sin x-\sin 2x=\cos x+3\cos 2x+1$. Or $\sin x(1-2\cos x)=\cos x+6\cos^2x-2$.

Either $\cos x =1/2$ or $\sin x=((6\cos^2x+\cos x-2)/(1-2\cos x))$. Or $\sin x=-3\cos x-2$. Squaring we get $10\cos^2x-12\cos x+3=0$. Thus $\cos x=((6\pm \sqrt{6})/(10))$.


Solutions of $q=\frac{x}{y} +\frac{y}{z} + \frac{z}{x}$ s.t. $q \geq 3$ Is it true that for every rational $q \geq 3$ , the following equation has a solution over $\mathbb N$ ?

$$q=\frac{x}{y} +\frac{y}{z} + \frac{z}{x}$$

The problem

$N=\frac{x}{y}+\frac{y}{z}+\frac{z}{x}$

with $N,x,y,z \in \mathbb{Z}$ was considered by Andrew Bremner and Richard Guy in "Two more representation problems" published in the Proceedings of the Edinburgh Mathematical Society, vol. 40 pp.1-17 in 1997. An online copy is available here. They showed solutions only occurred for those $N$ where the elliptic curve

$t^2=u^3+N^2u^2+8Nu+16$

has rank at least $1$.

For small $N>0$, the first solution is for $N=6$, with $x=18$, $y=4$ and $z=3$.

Maple code that searches for solutions in specific range :

for q from 4 to 30 do
for x from 1 to 200 do
for y from 1 to 200 do
for z from 1 to 200 do
if x/y+y/z+z/x=q then
print(q,x,y,z);
end if;
end do;
end do;
end do;
end do;

For $~q=9~$ ;$~(x,y,z)=(12,63,98)$


Good algebra book to cover these topics? I will be studying two algebra modules next year and I am looking for a comprehensive book that will cover both of them, however due to having very minmal exprience with algebra I am looking for your advice on this, so please don't delete this topic like some other topics I made (please?)

The topics we will cover in the first module (called "further linear algebra") are:

Revision of Linear Algebra, change of bases, characteristic polynomial, eigenvalues,
traces.
Hamilton-Cayley theorem, minimal polynomial, nilpotent matrices, Jordan
forms of nilpotent matrices, primary decomposition.
Jordan normal form, functions of matrices, norms and convergence.
Powers and exponents, applications to difference and differential equations.
Bilinear forms, Quadratic forms, GLn- classification.
Orthogonal matrices, Gramm-Schmidt process, O(n)-classification, normal matrices.
Spectral theorem and applications to hypersurfaces, classifications of quadrics
in R3
.
Hermitian, unitary and normal matrices, applications to quantum mechanics.
Linear algebra over Z, abelian groups, GLn(Z) and SLn(Z), examples of
quadratic forms over Z.
Echelon forms of matrices over Z, application to finitely generated abelian
groups.

And for the second algebra module ("groups and rings"):

Groups and subgroups, examples, rings and subrings, the issue of 1 in signature,
fields, examples.
Isomorphisms, cyclic groups, orders of an element in a group, generators and
relations, Cosets and Lagranges theorem, Eulers theorem, RSA, groups of
order 4 and p.
Homomorphisms, examples of homomorphisms, kernels and images, normal
subgroups and ideals, classification of groups of order 6 and 8.
Quotient groups and quotient rings, examples, isomorphism theorems, Cayley
Theorem, Chinese remainder theorem.
Group actions on sets, groups, rings, graphs etc., orbit-stabilizer theorem,
conjugacy classes, conjugacy classes in Dn, Sn and GLn, simplicity of A5.
Class equation, groups of order p
2
, semidirect products, classification of groups
of order 2p.
Sylows theorems, classification of groups of order 12 and 15.
Polynomial rings, Euclidian domains, PID-s, UFD-s, rings of integers.
Gauss theorem about polynomial rings, matrix rings.
Ideals in matrix rings, idempotents, cyclotomic polynomials, discrete Fourier
transform.

I am thinking that Artin is a good choice of book, since I have already covered the basics of groups, rings and linear algebra last year but I am more looking for a book that will cover the topics above most comprehensively (but not an encyclopedic one such as dummit/foote since I won't have too much time mainly due to work) in preference to a "fun" book if that makes sense (but of course if there is a fun book that covers most of the topics I would love to hear about it, especially if it is Artin as I read some and like it so far)

Thanks!

Module $2$ syllabus can be found in almost all books on Algebra but I recommend Dummit and Foote.

For Module $1$, I would recommend "Linear Algebra done right" by Axler for abstract approach avoiding Matrices, and "Linear algebra done wrong " by Sergie Treil (Google it for e-copy). These two books covers all topics espically Sergie's book, but Axler is great to understand proofs and concept. You can accompany Axler by "Linear algebra" by Friedberg and Spence. It is a good book too.

All the best.

Some topics wont be covered as it is impossible to have one or two books covering all the material up there. So google and read notes for them

Basic Mathematics by Serge Lang for a start in Algebra, his Linear Algebra book was okay, look up other math books by Springer (publisher).


For any two points, there exist a path of $n$ segments that connects them. Is there a name for this kind of set? We consider a set $A$. $A$ is called convex if for every $x,y\in A$, we have the line segment $xy$ is also in $A$.

I want to generalize this notion, such that instead of one line segment, there can be $n$ line segment, where $n$ is some fixed number. Formally, there exist $x=a_1,a_2,\ldots,a_{n-1},a_n=y$, such that all the segments $a_1a_2$, $a_2a_3$,...,$a_{n-1}a_n$ is in $A$.

Is there a name for such sets?

Assuming you allow arbitrarily large $n$, I believe the term is "polygonally connected".

It is called the link diameter. So those sets can be called as sets with link diameter k.

An efficient algorithm for link-distance problems


Is the function $z=x^{3}y$ even or odd? I just wanted to know because I am trying to calculate $\int_{-1}^{1}\int_{-1}^{1}x^{3}ydxdy$.

It is even because $(-x)^3(-y)=x^3y$.

Despite that, this is unnecessary for calculating your integral, where you can use  that $x^3$ and $y$ are odd, which results in the value $0$.

Parity is not defined for multivariate functions.

In your case, by Fubini

$$\iint f(x)\,g(x)\,dx\,dy=\int f(x)\,dx\int g(y)\,dy$$ and it suffices that any of the simple integrals vanishes.


How to find the infimum of $A=\{\frac{1}{n+10}\}_{n\in\mathbb N}$ I know the infimum is zero, and I know I need to find $a_{n}(\epsilon)\in A$ such that $\forall \epsilon>0:0+\epsilon>a_{n}(\epsilon)$. How do I go about finding $a_{n}(\epsilon)$?

Method 1:-

$\because$ Given sequence is monotonically decreasing, so, the sequence $\{\frac{1}{n+10}\}$ converges to its infimum.

Method 2:-
Infimum is the greatest lower bound. We know that $0<\frac{1}{n+10}$. $\therefore$ $0$ is the lower bound. Next, we need to prove $0$ is the greatest lower bound. Suppose $\epsilon>0$ be another lower bound of the set. $$\epsilon<\frac{1}{n+10}$$
$$\implies n<\frac{1}{\epsilon}-10$$. It is true for all Natural numbers. natural numbers are not bounded. So, $\epsilon$ cannot be the lower bound. $\epsilon$ was an arbitrary positive real number.

Start here $${1\over n + 10} < \epsilon.$$
Take reciprocals; note that this reverses order
$$n + 10 > {1\over \epsilon}$$


Prove that $\sum_{i=1}^{n}a_i^2\geq\frac{A^2}{n}$, where $a_1,...,a_n\in\mathbb Z^+,\sum_{i=1}^{n}a_i=A$. Suppose we have $n$ positive integers $a_1,...,a_n$ and $\sum_{i=1}^{n}a_i=A$. If we further assume $a_i<n$ for every $i$, then is the inequality $\sum_{i=1}^{n}a_i^2\geq\frac{A^2}{n}$ true? If yes, how can we prove it?

Just simply use the QM-AM inequality. It makes perfect sense as each $a_{i}$ $\in$ $\mathbb{N}$. Then by QM-AM inequality, we have that
\begin{align*}
&\sqrt{\frac{\sum_{i= 1}^{n} a_{i}^{2}}{n}} ~\geqslant~\frac{\sum_{i = 1}^{n}a_{i}}{n}\\
\Leftrightarrow~ & \frac{\sum_{i= 1}^{n} a_{i}^{2}}{n} ~\geqslant~ \frac{A^2}{n^2}\\
\Leftrightarrow~ & {\sum_{i= 1}^{n} a_{i}^{2}} ~\geqslant~ \frac{A^2}{n}
\end{align*}

By Jensen's inequality, since $\frac{d^2}{dx^2}x^2>0$,
$$\frac{\displaystyle\sum_{i=1}^na_i^2}n\ge \left(\frac{\displaystyle\sum_{i=1}^na_i}n\right)^2$$
, where $a_i\in\mathbb R,n\in\mathbb Z^+$.


Asymptotic expansion of the complete elliptic integral of the first kind The complete elliptic integral of the first kind is defined as $$K(k) = \int_0^{\pi/2} \frac{d x}{\sqrt{1 - k^2 \sin^2 x}}.$$ I would like to derive (at least the first term of) the asymptotic expansion for $k = 1 -\epsilon$. This is certainly not trivial due to lack of uniform convergence which implies that I can't use the Taylor expansion of the integrand. What is the best way to proceed?

You can employ the substitution $y=1- k^2 \sin^2x$ such that the integral can be written as
$$ K(k) = \int_{1-k^2}^1\!dy\,\frac{1}{2 \sqrt{y(1-y)(y-1 +k^2)}}.$$

Now, we can expand the integrand in $\epsilon = 1-k$ and obtain
$$ K(k) = \int_{1-k^2}^1\!dy\frac{1}{2 y \sqrt{1-y}} + O(1). \tag{1}$$
The estimate of the error term follows from the fact that expanding in $\eta =1-k^2$, we have
$$K(k) = \sum_{n=0}^\infty c_n \eta^n  \int_{\eta}^1\!dy\, y^{-1-n} (1-y)^{-1/2}$$
with $c_n$ some constants. Now, we have for $n>0$ that
$$ \int_{\eta}^1\!dy\, y^{-1-n} (1-y)^{-1/2} = O(\eta^{-n})$$
such that only the $n=0$ term diverges for $\eta \to 0$ (which corresponds to $\epsilon \to 0$).

It thus remains to estimate the first term in (1) for $k \to 1$ which is not that difficult:

In fact due to the $1/y$ behavior close to $y=0$, the integral is logarithmically divergent and we have that
$$K(k) = \frac12 \left|\log (1-k^2)\right| + O(1) = \frac12 \left|\log \epsilon\right| + O(1).$$

Using two point Taylor expansion around $x=(0 ,\frac{\pi}{2})$ series you could get the expression
$$\frac{\pi ^5 k^4}{2560}-\frac{\pi ^5 k^2}{5760}+\frac{\pi }{4 \sqrt{1-k^2}}+\frac{\pi }{4}$$ This has a better quadratic error than the other series near $k=1$.


Prove that [0,1] is equivalent to (0,1) and give an explicit description of a 1-1 function from [0,1] onto (0,1) The problem is stated as follows:

  Show that there is a one-to-one correspondence between the points of the closed interval $[0,1]$ and the points of the open interval $(0,1)$.  Give an explicit description of such a correspondence.

Now, I think I can prove the first part of the problem by demonstrating the following:

Define $f: (0,1) \to \mathbb{R}$ as follows.

For $n \in \mathbb{N}$, $n \geq 2$, $\space{ }f(\frac{1}{n}) = \frac{1}{n-1}$
and for all other $x \in (0,1)$, $\space{}f(x) = x$

Prove that $f$ is a $1-1$ function from $(0,1)$ onto $(0,1]$
Slightly modify the above function to prove that $[0,1)$ is equivalent to $[0,1]$
Prove that $[0,1)$ is equivalent to $(0,1]$

Since the "equivalent to" relation is both symmetric and transitive, it should follow that $[0,1]$ is equivalent to $(0,1)$.  Hence, there does exist a one-to-one correspondence between $[0,1]$ and $(0,1)$.

I have no trouble with the above.  My problem is in "finding an explicit description of such a correspondence."  Can I modify the above function, or will that not suffice?

Steps 2 and 3 are not necessary. The function $g:(0,1] \to [0,1]$ defined by $g(1) = 0$ and $g(x) = f(x)$ if $x \neq 1$ is a bijection. This shows that $(0,1]$ is equivalent to $[0,1]$ and, by transitivity, that $(0,1)$ is equivalent to $[0,1]$. Furthermore, the function $g \circ f$ is a one-to-one correspondence between $(0,1)$ and $[0,1]$ that you can describe explicitly.

It's far easier than step 1.

Step 2 : same definition than step 1, but with:
$\forall x∈[0,1), \space f(x)=x$
instead of $\forall x∈(0,1), \space f(x)=x$.

Step 3 : $\forall x, \space f(x)=1-x.$


Finding the curve $C$ whose every point in the first quadrant is the midpoint of $L(x,y) $? Let $C$ be a curve in $\Bbb{R}^2$  passing through $(3, 5)$ and $L(x, y)$ denote the segment of the tangent line to $C$ at $(x, y)$ lying in the first quadrant.
Assuming that each point $(x, y)$ of $C$ in the first quadrant is the
midpoint of $L(x,y)$.

So I know that $(3,5)$ satisfies the curve and that every point of $C$ in the first quadrant is the midpoint of $L(x,y)$. But how can I deduce from this information about the curve.

Is there some concept needs to be known which will make it trivial to find the curve?

Can we find the curve with these conditions?

The information is sufficient.  Let $x_0$ and $y_0$ be any point on C.  Then we have $$y-y_0 = (x-x_0)y'(x_0)$$

Then the x intercept of the line segment in the first quadrant is
$$(\frac{-y_0}{y'(x_0)}+x_0, 0)$$
The y intercept is $$(0, y_0-x_0y'(x_0))$$

As the mid point is $x_0$ and $y_0$ and replace $x_0$ by x and $y_0$ by y, we have xy' + y = 0 or $$xdy + ydx =0$$ or $$d(xy) = 0$$ or $xy=K$

Since (3, 5) is on C, we have K = 15.  Therefore the solution is xy = 15


Number of integers covered Let $x_{ij}\in\{0,1\}$ for ${i=1}$ to ${m}$ and for ${j=1}$ to $n$. How many different values does $$\prod_{i=1}^m\sum_{j=1}^nx_{ij}$$ cover?

Is there an $a_{ij}$ such that $$\prod_{i=1}^m\sum_{j=1}^na_{ij}x_{ij}$$ for which we cover least number of distinct values as compared to that values covered by first formula?

For $m=2$ this is one more (for the $0$) than the number of distinct products $s \cdot t$ with $s,t \in \{1,2,\ldots,n\}.$ The latter number is for $n=1,2,...$ the sequence $1,3,6,9,14,18$ which certainly no simple formula is known to produce. It is discussed as an o.e.i.s. sequence A027424.
For larger $m$ it would seem unlikely for there to be a closed formula, since even for $m=2$ there doesn't seem to be a known one. However note one of the links from the o.e.i.s. is to a paper discussing the more general case of an arbitrary number of factors. (Unfortunately that link only lets one see an outline, and refers to yet another site where the final paper might be found.)
Added note about the second part of the question about $a_{ij}$ choice: It may not be optimal, but choosing $a_{ij}=(-1)^{j+1}$ certainly brings the count down. Suppose $n=2k$ is even. Then the sums for a fixed row $j$ of the terms $a_{ij}x_{ij}=(-1)^{j+1}x_{ij}$ have as possible values the integers in the interval $[-k,k].$ If we omit the zero choice here (later one can add $1$ to the sum) then the possible values of the product are each up to a sign of $\pm 1$ the nonzero values of the original unaltered sum for parameters $(m,n)=(m,k).$ To state this more clearly, let $1+S(m,n)$ be the number covered as in the post, while $1+T(m,n)$ the number covered with our choice of coefficients $a_{ij}.$ Then what we have is $T(m,2k)=2 \cdot S(m,k).$ Even for $m=2$ this gives a significant decrease for the altered sum, comared to the unaltered one.

I suspect this is difficult.  But here is an upper bound.
Each expression is the product of $m$ numbers, each between $0$ and $n$.  If any number is zero, the product is zero.  Otherwise, there are $m$ numbers, each between $1$ and $n$.
By the standard 'stars and bars' argument, there are $m+n-1\choose n-1$ sets, so an upper bound is $$1+{m+n-1\choose n-1}$$
Of course, that doesn't take into account coincidences such as $1*4=2*2$.
A lower bound comes with all $a_i$ primes greater than $\sqrt{n}(i=1..m)$.  All these products are different.  Let $q=\pi(n)-\pi(\sqrt{n})\approx n/\log n$.  The lower bound is $$1+{m+q-1\choose m}$$


The probability that a candidate comes with all $3$ pens having the same color Question

Candidates were asked to come to an interview with $3$ pens each.
Black, blue, green and red were the permitted pen colours that the candidate could bring.
The probability that a candidate comes with all $3$ pens having the same color is _________.

My Approach
$\binom{4}{1}$ to select 1 colors from $4$ available colors and then probability of each color $ =\frac{1}{4}$
Reqd probability$  =4 \times (\frac{1}{4} )^{3}$
is it correct?

We can write down a sample space:

$$\Omega =\Big{\{}\{R,R,R\},\{R,R,B\}, \{R,R,G\},\{R,R,Y\},  \{R,Y,Y\}, \\ \{R,B,B\},  \{R,G,G\}, \{R,G,B\},  \{R,Y,B\}, \{R,Y,G\} , \\\{Y,Y,G\},  \{Y,Y,B\}, \{G,G,Y\},  \{G,G,B\}, \{B,B,G\},\\ \{B,G,Y\},\{B,B,Y\}, \{B,B,B\}, \{G,G,G\}, \{Y,Y,Y\}
\Big{\}}$$

So, if we assume that events are uniformly distiributed (since distribution is not mentioned in OP) we have: $$ P = {4\over |\Omega|} = {1\over 5}$$

Please, if somebody disagree with this solution then please write down your own  sample space and distribution and then tell me what is wrong.


Inequality between two functions I have a two functions defined for $x > 1$, and $c \in (0,1)$:
$$
f(x) = 1-\exp\left(-\frac{c}{x^2} \right),
$$
and
$$
g(x) = \exp\left(-\frac{x}{c} \right).
$$

From graphical tool (
https://www.desmos.com/calculator/hr8n8kkpym ), I know $f(x) > g(x)$. How can I prove this inequality analytically?

Calling

$$
\left\{ \begin{array}{rcl}
u & = & 1-e^{-\frac{c}{x^{2}}}\\
v & = & e^{-\frac{x}{c}}
\end{array}\right. (1)
$$

we have

$$
\left\{ \begin{array}{rcl}
\log(u) & = & \log(1-e^{-\frac{c}{x^{2}}})\\
\log(v) & = & -\frac{x}{c}
\end{array}\right. (2)
$$

and also

$$
\left\{ \begin{array}{rcl}
\frac{d}{dx}\log(u) & = & -\frac{1}{(e^{\frac{c}{x^{2}}}-1)x^{3}}\\
\frac{d}{dx}\log(u) & = & -\frac{1}{c}
\end{array}\right. (3)
$$

and for $x=1$ we have

$$
-\frac{1}{e^{c}-1}>-\frac{1}{c}
$$

and for $x>1$ and $c\in(0,1)$ we have

$$
-\frac{1}{(e^{\frac{c}{x^{2}}}-1)x^{3}}>-\frac{1}{c}
$$

Now ressuming, by (3) if $\log(1-e^{-c})>-\frac{1}{c}$ then $\log(u)$
and $\log(u)$ does not intersect and will remain $\log(u)>\log(v)$
all along $x\ge1$

Concluding, $\log$ is a strict monotonic increasing function for
$x>1$ hence $u > v$

As x tends to infinity f(x) tends to 0 in a slower rate which can be seen by differentiation compared to g(x)
You can prove it by finding inverse function of f and g which are simpler to prove than this


How to describe the locus of $z$ where $\arg \left(\frac{z-z_1}{z-z_2}\right)=\theta$. I need to describe the locus of $z$ which stisfies $\displaystyle\arg \left(\frac{z-z_1}{z-z_2}\right)=\theta$.

$\displaystyle\arg \left(\frac{z-z_1}{z-z_2}\right)=\theta \Rightarrow \arg(z-z_1)-\arg(z-z_2)=\theta$

So I got that  ,

$\arg(z-z_1)=\arg(z-z_2)+\theta$

So $z$ lies on an arc.

If  $0<\theta \leq\frac{\pi}{2}$ , $z$ lies on the major arc of a circle which passes through $z_1$ and $z_2.$ But there are two such circles ! How can we  identify the correct circle ?

If  $\frac{\pi}{2}<\theta < \pi$ , $z$ lies on the major arc of a circle which passes through $z_1$ and $z_2.$ But there are two such circles ! How can we  identify the correct circle ?

Since $\theta$ is a given real number, the arc is oriented. Only one of two possible circles has the right orientation of the angle from $z_1$ via $z$ to $z_2.$


Error in attempting the combinatorics problem In how many ways can $15$ identical blankets be distributed among $6$
  beggars such that everyone gets at least one blanket and two
  particular beggars get equal blankets and another three particular
  beggars get equal blankets.

Attempt:

First, $6$ blankets are given and then $9$ blankets are left.
After that, according to me there are only two possibilities:

Possibility $1$:

First set of particular beggars receives: $(3,3)$

Second set of particular beggars receives: $(1,1,1)$

Possibility $2$:

First set of particular beggars receives: $(0,0)$

Second set of  particular beggars receives: $(3,3,3)$

However answer given is $12$.

Please tell me my mistake.


Prove the closure is closed and is contained in every closed set Let $(X,d)$ be a metric space and $Y \subseteq X$ a subset. I want to show the following without using accumulation points or limit points at all.

  $\overline Y$ is a closed subset of $X$.
  $\overline Y$ is contained in every closed set which contains $Y$.

My definitions are:

The interior of $Y$ is

$$Y^\circ = \{y \in Y \mid \exists \varepsilon \gt 0:B_\varepsilon(y) \subseteq Y\}.$$

The boundary of $Y$ is

$$ \partial Y= \{x \in X \mid \forall \varepsilon \gt 0: B_\varepsilon(x) \cap Y \neq \emptyset \neq B_\varepsilon(x) \cap (X \setminus Y) \}.$$

And the closure of $Y$ is

$$\overline Y = Y \cup \partial Y.$$

I know also that a set $O \subseteq X$ is open in $X$ iff every convergent sequence with limit in $O$ has also almost all of its terms in $O$. And a set $A$ is closed in $X$ iff every convergent sequence with all of its terms in $A$ has also its limit in $A$.

I tried showing that $X \setminus \overline Y$ is open as well as using the above result for sequences but both times I got stuck in complicated set expressions resulting from figuring out what $X \setminus \overline Y$ is. Can you show me how to prove this?

I found only proofs using limit or accumulation points (links: here, here, here, here, here)

Asserting that $\overline Y$ is closed is the same thing as asserting that $X\setminus\overline Y$ is open. Take $x\in X\setminus\overline Y$. Since $x\notin\partial Y$, there is some $\varepsilon>0$ such that $B_\varepsilon(x)\cap Y=\emptyset$ or that $B_\varepsilon(x)\cap(X\setminus Y)=\emptyset$. But the second possibility cannot occur, since $x\in B_\varepsilon(x)\cap(X\setminus Y)$. Therefore, $B_\varepsilon(x)\cap Y=\emptyset$. Now, take $y\in B_\varepsilon(x)$ and take $\varepsilon'>0$ such that $B_{\varepsilon'}(y)\subset B_\varepsilon(x)$. Then $B_{\varepsilon'}(y)\subset Y\setminus X$, which proves that $y\notin\partial Y$. Therefore, $B_\varepsilon(x)\cap(Y\cup\partial Y)=\emptyset$. In other words, $B_\varepsilon(x)\cap\overline Y=\emptyset$. This proves that $B_\varepsilon(x)\subset X\setminus\overline Y$ and, since this happens for each $x\in X\setminus\overline Y$, $X\setminus\overline Y$ is open.
Let $F$ be a closed set such that $F\supset Y$. Let $y\in\partial Y$. Can we have $y\notin F$? No, because $X\setminus F$ is open and therefore there would be a $\varepsilon>0$ such that $B_\varepsilon(y)\subset X\setminus F$. In other words, $B_\varepsilon(y)\cap F=\emptyset$ and, in particular, $B_\varepsilon(y)\cap Y=\emptyset$. This is absurd, since $y\in\partial Y$. So, this proves that $F\supset Y\cup\partial Y=\overline Y$.

Let $\{x_n\}_n$ be a convergent sequence in $X$ with limit $x\in X\setminus\overline Y$.
In particular $x\notin\partial Y$, hence there exists $\epsilon>0$ such that one of $B_\epsilon(x)\cap Y$, $B_\epsilon(x)\cap (X\setminus Y)$ is empty. Of course the latter is wrong for $x\notin Y$. Hence $B_\epsilon(x)\cap Y=\emptyset$. Note that almost all $x_n$ are in $B_\epsilon(x)$ (and consequently in $X\setminus Y$). And for each such $x_n$, we have $B_{\epsilon-d(x_n,x)}(x_n)\subseteq B_\epsilon(x)$ and hence $B_{\epsilon-d(x_n,x)}(x_n)\cap Y=\emptyset$. We conclude that $x_n\notin \partial Y$. Together with $x_n\notin Y$, we get $x_n\in X\setminus \overline Y$, as desired.


Derivation of Fourier Transform? So from the fourier series, we can simplify it further and use trig identities to get the following:

$$ f(t) = \frac{a_0}{2} + \sum^{\infty}_{n=1}
\left(\frac{a_n}{2}+\frac{b_n}{2i}\right)e^{i n \omega t} +
\left(\frac{a_n}{2}-\frac{b_n}{2i}\right)e^{-i n \omega t} $$

So how do you go from the above line to

$$ f(t) = \sum^{\infty}_{n=-\infty} C_n e^{in\omega t} $$

and

$$ f(t) = \frac{1}{\sqrt{2\pi}} \int^{\infty}_{-\infty} F(\omega) e^{-i \omega t} d\omega $$

Thanks

There's a mistake. Sin formula is $\sin(\omega nt)=\frac{e^{i \omega nt}-e^{-i\omega nt}}{2i}$

so the serie become:

$ f(t) = \frac{a_0}{2} + \sum^{\infty}_{n=1}
\left(\frac{a_n}{2}+\frac{b_n}{2i}\right)e^{i n \omega t} +
\left(\frac{a_n}{2}-\frac{b_n}{2i}\right)e^{-i n \omega t} $

Well, you call $C_{n}=\left(\frac{a_n}{2}+\frac{b_n}{2i}\right)$ and $C_{-n}=\left(\frac{a_n}{2}-\frac{b_n}{2i}\right)$ and obtain:

$ f(t) = \frac{a_0}{2} + \sum^{\infty}_{n=1}
C_{n}e^{i n \omega t} +
C_{-n}e^{-i n \omega t} $

that can be re-written in

$ f(t) = \sum^{\infty}_{n=-\infty} C_n e^{in\omega t} $

because $e^{i\omega0t}=1$ and you can rename $\frac{a_0}{2}$ in ${c_0}e^{i\omega0t}$.
Notice that if f(t) is real $a_n$ and $b_n$ are reals. so $c_n=\overline c_{-n}$.

For Fourier Transform the question is a little more complicated, because Fourier Transform is an operator that you define in $L^1$, you extend it on $L^2$ and in temperate distributions.

Fourier serie is connected with  a function $L^1$on a interval (that become periodic on $\mathbb{R}$) or an hyperrectangle on $\mathbb{R}^n$.  Imagine you can extend this to all $R^n$...you obtain Fourier transform. The factor  $\frac{1}{2\pi}$ is arbitrary, there are other choices, but all they must be coherent with the choice of the factor of the inverse Fourier Transform. $\frac{1}{\sqrt {2\pi}}$ is the most natural, because it's very similar to normalized complex exponencials that you use in Fourier Serie for the finite interval.
And this choice makes Fourier Transform an Unitary Operator, because $F^{+}=F^{-1}$.

The proof of Fourier in based that integration $-\pi$ to $\pi$ $\cos(kt)\cos(nt)dt$ is  zero.
What we basically do is multiply the Fourier series with $\cos(nt)$ on both sides. And integrate both sides from $-\pi$ to $\pi$.
All other terms on the right hand side become zero except $\cos(nt)$(because $\cos(nt)^2 dt$ integrated from $-\pi$ to $\pi$ is not zero but $\pi$). I have derived the Fourier transform (in detail) in the post http://communicationsm.blogspot.in/.


The shortest path that meets every extended line that passes through a circle of radius $r$ Problem:
A straight underground pipe passes by a statue at a maximum distance of 100 meters. The exact whereabouts of this pipe is unknown. What is the shortest trench i must dig that will assure me to find the arbitrary laid pipe.

The answer appears to be a half circle with two lines of length 100 m aranged like a horse shoe magnet.
If this is the shortest trench path how could such be proven formally. e.g disproving all other trench paths.

the bold line represent the shortest trench path

is it worth investigating this problem for IB MATHS HL INTERNAL ASSESMENT?

Without typo, length is equal to $ 2 r + \pi r.$


Take a regular coordinate ball and you get a manifold with boundary. Suppose $M$ is a (topological) manifold of dimension $ n \geq 1$ and $B$, is a regular coordinate ball in $M$.
Show that $M\backslash B$ is an $n$-manifold with boundary and whose boundary is homeomorphic to $S^{n-1}$

Ok, so here is what I have. $M\backslash \overline{B}$ is an open space in $M$ and hence is locally euclidean. So what's left is to prove that $\partial B$, the boundary of $B$, is the boundary of $M\backslash B$.

Here is where we use the the fact that $B$ is a regular coordinate ball. We know there exists a a function $\phi: B' \to B_{r'}(x)$ such that $\phi(B)=B_{r}(x)$ and $\phi(\overline{B})=\overline{B_{r}(x)}$. With this information, we know that there is a neighborhood of the boundary of $B$, namely $B' \cap M\backslash B$, which is homeomorphic to a the closure of a a ball in $\mathbb{R}^n$. So we conclude that $M\backslash B$ is a manifold with boundary.

Moreover, the boundary of the manifold is the boundary of B, which is homeomorphic to $S^{n-1}$ because $\phi(\partial B)=\phi(\partial B_r(x))$.

Is this proof airtight? How can I make it more precise? For example exactly what would be the neighborhood of the boundary of B which works to prove that it is the boundary of the manifold?

No, your proof is insufficient (I will explain why in the end of the answer). You have to revise the notion of a coordinate ball in an $n$-dimensional manifold $M$ as follows:

A compact subset $B\subset M$ is called a coordinate ball if there exists an open subset $U\subset M$ containing $B$, and a homeomorphism $\phi: U\to U'\subset R^n$ (where $U'$ is necessarily open in $R^n$) such that the image $\phi(B)$ is the unit ball $B'=\{x: |x|\le 1\}$ in $R^n$.

With this definition the proof that $M-int(B)$ is a manifold
with boundary (equal to the topological frontier $fr(B)$ of $B$ in $M$) becomes effortless: The claim is equivalent to the assertion that $U-int(B)$ is a manifold with boundary (where the boundary equals to $fr(B)$), which is equivalent to the claim that $U'- int(B')$ is a manifold with boundary (whose boundary equals the sphere $S'=\{x: |x|=1\}$). The latter can be proven, for instance, by applying  stereographic projections (with centers at the north and the south poles of the sphere $S'$ sending $\{x: |x|>1\}$ to the upper half-space $\{x: x_n> 0\}$.

Now, an example showing that your notion of a coordinate ball is insufficient. There exist examples   of wild spheres $S\subset R^3$ which separate $R^3$ in two components: The closure $B$ of one component is homeomorphic to the closed unit ball. (Think of this as your "coordinate ball" in $R^3$.) On the other hand, the closure of the other component (I will call this closure $C$) is not a manifold with  boundary. The latter is because $S$ is "knotted". A precise way to state this knottedness property is in terms of the fundamental group. There exist points $x\in S$ such that for an arbitrarily small neighborhood $G$ of $x$ in $C$ the map
$$
\pi_1(G - S) \to \pi_1(int(C))
$$
is nontrivial. (There are loops in $int(C)$ arbitrarily close to $x$ which cannot be contracted to a point in $int(C)$.)  A manifold with boundary cannot have such pathological behavior.

I am answering here since the comment section does not allow so many characters, but I do not consider this as an answer, it's just a list of observations.

First of all your question is not well posed, or at least the thing you ask is false in general, i.e. consider the 2 sided cone without the vertex. This thing is even a smooth manifold, but if you remove a big enough ball with center near the vertex what you get is a bounduary which is homeomorphic to the disjoint union of 2 spheres.

Consider a hyperbolic surface of rotation https://en.wikipedia.org/wiki/Hyperboloid, once again if you choose the ball big enough, your claim is false.

Thus we reduce ourselves to wander if there is a small enough nhbd of points, on which removing a geodesic ball means getting a manifold with that bonduary (the shere $S^{n-1}$). This is given to you in smooth enoughmaifolds by the exponential map which in manifolds which are at least $C^2$ (as far as I Know), gives you a smooth local diffeomorphism between the tangent and the manifold, thus your claim follows easily by this observation.
In the general case you have that your manifold is just locally homeomorphic to some open set in $R^n$, but then you can use charts to perform the same argument as above.


Estimating the truncation error Consider the following method
$y_{n+1}=y_n+hf(t_{n+1},y_n+hf(t_n,y_n))$, the truncation error is given by
$T_n=\dfrac{y(t_{n+1})-y(t_n)}{h}-f(t_{n+1},y_n+hf(t_n,y_n)) $, Im having problem on how to estimate $T_n$ especially on how to apply Taylor series on $f(t_{n+1},y_n+hf(t_n,y_n))$. Any help I would appreciate.

suppose your $f$ to be an analytic function, then, given a numerical method $\mathcal{Y}(f,k,y(t_n))$, where $f$ is your function, $k$ is the step size (you call it $h$), and $y(t_n)$ is the analytical solution at the time $t_n$, then the local error is given by $y(t_{n+1})-\mathcal{Y}(f,k,y(t_n))$. In genereal therms, it's just the difference between the analytical solution at time $t_{n+1}$ and the numerical method applied to the analytical solution at time $t_n$.

Evaluating this difference, and remembering that $\vec f(t_n,y(t_n))=\vec y'(t_n)$, since we're considering first order systems of ODE, you have to center the Taylor series in the point $(t_n,y(t_n))$:
\begin{align}
&y(t_{n+1})-\mathcal{Y}(f,k,y(t_n))
\\
&= y(t_n) +ky'(t_n)+\frac{k^2}{2}y''(t_n) + \mathcal{O(k^3)}
\\&\qquad-y(t_n)-k[y'(t_n)+k\frac{df}{dt}(t_{n+1},y(t_n)) + ky'(t_n)\frac{df}{dy}(t_{n},y(t_n))+\mathcal{O(k^2)}]
\\
&=ky'(t_n)+\frac{k^2}{2}y''(t_n)+\mathcal{O(k^3)}-ky'(t_n)-k^2 \frac{df}{dt}(t_n,y(t_n))-k^2[y'(t_n) \frac{df}{dy}(t_n,y(t_n))]
\\
&=ky'(t_n)+\frac{k^2}{2}y''(t_n)+\mathcal{O(k^3)}-ky'(t_n)-k^2 \frac{df}{dt}(t_n,y(t_n))-k^2[y''(t_n)-\frac{df}{dt}(t_n,y(t_n))]
\\
&\approx \mathcal{O(k^3)}
\end{align}
Since we're integrating in an interval of length $k$, then this method is of the second order. (And btw it's a particular Runge-Kutta method)

Hope it will be useful for you


Locally Stable but Globally Unstable? In control theory, we talk about the direct and indirect Lyapunov methods, applied to the stability analysis of nonlinear systems.

There are systems that are locally unstable for specific operating points, but that are globally stable.

What i would like to know is if there is any system which is locally stable, for any linearized point, but is globally unstable.

In other words, I would like to know if a system that is guaranteed to be locally stable, thorough linearization, for any point, can be said to be globally stable as well.

In the case it is positive, I would like to know if uniform stability and exponential stability holds as well.

The answer is no. A system might be unstable for some initial conditions even if its linearization is Hurwitz for all $\boldsymbol{x}$.

An example is provided in this lecture, page 4 (Nonlinear Systems and Control — Spring 2015: State-Dependent Riccati Equation Method). Let

$$
\begin{bmatrix}
\dot{x}_1 \\
\dot{x}_2
\end{bmatrix} =
\begin{bmatrix}
\frac{-1}{1 + \epsilon\sqrt{x_1^2 + x_2^2}} & 1 + \sqrt{x_1^2 + x_2^2} \\
0 & \frac{-1}{1 + \epsilon\sqrt{x_1^2 + x_2^2}} \\
\end{bmatrix}
\begin{bmatrix}
x_1 \\
x_2
\end{bmatrix}
$$

with $\epsilon > 0.$ The eigenvalues are given by the matrix entries $a_{11}$ and $a_{22}$, which are obviously negative for any $\boldsymbol{x} \in \mathbb{R}^2$. Thus $A(\boldsymbol{x})$ is Hurwitz for all $\boldsymbol{x}  \in \mathbb{R}^2$.

However, take $\epsilon = 1$ and solve the ODE numerically, using for example the Runge-Kutta method, with initial conditions $\boldsymbol{x}_0 = \begin{bmatrix}1 & 1\end{bmatrix}^T$. You will observe that $x_1 \rightarrow \infty$ and $x_2 \rightarrow \text{const}$.


What is the nature of $e^{ix}$, real or complex? Since, $e^{i\pi}=-1$,

right part of the equation ($-1$) is real and left part is seemingly complex as

$e^{ix}=\cos(x)+i\sin(x)$ which is a complex number( I am not sure though).

I am a freshman, and don't know much regarding Euler's equations and advanced Mathematics.

Real numbers are also complex numbers; their imaginary part is $0$.  If $x$ is a real number then $e^{ix}$ is, in general, complex, but, in special cases, it can be real.  The only real numbers of the form $e^{ix}$ ($x$ real) are $1$ and $-1$.

Because $e^{ix}=\cos(x) + i\sin(x)$, it is only "real" when $\sin(x) = 0$, namely when $x = n \pi$ for $n \in \mathbb{Z}$. For every other value of $x$, $e^{ix}$ has a complex component. Note that I put "real" with quotes because $\mathbb{R} \subset \mathbb{C}$.