Question about the famous 'statistics in medicine' puzzle Context of the question:

Given a test that predicts a disease with 99% accuracy. Furthermore it is know that one in 10.000 people suffer from this disease. If we assume that I test positive, then the question of the riddle is: "what are the odds that I am actually sick".

One can make a table like this:

$$
\begin{array}{c|lcr}
& \text{Infected} & \text{Healthy} \\
\hline
\text{Positive test} & \frac{1}{10000}\times\frac{99}{100} & \frac{9999}{10000}\times\frac{1}{100} \\
\text{Negative test} & \frac{1}{10000}\times\frac{1}{100} & \frac{9999}{10000}\times\frac{99}{100}\\
\end{array}
$$

The answer of the riddle is: $$\frac{\frac{1}{10000}\times\frac{99}{100}}{\frac{1}{10000}\times\frac{99}{100}+\frac{9999}{10000}\times\frac{1}{100}}=\frac{1}{102}$$

Now here is my question:

If I test positive, then I am either infected, or I am healthy. There are no other options. Therefore I would think that their combined odds add up to one. Then how come that the odds of postive test/infected + positive test/healthy (the values of the first row) do not add up to 1?

Thanks!

You are making 2 mistakes:

First, while you are right that if I test positive, then there is a 100% chance that I am either healthy or infected, but what that means is that $P(healthy|positive)+P(infected|positive)=1$, and you are describing this as $P(positive| healthy)$ + P(positive |infected)$

Second, you point to the top row as the values of $P(healthy|positive)$ and $P(infected|positive)$, but those are the values of $P(healthy \land positive)$ and $P(infected \land positive)$, and those will only add up to $P(positive)$, which is indeed not equal to $1$

Finally, note that what you calculated as $\frac{1}{102}$ is $P(infected|positive)$. So, use a similar method to compute $P(healthy|positive)$ (change the numerator to $\frac{9999}{10000} \times \frac{1}{100}$ to do this ... The denominator (which is $P(positive)$ stays the same), and you wil find that $P(healthy|positive)=\frac{101}{102}$, and now you see that they do add up to $1$.

In short: $P(positive| healthy)$, $P(healty|positive)$, and $P(positive \land healthy)$ are $3$ different things! (Though we do have that $P(positive \land healthy)=P(healthy \land positive)$)

The statement "If I test positive, then I am either infected, or I am healthy" corresponds to the probabilities of infected / positive + healthy / positive = 1; the event that you're conditioning on is "positive" not infected and healthy.


Integrating with respect to a linear combination of two signed measures Let $(X, d)$ be a metric space and $\mathcal{B}(X)$ the Borel $\sigma$-algebra of X. Let $\mu, \nu$ be two real-valued signed measures defined on $(X, \mathcal{B}(X))$ and $f : X \to \mathbb{R}$ Borel measurable, $\alpha, \beta \in \mathbb{R}$.

Can anyone explain to me if the following takes place
$$\int f(x) d(\alpha \mu + \beta \nu)(x) = \alpha \int f(x) d\mu(x) + \beta \int f(x) d\nu(x)?$$

Thank you!

Yes. By definition if $E$ is measurable and $\mu(E)$, $\nu(E)$ are finite then
$$ (\alpha \mu + \beta \nu)(E) = \alpha \mu(E) + \beta \nu(E).$$

In integral notation this becomes
$$\int \chi_E d(\alpha \mu + \beta \nu) = \alpha \int \chi_E \, d\mu + \beta \int \chi_E \, d\nu.$$
From here you can prove the result for simple $f$ and from there for (sufficiently integrable) measurable $f$ via the usual limiting processes.

Ok. Thanks for the reminder, Leucippus.

Dealing with this argument, as i said, you have to take care, because you don't know yet, if the monotone convergence theorem is still valid for signed measure.

Despite of this, we are able to prove this using only the definition of integral.

By definition,

$\int f d\mu := \sup \{\sum_{i=1}^N \inf_{E_i} \{f(x)\} \mu(E_i)\} \text{ ,}$

where the sup is taken over all partitions

But we already now that if $A,B \subset \mathbb{R}$, then $\sup A+B = \sup A + \sup B$

So, applying the equallity above to $\int f(x) d(\alpha \mu + \beta \nu)(x) = \alpha \int f(x) d\mu(x) + \beta \int f(x) d\nu(x)$, we obtain the required identity.

Now, using the Hahn Decomposition Theorem you can prove the fact that te monotone convergence theorem also works for signed measures


Prove that if $3|(a^2+b^2)$, then $3|a$ and $3|b$, where $a, b$ are integers I would like to know how to prove the above statement by contradition. Somebody said that one should prove it by this method but I have no idea what it is.

For any integer, there are precisely three options for its reminder when divided by $3$, i.e. $0,1,2$. So suppose that $x_i=3k+i$ for $i=0,1,2$. Then
$$x_i^2=9k^2+6k+i^2=\begin{cases}3t & i=0\\ 3t+1& i=1,i=2\end{cases}$$
So, for any integer $x$, the reminder of $x^2$ when divided by $3$ is either $0$ or $1$.  Now assume, by the way of contradiction, that $a$ or $b$ are not divisible by $3$. Then what can you say about $a^2+b^2$?

suppose that a=3k+r  , b=3q+r'  r,r'=0,1,2

if a=3k , b=3q its ok

if not r=1,2 a=3k+1 or 3k+2

so a^2=9k^2+6k+1=3k'+1  ,or 9k^2+12k +4 =3k'+1

and for b like that

so a^2 mod 3=1 and b^2 mod3 = 1 and it is impossible

then the only condition that's possible is  a=3k , b=3q

proof complete


Show that $\sqrt{n}(\ln(2\overline{X}_{n}-\ln(\theta))\overset{\mathcal{D}} \to N(0,3^{-1})$ Let $(X_n)$ be a sequence of independent, identically distributed random variables, with $X_n \sim \mathcal{U}[0,\theta]$ for some $\theta > 0$.  Show that $\sqrt{n}(\ln(2\overline{X}_{n})-\ln(\theta))\overset{\mathcal{D}} \to N(0,3^{-1})$.

Im not being able to solve this problem, Im stucked in the central limit theorem with

$$\sqrt{n}\frac{\overline{X}_{n}-\frac{\theta}{2}}{\frac{\theta}{\sqrt{12}}}\overset{\mathcal{D}} \to N(0,1)$$

And I've already proved various results regarding continuous functions and convergence in distribution. But Im not being able to take the last formula to the one in the problem. Can you just give me a hint?

$\DeclareMathOperator{\Pr}{\mathbb{P}}$
Let $Z_n = \dfrac{X_n}{\theta}$ then $Z_n \sim \mathcal{U}[0,1]$, and let $S_n = Z_1 + \dots + Z_n.$

Choose any real $t$, we have

$
\begin{split}
\Pr(\sqrt{n}(\ln(2\bar{X}_n) - \ln(\theta)) \le t) = \Pr(\sqrt{n}(\ln(2\frac{\bar{X}_n}{\theta}) ) \le t)\\ =\Pr(\sqrt{n}(\ln(2nS_n) ) \le t) \\=\Pr\left(\dfrac{S_n - \frac{n}{2}}{\sqrt{\frac{n}{12}}} \leq \sqrt{3n}\left(\exp(\dfrac{t}{\sqrt{n}})-1\right)\right)\\
=\Pr\left(\dfrac{S_n - \frac{n}{2}}{\sqrt{\frac{n}{12}}} \leq t\sqrt{3} +\sqrt{3n}\left(\exp(\dfrac{t}{\sqrt{n}})-1-\frac{t}{\sqrt{n}}\right)\right)
\end{split}
$

Since $\exp(x) \geq 1 + x$ for all real $x$ we have   $0 \leq \sqrt{3n}\left(\exp(\dfrac{t}{\sqrt{n}})-1-\frac{t}{\sqrt{n}}\right) = \sqrt{3n}(\dfrac{t^2}{2n} + \text{O}(\dfrac{1}{n\sqrt{n}})) = \text{O}(\dfrac{1}{\sqrt{n}})$.

So for sufficiently large $n$ and any $ \epsilon > 0$ we have $0 \leq \sqrt{3n}\left(\exp(\dfrac{t}{\sqrt{n}})-1-\frac{t}{\sqrt{n}}\right) \leq \epsilon$
so $\Pr\left(\dfrac{S_n - \frac{n}{2}}{\sqrt{\frac{n}{12}}} \leq t\sqrt{3}\right) \leq \Pr(\sqrt{n}(\ln(2\bar{X}_n) - \ln(\theta)) \le t) \leq \Pr\left(\dfrac{S_n - \frac{n}{2}}{\sqrt{\frac{n}{12}}} \leq t\sqrt{3} + \epsilon\right).$

So letting $n \to \infty$ and using CLT we have
$$\Phi(t\sqrt{3}) \leq \lim_{n\to\infty} \Pr(\sqrt{n}(\ln(2\bar{X}_n) - \ln(\theta)\leq t)  \leq \Phi(t+\epsilon),$$ where $\Phi$ is the CDF of the standard normal distribution. Since $\epsilon > 0$ artibtrary and $\Phi$ is continuous we have $\lim_{n\to\infty} \Pr(\sqrt{n}(\ln(2\bar{X}_n) - \ln(\theta) \leq t) = \Phi(t\sqrt{3})$ and we are done.


Cantor Sets/nonempty/cardinality Let $S_0=[0,1]$ and define every $S_k$ for $k\geq 1$
\begin{align*}
S_1&=\left[0,\frac{1}{3}\right]\cup\left[\frac{2}{3}, 1\right],\\
S_2&=\left[0,\frac{1}{9}\right]\cup\left[\frac{2}{9}, \frac{3}{9}\right]\cup\left[\frac{6}{9}, \frac{7}{9}\right]\cup\left[\frac{8}{9},1\right],\\
S_3&=\left[0, \frac{1}{27}\right]\cup\left[ \frac{2}{27}, \frac{3}{27}\right]\cup\left[ \frac{6}{27}, \frac{7}{27}\right]\cup\left[ \frac{8}{27}, \frac{9}{27}\right]\cup\left[ \frac{18}{27}, \frac{19}{27}\right]\cup\left[ \frac{20}{27}, \frac{21}{27}\right]\cup\left[ \frac{24}{25}, \frac{25}{27}\right]\cup\left[ \frac{26}{27},  1\right]\\
\vdots
\end{align*}

Now let $C=\bigcap_{k=0}^\infty S_k$. This is known as a Cantor set.
I have to prove that C is non-empty, by showing that 1/4 is in C. I also have find the cardinal of C and prove my answer.

I have some ideas how to prove non-emptiness, but I am stuck as to how to specifically prove that 1/4 is in C. Also I think that the cardinal is equal to the cardinal of the reals $2^{\aleph_0}$. I probably need to find a bijection to prove this right?

One way to think of the cantor set is in terms of ternary expansions. We typically write a real number $r \in [0, 1]$ in decimal notation as
$$
r = 0.a_1a_2a_3a_4\cdots
$$
where $a_i \in \{0, \ldots, 9\}$. But we can pick other bases such as base 2 ($a_i \in \{0, 1\}$) or for our purposes, base 3 ($a_i \in \{0, 1, 2\}$).

Now if you think of numbers in terms of their ternary expansions, you should note that the Cantor set simply removes any number whose ternary expansion includes the digit 1 (that is, essentially, what throwing out the middle interval does). So to show that $1/4$ is in the Cantor set, it would suffice to show that the ternary expansion of $1/4$ does not include a 1.

Furthermore, if you think about this a little more, you should be able to adapt this to showing the uncountability of the cantor set. Remember that every number in the unit interval can be written in a binary expansion, which only involves 0 and 1...

Let's try the coding approach. We will take $0$ for left and $2$ for right. Then

$$C_0:=[0,1/3] \text{  and  } C_2:=[2/3,1]$$

and $S_1 = C_0\cup C_2$. Similarly the left and right subintervals of $C_0$ are coded $C_{00}$ and $C_{02}$ and for $C_2$ we have $C_{20}$ and $C_{22}$; for example $C_{02}:=[2/9,3/9]$. This gives us

$$S_2=C_{00}\cup C_{02}\cup C_{20}\cup C_{22}$$

The intervals in $S_3$ are specified for string of length three. In general we can use induction and shows that the intervals of $S_n$ are specified for string of $n$ elements each of which is a chain of $0$ and $2$'s. Now if we have an infinite string  of $0$´s and $2$'s, i.e., $a_1a_2a_3a_4\ldots$we have the intervals

$$C_{a_1}\supset C_{a_1a_2}\supset C_{a_1a_2a_3}\supset \ldots C_{a_1\ldots a_n}\supset \ldots$$

The intersection of which is a point in the Cantor set. So any infinite address string defines a point in $C$. Conversely any point $p$ in $C$ defines an infinite address, where its first $n$ symbols are specified by its location in $S_n$. Notice that this addressing is in fact injective, any other point $q$ in the Cantor set  has different address, because at some $n$, $p$ and $q$ are in different $C_{\alpha}$ and $C_{\beta}$ in $S_n$. Thus each point in $C$ is in a 1-1 correspondence with the collection of all address.

If each $2$ is replaced by $1$, we have a base two expansion of some $x\in [0,1]$. So as there are uncountable many points in $[0,1]$, uncountable many base two expansion and uncountable many addresses so uncountable many points in $C$.

Now you can interpret the addresses as a base three expansion of points $x\in [0,1]$ as we have shown there is a 1-1 correspondence between the Cantor set and the address and the base three expansion of $1/4_3= .02020202\ldots_3$, so is in a point in $C$.


Probability that each bucket has $\geq 3$ balls There are $30$ buckets. John throws $20$ balls, each time landing uniformly among the buckets. What is the probability that no bucket contains $\geq 3$ balls?

If the question were $\geq 2$ balls, we could calculate the probability that each bucket has at most $1$ ball, which is $\dfrac{29}{30}\dfrac{28}{30}\cdot\dfrac{11}{30}$.

You must to know the probability for at most a pair of balls in some bucket. For this you must calculate the probabilities for the different distributions with some pair.

For a pair and isolated balls the probability will be $P_2(1)=\frac{(30)_{19}}{30^{20}}\sum_{k=1}^{19} k=\frac{(30)_{19}(20)_2}{30^{20}2}$.

For two pairs and isolated balls $$P_2(2)=\frac{(30)_{18}}{30^{20}}\sum_{k=1}^{18} k\left(\sum_{j=1}^{k-1}j\right)=\frac{(30)_{18}}{30^{20}2}\sum_{k=2}^{18} k^2(k-1)$$

In general

$$P_2(h)=\frac{(30)_{20-h}}{30^{20}}\sum_{k_1=1}^{20-h}k_1\sum_{k_2=1}^{k_1-1}k_2\sum_{k_3=1}^{k_2-1}k_3\sum...;\ \#\{k_i\}=h\in[0,10]_{\Bbb N}$$

We can continue the same for any amount of pairs to the maximum of 10 and after sum all to know the probability to at most pairs, i.e. $P_2=\sum_{h=0}^{10}P_2(h)$. The problem is that this lead to very ugly equations with sums of powers (i.e. need of Falhauber formula and Bernoulli numbers).

You can try to compute this in a CAS or try a Monte-Carlo simulation.

EDIT: an alternative approach maybe evaluate the total number of possible distributions of 20 in 30 positions and after divide the numbers of ways of distribute using at most pairs by this total.

This can be done calculating the coefficient for 20 of this generating function that represent choosing between 0 and 20 balls for every position of 30 different positions

$$f(x)=(1+x+x^2+...+x^{20})^{30}=\left(\sum_{k=0}^{20}x^k\right)^{30}=\left(\frac{1-x^{21}}{1-x}\right)^{30}=\\=\sum_{k=0}^{30}(-1)^k\binom{30}{k}x^{21k}\sum_{j=0}^{\infty}\binom{29+j}{j}x^j$$

The coefficient of 20 is when $20=21k+j\to (k=0) \to j=20$ i.e. the total of different distributions of 20 balls in 30 places is $T=\binom{49}{20}$.

Now I need to know the number of different ways to distribute 20 balls on 30 positions where at most it will be 2 balls in a bucket, i.e., the permutations of three groups: groups of pairs, groups of isolated balls and group of voids buckets. They are

$$C_2=\sum_{k=0}^{10}\binom{30}{10+k,20-2k,k}$$

So from this we have that $1-\frac{C_2}{T}$ is the answer to the question.

EDIT 2: it seems that there is a mistake somewhere because if I evaluate the probability for isolated balls from one approach or the other the number is different. I mean

$$\frac{(30)_{20}}{30^{20}}\neq \frac{(30)_{20}}{(49)_{20}}$$

Hint:

You could have 20 buckets with 1 ball

or 18 buckets with 1 ball and 1 bucket with 2 balls

or 16 buckets with 1 ball and 2 buckets with 2 balls

etc.  Sum it up like in any probability question.

EDIT:
I don't see what's wrong with what I have above.  The answer I am pointing out is $$\frac{\sum\limits_{k=0}^{10}\left(\matrix{30\\k}\right)\left(\matrix{30-k\\20-2k}\right)}{\left(\matrix{30+20-1\\ 20}\right)}$$

EDIT 2:
For the isolated balls we just need to choose which of the 20 bins to put a ball in, it is simple inclusion-exclusion, so the answer is
$$\frac{30\choose 20}{49\choose 20}$$

EDIT 3:
I have just written a program to calculate these numbers and have numerically verified that my above reasoning is correct

(define (fact a)
  (if (= a 0) 1
      (* a (fact(- a 1)))))

(define (choose a b)
  (/ (fact a)
     (* (fact b) (fact (- a b)))))

(define (sum a b f)
  (if (= a b) (f a)
      (+ (f a) (sum (+ 1 a) b f))))

(define (how a b n)
  (if (= n 1) (choose a b)
              (sum 0 (floor (/ b n))
                     (lambda (c) (* (choose a c)
                                    (how
                                      (- a c)
                                      (- b (* c n))
                                      (- n 1)))))))

Here are the results that verify correctness:

> (choose 49 20)
28277527346376

> (how 30 20 20)
28277527346376

> (- (choose 49 20) (how 30 20 20))
0

as far as the original question

> (how 30 20 2)
1514286194715

> (/ (how 30 20 2) (choose 49 20))
0.0535508701367790686046560...


Expected number of pieces of a chessboard If n squares are randomly removed from a $p \ \cdot \ q$ chessboard, what will be the expected number of pieces the chessboard is divided into?

Can anybody please provide how can I approach the problem? There are numerous cases and when I go through case consideration it becomes extremely complex.

I am posting this as an answer to make it more visible. This is currently a Brilliant.org problem, in which the specific case $ 2 \times 500 $ is asked for. While it is an interesting problem meriting discussion, that should wait until after 8:00 PM EST 5/19 when the problem set is closed.

2 =         = 2 piece
3 = (3+1)/2 = 2 piece
4 =         = 2 piece
5 = (5+1)/2 = 3 piece
6           = 3 piece
7 = (7+1)/2 = 4 piece
8 =         = 4 piece

Series = (n+1)/2 pieces

Regards,
Yuvaraj


Equivalence of Definitions of lim inf of Sequence of Sets Prove : $\{w : w \in A_n \text{ for all $n$ except a finite number}\}= \bigcup_{n=1}^{\infty}\bigcap_{k=n}^{\infty}A_k$.
I am trying to prove these two definitions are equivalent but I am having trouble.

Proof so far: Let $x \in \{w : w \in A_n \text{ for all $n$ except a finite number}\}$. Then  $x$ is in all but a finite number of $A_n$. Now pick the largest $m$ such that $x$ is not in $A_m$. Then as $x \in A_j$ for all $j >m$. So $ x\in \bigcap_{j=m}^{\infty} A_j$. It follows that $x \in \bigcup_{n=1}^{\infty}\bigcap_{j=n}^{\infty}A_j$

Your proof so far is perfectly fine.  All you need to do is prove the reverse inclusion and you'll be set.

Interesting. I always thought this was the definition of $\liminf$. The proof so far is right. I would prefer to add to the last sentence 'Since this holds for some $m \ge 1$' and then use m in the indices instead of n.

The reverse inclusion can be proven as follows:

Suppose $\omega \in \bigcup_{m\ge1}\bigcap_{n\ge m} A_n$.

Then $\omega \in \bigcap_{n\ge 1} A_n$ or $\omega \in \bigcap_{n\ge 2} A_n$ or $\omega \in \bigcap_{n\ge 3} A_n$...

Case 1: $\omega \in \bigcap_{n\ge 1} A_n$

Then $\omega \in A_n \forall n$ except for a finite number of $A_n$'s i.e. 0

Case 2: $\omega \in \bigcap_{n\ge 2} A_n$

Then $\omega \in A_n \forall n$ except for a finite number of $A_n$'s i.e. 1

$$\vdots$$

Case m: $\omega \in \bigcap_{n\ge m} A_n$

Then $\omega \in A_n \forall n$ except for a finite number of $A_n$'s i.e. $m-1$

$$\vdots$$

QED


Prob. 15, Chap. 4 in Baby Rudin: Every continuous open mapping of $\mathbb{R}$ into $\mathbb{R}$ is monotonic Here is Prob. 15, Chap. 4 in the book Principles of Mathematical Analysis by Walter Rudin, 3rd edition:

  Call a mapping of $X$ into $Y$ open if $f(V)$ is an open set in $Y$ whenever $V$ is an open set in $X$.

  Prove that every continuous open mapping of $\mathbb{R}^1$ into $\mathbb{R}^1$ is monotonic.

My effort:

  Suppose $f$ is a continuous open mapping of $\mathbb{R}^1$ into $\mathbb{R}^1$. If $f$ is not monotonic, then there exist real numbers $x$, $y$, and $z$ such that
  $$x < y < z,$$
  and
  $$\mbox{ either } f(x) < f(y) \ \mbox{ and } \ f(y) >  f(z), \ \ \mbox{ or } \ \ f(x) >  f(y) \ \mbox{ and } \ f(y) < f(z). $$

  Now as $f$ is continuous on the closed interval $[x, z]$ and as $[x, z]$ is a compact subset of $\mathbb{R}$, so the image $f\left( [x, z] \right)$ is also a compact --- and hence closed and bounded --- subset of $\mathbb{R}$.

  Thus the set $f\left( [x, z] \right)$ has a maximum element $M$ and a minimum element $m$.

  First, assume that $$x < y < z, \  \ f(x) <  f(y), \ \mbox{ and } \ f(y) > f(z). $$
  Then $$ f(y) >  \max \left\{ f(x), f(z) \right\}. \ \tag{1}$$

  But in view of (1) above, we can conclude that the maximum $M$ of $f\left( [x, z] \right)$ is attained at some interior point $p$, say, of $[x, z]$.

  Then we can conclude that the image set $f\left( (x, z ) \right)$ of the open set $(x, z)$ in the domain space $\mathbb{R}^1$ has a maximum element $M$ and therefore cannot be open in the codomain space $\mathbb{R}^1$; for no $\delta > 0$ can the open interval $( M-\delta, M+\delta)$ be contained in $f\left( (x, z ) \right)$, which gives rise to a contradiction.

  So we assume that $$ x < y < z, \ \ f(x) > f(y), \ \mbox{ and } \ f(y) < f(z). $$
  Then
  $$f(y) < \min \left\{ f(x), f(z) \right\}. \ \tag{2}$$
  Then the minimum $m$ of the set $f \left( [x, z] \right)$ is attained at some interior point $q$, say, of $[x, z]$, which implies that the image under $f$ of the open set $(x, z)$ fails to be open because this image set contains $m$ but fails to contain the open interval $(m-\delta, m+\delta)$ for any real number $\delta > 0$, which is a contradiction.

  Hence every continuous open mapping $f$ of $\mathbb{R}^1$ into $\mathbb{R}^1$ is monotonic.

Is this proof correct? If so, then what about the presentation? Is the presentation lucid enough too? If not, then where does the problem lie?

Hint: Let $a<b$. Since $f(x)$ is continuous and open, then
$$ f((a,b))=(c,d) $$
for some $c<d$. Try to show $f({a,b})={c,d}$. From this, either $f(a)=c,f(b)=d$ or $f(a)=d,f(b)=c$. Namely $f(a)<f(b)$ or $f(a)>f(b)$.

I think it has a minor problem, since this problems wants you to prove that f is strictly monotomic. If f is not strictly monotonic, we can find three points x


Is the closure of a compact set compact? While looking up information on compact operators I came across these two conflicting posts.

If a set is compact then it is closed
Topology: Example of a compact set but its closure not compact

So the first link says that if a set $U$ is compact then it is closed. $U$ closed means $U = \overline{U}$ and hence $\overline{U}$ is compact. This seems to be in direct contradiction with the second post?

This is not a contradiction, because the main property is:

  A compact subspace $K$ of a Hausdorff space $X$ is closed.

Indeed, we show that for every $x \in X \setminus K$, there is an open set $U$ such that $x \in U \subset X \setminus K$. Fix such an $x$.

As $X$ is Hausdorff ($T2$), for every $y \in K$, there are disjoint open sets $U_y,V_y$ such that $x \in U_y$ and $y \in V_y$. Now you can use the compactness of $K \subset \bigcup_{y \in K} V_y$, so that
$$K \subset \bigcup_{y \in E} V_y$$
for some finite subset $E \subset K$.

Define $U = \bigcap_{y \in E} U_y$, which is an open set because $E$ is finite. You can finally show that $U$ satisfies the desired conditions. I leave it to!

Actually, the co-finite topology on R, the set of real numbers. Every subset of R is compact. However, it is not every subset of R is closed. There is a special spaces that assumed that every compact set is closed. Certainly, in Hausdorff spaces, every compact set is closed. For your the second question, Take X=R, the set of real numbers. Define topology on X by U is a subset of X is open if and only if U contains 1 or U is the empty set.Then K={1,2} is compact, but cl(K)=R which is not compact with this topology. Actually, every point x in not K, and any open subset U contain x then 1 is also in U. So, the intersection of U and K is not empty, then x is in the closure of K. Therefore, cl(K)=R. The collection U(x)={1,x} forms an open cover of X, obvious there is no finite sub-collection of it cover X.


PDF for $\xi + \eta$ Let $\xi $ and $ \eta$ be independent random  variables, $\xi$ is $U[0,1]$ , $\eta$ takes values $1/3$ and $2/3$ with probability $1/2$.
Find the PDF for $\xi + \eta$

It is easy for two continuous random variables, but what should I do in this situation?

Preliminary Lemma: if $f$ is a pdf, $$f(x)*\delta_a(x)=f(x-a).$$

Proof: Let $X$ a random variable with pdf $f$ and $Y=a$ (constant), thus with pdf $\delta_a$, ("a probability peak of value 1 concentrated in position $a$"). Then the pdf of random variable $X+Y=X+a$, which is clearly the shifted version: $x\mapsto f(x-a)$, is  besides, the pdf of a sum of random variables, thus the convolution of the associated pdfs.

The pdf of Random Variable $X$ is $\Pi_{[0,1]}$, the characteristic function of interval $[0,1]$.

The pdf of Random Variable $Y$ is $\frac12(\delta_{1/3}+\delta_{2/3})$.

The pdf of $X+Y$ is the convolution:

$$g=\Pi_{[0,1]} * \frac12(\delta_{1/3}+\delta_{2/3})$$

By distributivity property:

$$g=\frac12\Pi_{[0,1]} * \delta_{1/3}+\frac12\Pi_{[0,1]} * \delta_{2/3}$$

Using the Lemma, one can write:

$$g=\frac12\Pi_{[1/3,4/3]}+\frac12\Pi_{[2/3,5/3]}$$

which can be also written:

$$g=\frac12\Pi_{[1/3,2/3]}+\Pi_{[2/3,4/3]}+\frac12\Pi_{[4/3,5/3]}$$

The graphical representation of this pdf is:

Comment: Here is a simulation in R statistical software that may help you verify the answer when
you get it. I used $X \sim Unif(0,1)$ and $Y$ taking values $1/3$ and $2/3$
each with probability $1/2,$ and $S = X + Y.$ With a million realizations
of each random variable simulated means and SDs should be accurate to about
three places.

m = 10^6;  x = runif(m)
y = sample(c(1/3,2/3), m, repl=T)
s = x + y
cut = seq(1/3, 5/3, 1/18)
hist(s, prob=T, br=cut, col="wheat")
mean(x); mean(y);  mean(s)
## 0.4995831      # aprx E(X) = 1/2
## 0.500128       # aprx E(Y) = 1/2
$$ 0.9997111      # aprx E(S) = 1
sd(x); sd(y); sd(s)
## 0.28881
## 0.1666667
## 0.3337064
summary(s)
##    Min. 1st Qu.  Median    Mean 3rd Qu.    Max.
##  0.3333  0.7492  0.9993  0.9997  1.2500  1.6670
quantile(s, (1:6)/6)
## 16.66667% 33.33333%       50% 66.66667% 83.33333%      100%
## 0.6659555 0.8329647 0.9993092 1.1662825 1.3340041 1.6666663

You can view this as a 50:50 mixture of two uniform distributions,
one of them on the interval $(1/3, 4/3).$


Find probability that random triangle covers centre of circumscribed circle We are given the equilateral triangle A.
 On each edge of the triangle we pick a point:

randomly (probability distribution is uniform)
independently of others

We construct new triangle B from randomized points.
Task is to find the chance of B containing the centre of circle circumscribed around triangle A.

I would appreciate hints and pointers. Thanks!

Edit
Suppose $P_0$ is chosen on the bottom line of triangle A. $P_O = (x_0,0), x_0 \in [0,1]$
Let's find geometrical place of points $P_1, P_2$.
If $P_1$ is on the upper left edge, that $P_1 = (x_1,\sqrt 3 x_1), x_1 \in [0, \frac 1 2]$.
Then $P_2$ is upper right edge, and $P_2 = (x_2, \sqrt 3 (1 - x_2)), x_2 \in [\frac 1 2, 1]$.

Centre point is $C = (\frac 1 2, \frac 1 {2 \sqrt 3 })$.
Now, we shall find the constraints for points.
For fixed $x_0$, $P_1$ and $P_2$ must lay below the line between $P_0$ and $C$.
But I wonder, if there is a way to describe constraint better?
Since this most trivial way suffers from situation when line $CP_0$ and one of the edges intersects outside of range $x_m$.
Edit 2
I have listened to the useful hints about advantages of angle-based view on the problem, and believe that now I have the geometry part figured:
As earlier, We fixed $P_0$ on the bottom edge and on the step 1 we pick $P_1$ on the left edge.
Consider $\angle \alpha$ line between bottom edge and line $CP_0$. Then $\alpha \in [\frac \pi 6, \frac \pi 2]$.
Then prohibited sector for left edge has angle $\frac \pi 2 - \alpha$.
Using same logic on the step 2 we obtain prohibited sector for point $P_2$.

And now it looks like I have to get down to all that integration, so your suggestions are welcome!

Let $a_1, a_2, a_3$ be the vertices of $A$.
Choose a coordinate system such that
$$a_1 = (0,0),\quad a_2 = (1,0),\quad a_3 = \left(\frac12,\frac{\sqrt{3}}{2}\right)$$
The centroid of triangle $A$ will be located at
$$c = \frac{a_1 + a_2 + a_3}{3} = \left(\frac12, \frac{1}{2\sqrt{3}}\right)$$

Let $b_1, b_2, b_3$ be the vertices of $B$. By definition of triangle $B$, there exists
three uniform random numbers $u_1, u_2, u_3$ taking values over $[0,1]$ such that
$$b_1 = (1-u_1) a_1 + u_1 a_2,\quad b_2 = (1-u_2) a_2 + u_2 a_3,\quad b_3 = (1-u_3) a_3 + u_3 a_1$$

Consider following 4 events concerning the relative position of $c$ with respect to these 6 vertices.
$$\begin{cases}
\mathcal{E}_0 &\stackrel{def}{=} \text{Event}( c \in B = \triangle b_1 b_2 b_3 )\\
\mathcal{E}_1 &\stackrel{def}{=} \text{Event}( c \in \triangle a_1 b_1 b_3 )\\
\mathcal{E}_2 &\stackrel{def}{=} \text{Event}( c \in \triangle a_2 b_2 b_1 )\\
\mathcal{E}_3 &\stackrel{def}{=} \text{Event}( c \in \triangle a_3 b_3 b_2 )\\
\end{cases}$$
With probability one, these $4$ events are mutually disjoint. More precisely,
$$\text{Prob}( \mathcal{E}_i \,\cap\, \mathcal{E}_j ) = 0\quad\text{ for } 0 \le i < j \le 3$$
Notice $\text{Prob}( \bigcup_{i=0}^3 \mathcal{E}_i ) = 1$ and by symmetry,
$$\text{Prob}( \mathcal{E}_1 ) = \text{Prob}( \mathcal{E}_2 ) = \text{Prob}( \mathcal{E}_3 )$$
The probability we want is given by
$$\text{Prob}( c \in B ) = \text{Prob}(\mathcal{E}_0) = 1 - 3\text{Prob}(\mathcal{E}_1)$$

Let $u = u_1$ and $v = 1-u_3$, we have
$b_1 = (u,0), b_3 = \left(\frac{v}{2},\frac{v\sqrt{3}}{2}\right)$.
The line passing through $b_1, b_3$ is given by
$$\left|\begin{matrix}
x & y & 1\\
u & 0 & 1\\
\frac{v}{2} & \frac{v\sqrt{3}}{2} & 1\\
\end{matrix}\right| = 0$$
Plugging the coordinates of $c$ into LHS and treat the resulting expression as a function
of $u,v$. It is not hard to see the condition for $c \in \triangle a_1 b_1 b_3$ is simply

$$\left|\begin{matrix}
\frac12 & \frac{1}{2\sqrt{3}} & 1\\
u & 0 & 1\\
\frac{v}{2} & \frac{v\sqrt{3}}{2} & 1\\
\end{matrix}\right| \ge 0
\quad\iff\quad
3uv - u - v \ge 0
\quad\iff\quad
u \ge \frac12 \,\land\, v \ge \frac{u}{3u-1}
$$
This leads to
$$\text{Prob}(\mathcal{E}_1) = \int_{\frac12}^1 \left(1 - \frac{u}{3u-1}\right) du
= \frac{3 - \log 4}{9}$$
and hence
$$\text{Prob}( c \in B ) = 1 - 3\left(\frac{3 - \log 4}{9}\right) = \frac{\log 4}{3}
\approx 0.4620981203733$$

As a double check, I have performed a numerical simulation of this problem.
Out of $10^9$ copies of triangle $B$ generated, $462114483$ of them contains $c$.
This leads to an estimate of the probability $c \in B$ at
$0.462114 \pm 0.000016\;(\leftarrow 1 \sigma)$.
This is consistent with what we have just derived.

I may be mistaken but I dont believe there is a solution to this problem. It depends entirely on how you construct the triangle. Possibly. Have you heard of Bertrand's Paradox? Its a paradox in probability theory that asks a similar question: Take an equilateral triangle inscribed in a circle, what is the probability that a random chord of the circle is longer than the side of the triangle?  Its simple enough to understand, and easy enough to answer, but it can be shown that you get drastically distinct answers depending on how you construct the chord.

Maybe Im wrong though. Maybe this is nothing like Bertrand's Paradox.


Histogram of random numbers from normal distribution If I generate, say, 10000 numbers from the normal distribution (in Matlab) and want to draw a histogram with 10 bins, it resembles the normal distribution pretty accurately. However, if I decide to draw a histogram with 100 bins or 1000 bins, it doesn't look like a normal distribution anymore (it looks more noisy, with some peaks and valleys in different places). It would get better if I had a billion numbers.

Why does it work that way? How to explain it mathematically? My intuition is that as the number of histogram bins gets larger, the probability that the generated number belongs to such a small interval (we have large number of bins) gets smaller. But it's just intuition, not a mathematical explanation.

Your intuition is exactly correct. If you make the bins small enough then there is effectively a zero probability of seeing a sample in that exact bin. Hence for bins approaching zero width you get either a count of one or zero for your histogram, which is why it looks nothing like your continuous density function.

Mathematically we have the following set-up. Let $X_1\,\ldots X_N$ be i.i.d. random variables from some distribution, for which we have a cumulative distribution function (CDF); $\mathbb{P}( X_i\leq a) : \mathbb{R}\rightarrow[0,1]$. For some semi-closed interval $[a,b)$, let $\mathcal{X}=\{X_i : X_i \in [a,b)\}$ and further, let $M = |\mathcal{X}| \leq N$ be the number of draws which happen to fall in our interval $[a,b)$. Note that $M$ is a random variable.

We can see intuitively that
$$\lim_{N\rightarrow\infty} \frac{M}{N} \xrightarrow{p} \mathbb P(X\in[a,b)),$$
that is, if we have infinite samples the fraction falling in our interval is exactly equal to the probability mass between $a$ and $b$. Further one can show that even for finite $N$ we have
$$\mathbb{E}\left[\frac{M}{N}\right] = \mathbb{P}(X\in [a,b))$$

To show these facts we wish to know what the distribution of the random variable $M$ is. In this case because $M$ is discrete, it has a probability mass function (not a density) $\mathbb P(M=m)$.

For any one draw $X_i$, the probability of being in our interval is just $\mathbb P(X\in[a,b)):=p$, and the probability of being not in our interval is just $(1-p)$, i.e. a Bernoulli RV. Thus for $N$ draws the probability of having $m$ in our interval is given by
$$\mathbb P(M=m) = \binom{N}{m} \,p^m \,(1-p)^{N-m}$$
which should be recognizable as the Binomial distribution. The binomial distribution has mean (expectation) of $N\,p$ and thus we get the second equation from above directly;
$$\mathbb E \left[\frac{M}{N}\right] = \frac{1}{N} \mathbb E[ M ] = \frac{N\,p}{N} = p$$
This tells us that the expected value of $M/N$ is indeed the mass associated with our interval. To show the limit above holds we just need to show that as $N\rightarrow \infty$ the variance becomes zero. This also follows easily from the fact $M$ is binomially distributed;
$$\mathrm{Var}\left(\frac{M}{N}\right) = \frac{1}{N^2}\mathrm{Var}(M) = \frac{Np(1-p)}{N^2} = \frac{p(1-p)}{N}$$
and we're done.

As an example, lets consider RVs drawn from $\mathcal N(0;\sigma^2)$ and the interval $[0,\infty)$. Clearly in this case we have $p=\mathbb P(X\in[0,\infty)) = 1/2$ and thus the variance of the number of samples in our bin should go as $\frac{1}{4 N}$ with the number of draws. Indeed, this is exactly what we see if we test this hypothesis;

So, how does this relate to your exact question? Well we can see that if we increase $N$ while keeping the bins the same, the variance in our histogram bars will decrease in keeping with our intuition. As for the bin widths, it might seem like we expect the variance to increase if we decrease the width a lot but actually if $p<\frac{1}{2}$ then in fact decreasing the bin width, and with it $p$, leads to a decrease in the variance of $M$. This can be explained by the fact that for sufficiently small $p$, the most likely outcomes are $M=0$ and thus the variance becomes small again because the probability of seeing no (or few) samples in our bin dominates and thus pulls the variance down; remember that
$$\mathrm{Var}(M)\sim \sum_{m}m^2\,\mathbb P(M=m).$$
Unfortunately this very effect makes it difficult to define an 'optimum' number of bins because we can either minimize the variance by choosing the whole real line as our bin (in which case $(1-p)=0$ and our variance is zero) or by letting the number of bins go to infinity so that $\min_{\mathrm{bins}}\{p\}\rightarrow 0$.

The Normal distribution is a probability density function. You are correct in your intuition.

This  Wikipedia article explains it well - in particular the section entitled "Absolutely continuous univariate distributions"

If the $i$-th bin is $x_i \leq x\leq x_{i+1}$ then the probability, hence the proportion of points, in that bin is

$$Pr[x_i\leq X\leq x_{i+1}]=\int_{x_i}^{x_{i+1}} f_X(x)\mathop{dx}$$

As the size of the bin decreases this gets smaller and you need a larger number of draws overall to achieve a smooth histogram. By "number of draws" I mean the 10000 at the top of your question.

Non-formal argument : each bin is a sample from the parent Normal distribution and will look like the parent distribution as the size of the sample gets larger. For smaller samples (read : larger number of bins relative to number of draws or looking at bins close to the extremes of the histogram) the sample error will be relatively larger and the bin will look less like you expect and the histogram more "noisy" overall.


Question about asymptotes. Let $f(x) = 3\cdot\frac{x^4+x^3+x^2+1}{x^2+x-2}.$ Give a polynomial
$g(x)$ so that $f(x) + g(x)$ has a horizontal asymptote of $0$ as $x$
approaches positive infinity.

I've tried using that if the degree of the denominator is bigger than the degree of the numerator, the horizontal asymptote is $y = 0$, but I couldn't get anywhere. How should I solve it?

Perform the Euclidean division of the numerator by the denominator:
$$x^4+x^3+x^2+1=(x^2+3)(x^2+x-2) -3x+7. $$
We deduce
$$f(x)=3(x^2+3)+3\frac{-3x+7}{x^2+x-2}.$$
The fraction tends to $0$ as $x\to\infty$, hence we can take $\:g(x)=-3(x^2+3)$.

There are many possibilities. If the given polynomial is $f(x)$
then addition of function  $ g(x)=u(x)-f(x)$
$$ u(x)= f(x) e^{-x^2} $$

Instead of the Bell curve any such polynomial $(x \rightarrow\pm \infty$ , $y\rightarrow 0,$ and symmetric to y-axis) can also be employed.


Show that the cube roots of three distinct prime numbers cannot be three terms (not necessarily consecutive) of an arithmetic progression I'm thinking we could do a contradiction, maybe showing that one of the primes is a composite number if they are in a sequence, but I'm not sure how to finish this up.
I had this as a math problem in a number theory class, but the deadline has passed so I think it is okay ask.

Suppose not. (We take the negation of the theorem and suppose it to be true.) Suppose there are three cube roots who are in arithmetic progression although not necessarily consecutively.
Let our radicands be the primes $p, q,$ and $r$ and the difference between terms in the arithmetic progression be $d$. $k_1$ and $k_2$ respectively represent the number of common differences. WLOG $p<q<r$.
We can deduce the following statements:
$$\sqrt[3]{r}-\sqrt[3]{q}=k_1d$$
$$\sqrt[3]{q}-\sqrt[3]{p}=k_2d$$
Dividing the first and second equations gives us $k_1\sqrt[3]{p}+k_2\sqrt[3]{r}=(k_1+k_2)\sqrt[3]{q}$. Cubing this equation gives us at least one term with a cube root of $pq$ by the binomial theorem on the LHS, but there are no cube roots on the RHS. This is impossible to satisfy becaues all variables are positive and integral. Because this equation can never be true for primes $p, q$ and $r$, the supposition is false and the statement is true, and we are done.


Finding the number of possible right angled triangles If the inradius=$2013$ of a right angled triangle with integer sides. Find the no. of possible right angled triangles that can be formed using the above information.
I have tried $r(a+b+c)=ab$ and $a^2+b^2=c^2$ , but couldn't reach further.
Thanks in anticipation

One thing we can get is

$CF=CD=b-r$ and $BE=BD=c-r$. So, $BC=a=b+c-2r\quad (1)$.

By Euclide's formula we can look for a solution such that $\gcd(a,b,c)=1$ and $b=m^2-n^2$, $c=2mn$ and $a=m^2+n^2$ with, $\gcd(m,n)=1$ and not both odd.

We can then back to $(1)$ and get

$$a+2\cdot2013=b+c\\
m^2+n^2+2\cdot2013=m^2-n^2+2mn\\
n^2+2013=mn\to n(m-n)=2013$$

so, $n|2013$ and once $2013=3\cdot11\cdot61$ then we have the possibilities:

$$n\in\{3,11,61,3\cdot11,3\cdot61,11\cdot61,3\cdot11\cdot61\}$$

So you just have to try all possibilities for $n$ and find a suitable $m$.

For example, for $n=3$ you will find $m=674$ and then you have one solution which is $(a,b,c)=(674^2+3^2,674^2-3^2,2\cdot3\cdot674)$.

Now just find the others.

Answer to the question is easy: 8 primitive Pythagorean triples are there for the inradius 2013. There are no other triples (non-primitive Pythagorean triples).


We can factor out the constant term to make life easier (and just multiply by that $-48$ at the end of our calculation) as:

$$-48 \dfrac{x}{(x^2+12)^2}$$

This makes it easier to use the quotient and chain rule (I will assume you know these).

The derivative of $\ {dy \over dx} = {-48x \over (x^2+12)^2} $, using the quotient and chain rule is:

$\dfrac{d^2y}{dx^2} = -48 \dfrac{(1)(x^2+12)^2 - 2(x^2+12)(2x)x}{(x^2+12)^4} = -48 \dfrac{(x^2+12) - 2(2x)x}{(x^2+12)^3} = \dfrac{-144(4-x^2)}{(x^2+12)^3}$

Recall the Quotient Rule:

$$(\frac{f}{g})' = \frac{f'\cdot g - g' \cdot f}{g^2}$$

So you have the first derivative, $\frac{dy}{dx}= \frac{-48x}{(x^2+12)^2}$. Just apply the quotient rule:

Let $f(x) = -48x$. So $f'(x) = -48$.
Let $g(x) = (x^2 + 12)^2$. So, by the chain rule, $g'(x) = 2(x^2 + 12) \cdot 2x = 4x(x^2+12)$.

$$\frac{d^2y}{dx^2} = \frac{(-48 \cdot (x^2+12)^2) - (4x(x^2+12) \cdot -48x)}{((x^2+12)^2)^2}$$
Notice that both elements of the numerator contain $-48$. We can factor that out.
$$\frac{d^2y}{dx^2} = \frac{-48 \cdot ((x^2+12) - (4x^2(x^2+12)))}{(x^2+12)^4}$$
We also can factor out the $(x^2+12)^2$ term:
$$\frac{d^2y}{dx^2} = \frac{-48 \cdot (x^2+12) \cdot ((x^2+12) - 4x^2)}{(x^2+12)^4}$$
And now we can cancel:
$$\frac{d^2y}{dx^2} = \frac{-48 \cdot (12 - 3x^2)}{(x^2+12)^3}$$
Observe that we can factor the numerator some more:
$$\frac{d^2y}{dx^2} = \frac{-48 \cdot 3 \cdot (4 - x^2)}{(x^2+12)^3}$$
And there we are:
$$\frac{d^2y}{dx^2} = \frac{-144 \cdot (4 - x^2)}{(x^2+12)^3}$$


Gallian, D&F, or something else? I will be taking my university's Honors Abstract Algebra sequence next year, and it uses Dummit & Foote. I was looking for a book to begin studying from this summer, and a friend recommended Gallian's book on the subject. Thoughts?

D&F was used in the course Honors Basic Algebra that I took my second year. I think it is generally an excellent book, although it contains no category theory. A First Course in Abstract Algebra by Fraleigh would be an excellent companion to introduce you to category theory.

These two are best out and out and serve varying purposes gallian for easy read and first understanding. Dummit Foote for expansive knowledge and and thoroughness.You may choose pinter in pplace of gallian as well.


Example of a set whose interior is countably infinite. Our teacher asked to give an example of set whose interior is countably infinite. I can't construct it.

In the metric space $\mathbb R$ every open set is either empty or uncountable. This is because open sets are unions of open intervals, and the non-empty open intervals are all uncountable. So no subset of $\mathbb R$ is countably infinite.

However there are some other metric spaces in which countably infinite open sets exist. It suffices to take a metric space that is itself countably infinfite. An example of such a metric space is $\mathbb Z$ with the euclidean norm. Notice that in this metric space $\mathbb Z$ itself is an open set that is countably infinite (so its interior is itself and is countably infinite).

In $R$ under the usual metric it can't be since if it is an interior then it is open and if it is open then it should contain a nbd of each of it's points and nbds in$ R $are uncountable.and a countable set cant contain an uncountable subset


What are good questions that could be used to demonstrate the nature of mathematics study? Although I try to tell people about what we actually do in mathematics, sometimes I can't translate the activity very well. For example, when telling them that we study were do mathematical ideas come from, they usually understand it with an arbitrary self-explanation, for example: When someone learns the derivative, they are able to see that it comes from the concept of limit. But that wasn't really my point, for the concept of limits, there are some other concepts that ought to be learned in order to give the limits genesis and, who knows what other kinds of concepts could form a deeper construction that, perhaps, could be used to construct all mathematics?

Until the present date, I have only a handful of examples of mathematical phenomena and I guess that the most interesting question I've been able to produce is:

  Are there simple objects which could be used in arbitrary combinations to build all known mathematics? If yes, what is the proof for that?

I have some questions now:

What are good questions that could be used to demonstrate the nature of mathematics study?
Is that question I've formulated suitable for my purpose? I believe it is but I'd like to hear what other people with more competence than me could think about it.

Contrary to the gospel, I submit that mathematics is the physical science of the amazing variety and number of aspects and implications of quantity. I find it very disturbing that mathematics is transmogrified as abstract, unworldly, about logic and thinking and cognition and skill and philosophy and stylized as so-called "definitions" that do not define in the sense of inform, and "theorems" that give no clue as to their origin and relevance. The objects of mathematics are even more real than the objects of the other sciences. The fact that we can prove is due to the fact that the simplest variety of quantity is immediately discernible to us and many if not all other animals without language or training or mathematics. The objects of the other sciences are not directly observable so are not concrete and immediate. In this sense mathematics is the most concrete of the sciences. Mathematics is habitually disguised as the emperor in the invisible clothes of esoterica. For example, contrary to your (and everyone else's) mantra, calculus is not about limits. The limit is merely a device that enables us to obtain the derivative or integral of a particular function. It is not at all what calculus is about. Differential calculus is about a very interesting and useful point property of functions called the instantaneous rate of change that is characterized as slope or grade. Integral calculus is about the very curious antiderivative of a function which is a point property that very simply gives an interval property, a remarkable thing indeed. I'm sure that if we divest mathematics of its esoteric and procedural draperies we could grasp ten times as much in one tenth of the time and appreciate it accordingly.


Number of roots of special polynomials in interval $(0,1)$ I proved that some special class of polynomials has only one root in interval (0,1)

For example:
$x^{20}+x^{10}+3x^5-30x^{4}+2x+3=0$

Or

$2x^7-10x^5+2x^3+3x+2=0$

Or

$8x^6-2x^5-2x^4-2x^2-1=0$

My question: Is this an interesting result? How can I check for which class of polynomials we know such property?

Note that any polynomial (with real coefficients) which has a (real) root can be scaled and translated so that it has just one root in any given interval [may be a multiple root if the only roots of the original polynomial are multiple roots]. So there has to be some additional feature to make the observation interesting mathematically.

It would be interesting to know what is special about the class you have identified. [It won't form a linear space - the sum of two such polynomials need not have the same property]

Maybe if you say what special kind of polynomials could be an interesting topic ... but my first impression is that the topic looks promising


Calculate $\frac{3\sin\alpha+2\tan\alpha}{\cot\alpha}$ if $\cos\alpha=\frac13$ Calculate $\dfrac{3\sin\alpha+2\tan\alpha}{\cot\alpha}$ if $\cos\alpha=\dfrac13$ where $\alpha$ is an acute angle.

So I am trying to simplify the given expressions but I am stuck on $\dfrac{\sin^2\alpha(3\cos\alpha+2)}{\cos^2\alpha}.$ Can you give me a hint?
If we use that $\sin^2\alpha+\cos^2\alpha=1$ we can find $\sin\alpha,\cos\alpha,\tan\alpha,\cot\alpha$ but this is not the most rational approach, I think.

$$\dfrac{3\sin\alpha+2\tan\alpha}{\cot\alpha} = \dfrac{3\sin^2\alpha \cos \alpha+2\sin^2\alpha}{\cos^2\alpha} = \dfrac{3(1- \cos^2 \alpha) \cos \alpha+2(1- \cos^2 \alpha)}{\cos^2\alpha}$$
You should be able to conclude.

Hint : $\dfrac{3\sin\alpha+2\tan\alpha}{\cot\alpha}=\dfrac{3\sin\alpha}{\cot\alpha}+\dfrac{2\tan\alpha}{\cot\alpha}$
But $\dfrac{\sin\alpha}{\cot\alpha}=\frac{1-\cos(\alpha)^2}{\cos(\alpha)}$ and $\dfrac{\tan\alpha}{\cot\alpha}=\frac{1-\cos(\alpha)^2}{\cos(\alpha)^2}$ (why ?)


Prove: ${n\choose1} +2{n\choose2} +3{n\choose3} +\cdots+n{n\choose n} = n\cdot2^{n-1}$ Prove that

$$
{n\choose1} +2{n\choose2} +3{n\choose3} +\cdots+n{n\choose n} = n\cdot2^{n-1}
$$

Now I can prove this using,

$$
{n\choose0}+{n\choose1}x+{n\choose2}x^2 + {n\choose 3}x^3 + \cdots+{n\choose n}x^n = (1+x)^{n}
$$

Differentiating both sides,

$$
{n\choose1} +2{n\choose2}x +3{n\choose3}x^2 +\cdots+n{n\choose n}x^{n-1} = n\cdot(1+x)^{n-1}
$$

Substituting $x=1$, the given identity is proved.

My exercise specifically asks for a combinatorial proof though, by double counting.  Any help?

Consider the number of ways to choose a team (with $k\ge 1$ members) and choosing the team's captain. The left hand side counts this based on $k$ and adds the values up (for each $k$, we first choose the team members and then choose one of them as captain). Alternatively, we can first pick one person as the captain ($n$ choices) and then choose the rest of the team (zero or more members out of the remaining $n-1$ people).


Linearize differential equation Given the following differential equation:
$$
\dot{y}_1 + p(t)f(y_1) = g(t)
$$
then I want to linearize it in the sense that it becomes a system of equations on the form:
$$
\dot{y}_1 + p_1(t)y_1(t) = g_1(t)
$$
$$
\dot{y}_2 + p_2(t)y_2(t) = g_2(t)
$$
$$ \vdots $$
$$
\dot{y}_n + p_n(t)y_n(t) = g_n(t)
$$
where I allow for any finite number of additional linear differential equations, also the functions $g_i(t)$ and $p_i(t)$ may depend on any $y_j$ where $i\neq j$

Is this possible? preferably exact (of course) or at least as an approximation where it converges towards the exact case for $n\to \infty$ Any ideas or references are highly appreciated.

No, that is in general not possible. It would be a miracle making decades of research into numerical methods worthless.

What you can do is linearize along a solution, so that a close solution $\bar y=y_0+εy_1$ satisfies in the first order the linearized system
\begin{align}
\dot y_0+f(y_0)&=g\\
\dot y_1+f'(y_0)·y_1&=0\\
\end{align}
One might expand to higher order terms by including higher orders in the perturbation parameter $ε$ in $\bar y=y_0+εy_1+ε^2y_2+…$ which then includes non-linear terms in the lower order coefficients.

Warning: this answer may contain traces of engineering language and therefore lack of rigor.

One way to do this would be to split the functions into representations which are valid at different intervals where we assume the function to be nice in each local neighbourhood. Then we require the function values at the boundaries between these sets to be smooth. We could for instance be using polynomials windowed with a function from the exponential family.

Easiest example would probably be a gaussian which windows a set of polynomials. They are easy to differentiate so we can easily get linear constraints on the polynomial coefficients.


Find the function which satisfies the given conditions Let $f :R \to R$  is a function satisfying $f(10-x) = f(x)$ and $f(2-x) = f(2+x)$

for all $x$ belonging to $R$.If $f(0) = 101$, then find the minimum possible no.

of values of $x$ satisfying $f(x) = 101$ for $x$ belonging to $[0,30]$.

As $f(2-x) = f(2+x)$, hence $f(x)$ is symmetric about the line $x=2$ but I am not able to understand how to use other information. Could someone give me slight hint?

Hint (as requested): What can you say about  $f(5)$, using the first condition?

What about $f(4)$? What's the second condition tell you about $f(0)$? What's the first condition tell you about $f(10)$? Do you notice a pattern here?

Your question title is a bad one, because you cannot find "the function" satisfying these conditions -- there may be many many such functions.

By condition $f(2-x)=f(2+x)$ we can say $f$ is periodic with period $4$ and with condition $f(10-x)=f(x)$ means $f$ is periodic with period $10$ so as $f$ is the function satisfying both the properties so its period should divide $4$ and $10$ so period of $f$ will be either $1$, $2$ or any number. But as we want the minimum so period should be the maximum of all these so the period should be $2$ which implies it has $16$ values of in $[0,30]$ satisfying given conditions.


The Radical conjugate Roots Theorem If the coefficients of $$p(x)=a_nx^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0$$ are rational, the Conjugate Radical Roots theorem states that if the equation $p(x)=0$ has a root of the form $x=s+t\sqrt{u}$ where $\sqrt{u}$ is rational, then the equation must also have the conjugate radical, $x=s-t\sqrt{u}$, as a root.

How to prove that statement? Can anyone show it please?

Suppose that $s+t\sqrt u$ is a root of $p(x)$ and that $\sqrt u\notin\mathbb Q$. Then $s+t\sqrt u$ is also a root of the polynomial$$q(x)=x^2-2sx+s^2-t^2u,$$ since$$q(x)=\bigl(x-s-t\sqrt u)\bigl(x-s+t\sqrt u\bigr).$$By polynomial long division, you can express $p(x)$ as $q(x)q^\star(x)+r(x)$, where $q^\star(x),r(x)\in\mathbb{Q}[x]$ and the degree of $r(x)$ is smaller that $2$ (which is the degree of $q(x)$). But, since $s+t\sqrt u$ is a root of both $p(x)$ and $q(x)$, it must be a root of $r(x)$ too, and the only polynomial in $\mathbb{Q}[x]$ with degree smaller than $2$ with an irrational root is the null polynomial. Therefore, $q(x)\mid p(x)$, and, since $s-t\sqrt u$ is root of $q(x)$, this implies that $s-t\sqrt u$ is also a root of $p(x)$.

Will represent a polynomial with rational coefficients:

$p_n(x)=\sum_{i=0}^{n} {a_i x^i},{a_i}\epsilon\mathbb{Q},a_n \ne{0},i\epsilon\overline{1,n},n\epsilon\mathbb{N},n\ge{2}$

that has an irrational root:
$p_n(a+\sqrt{b})=0, a,b\epsilon\mathbb{Q},b\gt0$

This polynomial solution becomes:
$p_n(a+\sqrt{b})=a_0+a_1(a+\sqrt{b})^1+...+a_n(a+\sqrt{b})^n =0$

Analyzing the irrational factor in the k-term:

$(a+\sqrt{b})^k=s_k+q_k\sqrt{b}, s_k, q_k\epsilon\mathbb{Q}$

then a conjugated form could be deducted for the conjugate:

$(a-\sqrt{b})^k=s_k-q_k\sqrt{b}$

This result could be applied to the solution:

$p_n(a+\sqrt{b})=S_n+Q_n\sqrt{b}=0 {\Rightarrow}S_n=Q_n=0$

However,

$\left.\begin{matrix}p_n(a-\sqrt{b})=S_n-Q_n\sqrt{b}\\S_n=Q_n=0\end{matrix}\right\}{\Rightarrow}p_n(a-\sqrt{b})=0(\therefore)$


Normal subgroups in Permutation groups I am given the question like this:

Let $N = \{a \in S_4 : a(4) = 4\}$. Explain why $N$ is a subgroup of $S_4$ that is isomorphic to $S_3$. Is $N$ normal in $S_4$? Generalize.

My answer:

Since we already fix $a(4) = 4$, so $N$ is basically isomorphic to group of permutations of 3 symbols which is $S_3$.

For the second question, giving counter example: $(14)(12)(41) = (1)(24) \not \in N$ where $(14)$ and $(14)^{-1} = (41) \in S_4$ and $(12) \in N$, therefore, $N$ is not normal in $S_4$.

Can someone please verify if this answer is correct? Thanks a lot.

You also have that $N$ cannot be normal in $S_4$ because a non-trivial normal subgroup of a primitive group is transitive.


Prove that first exit time is a stopping time I have seen several questions on the site similar to the one I am about the ask, however none of the previous answers was exactly considering the kind of problem I am having. Therefore, I would appreciate if someone explained the following puzzle to me. I am following the proof given in Klebaner, Fima: Introduction to Stochastic Calculus with Applications, 2nd edition, pp. 52.

Let $S(t)$ be a continuous stochastic process adapted to the filtered probability space ($\Omega$, $F$,  $\mathcal{F}$, $P$), where $\Omega$ is the sample space, $F$ is a $\sigma$-field of subsets of $\Omega$, $P$ is a probability measure defined on the elements of $F$ and $\mathcal{F}$ is a filtration of increasing $\sigma$-fields, such that $F_t$ $\subset$ $F$.

We want to prove that for any interval $D = (a,b)$ on the real line a first exit time (defined as $\tau_d =$ inf $\{t \geq 0: S(t) \notin D \}$) is also a stopping time.

The proof: We know that the following sets are equal: $\{\tau_d > t\} = \{S(u) \in D, \text{ for all } u \leq t \} = \cap_{0 \leq u \leq t} \{S(u) \in D\}$. Although we know that each of the elements in the last set belong to $F_t $, we are not able to use this knowledge, because by definition $\sigma$-algebra is closed only under countable intersections. Therefore we have to transform this set to a countable set. And now comes the argument where I am stuck. The author further argues: "Due to the continuity of $S(u)$ and $D$ being open, for any irrational $u$ with $S(u) \in D$ there is a rational $q$ with $S(q) \in D$. Therefore:

$\bigcap\limits_{0 \leq u \leq t} \{S(u) \in D\}  = \bigcap\limits_{0 \leq q \leq t} \{S(q) \in D\}.$"

I understand that the last term is a countable intersection of events in $F_t$ and therefore the event $\{\tau_d > t\}$ belongs to $F_t$ and is therefore a stopping time.

Therefore my question is:

Why is the set $\cap_{0 \leq u \leq t} \{S(u) \in D\}$ substitutable by $\bigcap\limits_{0 \leq u \leq t} \{S(u) \in D\}$ encompassing only irrational numbers and further why is this set $ \bigcap\limits_{0 \leq q \leq t} \{S(q) \in D\}$ encompassing only rational numbers identical to the previous one?

Every irrational number can be approximated by a rational number.

Source: https://www.google.co.in/url?sa=t&source=web&rct=j&url=https://people.eecs.berkeley.edu/~wkahan/Math55/p249n17.pdf&ved=0ahUKEwjc-OTgyYjSAhWBLY8KHe2jDQgQFggfMAE&usg=AFQjCNHUzr1RgddnCUCEsvGaSJ5Y1XT3sA

Now because S is continuous and since D is open (as the book said), the 2 intersection sets are equal (using the above facts, prove this by showing each is a subset of the other.)


How do I find the equation of the normal to $y = x^2 - 3x +2$ The question I got first asked me to solve the simultaneous equation:
$y = x^2 - 3x +2$ and
$y = 3x - 7$

then hence or otherwise find the equation of the normal of $y = x^2 - 3x +2$ at one point $(3,2)$. Giving the answer in the form of $ax + by + c =0$

How do I solve this?

When solving the system of equations, you'll obtain a quadratic with exactly one real solution at $x = 3$. This implies that the given line is actually tangent to the given parabola.

Coincidentally, the second part of the question asks for the normal line at $x = 3$. Since perpendicular lines have slopes that multiply to be $-1$, and the slope of the tangent line is $3$, it follows that the slope of the normal line is its negative reciprocal, $-1/3$. Using the point-slope formula, we have:
$$
y - 2 = \frac{-1}{3}(x - 3)
$$
It remains to convert this to the form $ax + by + c = 0$ by clearing the fraction and bringing all terms to one side.

I assume that ''2Y'' part is a typo? Solving that equation is as follows:

x^2-6x+9=0
(x-3)^2 = 0
x=3

For the hence/otherwise part

dy/dx = 2x - 3
Substitute in the x co-ordinate to get the gradient of it at the specified point. (which is x = 3)
M(Tangent) = 3, therefore, M(Normal) = -1/3
y= -1/3x + c, sub in point
c= 3
Therefore -> y= -1/3x + 3 (Multiply both sides by 3)
3y= -x + 9
3y + x - 9 = 0 (as specified in the question)

Therefore, equation to the normal is 3y + x - 9 = 0


Can a bipartite graph have many Hamiltonian paths but no Hamiltonian cycle? Can a bipartite graph with at least three vertices have the following properties simultaneously:

Every vertex is the initial vertex of some Hamiltonian path.
The graph contains no Hamiltonian cycle.

I found out that this is only possible if the number of vertices is even and if the partitions have equal size. My conjecture is that there is no bipartite graph with the desired properties, but I cannot prove it.

Yes! The Georges graph has this property. This graph is the smallest known counterexample to the Tutte conjecture that any 3-connected bipartite 3-regular graph is Hamiltonian. It has 50 vertices and 75 edges.

I verified that every vertex is the initial vertex of a Hamiltonian path with some Sage code:

def homtrac_orbits(g):
    orbits = g.automorphism_group(return_group=False, orbits=True)
    return all(g.hamiltonian_path(o[0]) is not None for o in orbits)

def report(G):
    print G.name() + ':', homtrac_orbits(G), G.is_hamiltonian(), G.is_bipartite()

report(Graph(">>graph6<<qO`a`_WO?O?_?_?OO???K?@O?B_?@??__?_I?A?_?_????@?A?@???\
O????g???`????C????????A?????O????C??????????K?????K?????S?????K?????\
B????OO?????__??????C??????C??????A?????C????????D_??????I???????[\
??????AG??????GA_", name="Georges"))

You can see it on SageCell but it takes some time to run. It prints "Georges: True False True" when complete.

The Ellingham-Horton graphs on 54 and 78 vertices are also homogeneously traceable but non-Hamiltonian. The 54-graph can be checked with

report(graphs.EllinghamHorton54Graph())

(included in the above SageCell). I also checked the 78-graph but this is too slow to include in the SageCell.

It seems reasonable to guess that the other known counterexamples to Tutte's conjecture—namely the Owens graph, the Horton 92-graph, and the Horton graph—are also homogeneously traceable, but checking in Sage is too slow for those to verify that way.

a bipartite graph is acyclic, all vertices of two sets must have at least one connection, graph is two colorable, has distinct edge


Pullback of sheaves. Let $ f : X \to Y$ be a continuous map of topological spaces.. Let $ \mathcal{F} = \underline{\mathbb{R}} $ be a sheaf of locally constant function from Y to $\mathbb{R}$ on Y. Let $ f^* \mathcal{F}$ be the pullback of the sheaf on X. Is the pullback of $ \mathcal{F}$, the sheaf of locally constant functions on X?

The answer is yes. This is one of the cases where it is better to think as sheaves as étalé spaces.

The constant sheaf $\mathcal F=\mathbb R_Y$on $Y$ with stalk $\mathbb R$ corresponds to the étalé space $Y\times \mathbb R_{disc}$, where $\mathbb R_{disc}$ denotes $\mathbb R$ endowed with the discrete topology.
The étalé space corresponding to $f^*(\mathcal F)$ is the fibre product $X\times_Y (Y\times \mathbb R_{disc})$ and this fibre product is homeomorphic to $X\times \mathbb R_{disc}$.
Now, the étalé space $X\times \mathbb R_{disc}$ on $X$ corresponds to the constant sheaf $\mathbb R_X$ and this  shows what you wanted to know: $$  f^*(\mathcal F) =f^*(\mathbb R_Y) =     \mathbb R_X      $$

No, suppose that $Y$ is a point. The pullback $f^*{\cal F}$ is a constant sheaf, and the sheaf of locally functions defined on $X$ is not always the constant sheaf.


Determinant of complex matrix with almost constant lines Let $0\neq c\in\mathbb{C}$. Take the matrix
$$A_C=\begin{pmatrix}
n&c&\dots&c&c \\
c&n&c &\dots & c\\
c &c & n &c &\dots\\
\vdots &\vdots&\vdots&\ddots & \vdots \\
c & \dots &\dots &c& n
\end{pmatrix}\in \mathbb{C}^{n \times n}.$$

I want to show that $A_C$ invertible. It seems like one can compute the determinant explicitly but I messed with it a bit.

How can I prove $\det A_C\neq 0$?

EDIT: We assume $c\neq n$.

This is a circulant matrix and as such has normalized eigenvectors
$$v_j=\frac{1}{\sqrt{n}}(1,\omega_j,\omega_j^2,\ldots,\omega_j^{n-1})$$
where
$$\omega_j=exp\left(\frac{2\pi i j}{n}\right)$$
The eigenvalues are
$$\lambda_j=n+c\omega_j+c\omega_j^2+\cdots+c\omega_j^{n-1}$$
Taking the product of the eigen values gives the determinant. Now since the $\omega_j$ are roots of unity we have
$$\sum_{k=1}^{n-1}\omega_j^k=-1$$
which gives
$\lambda_j=n-c$. Since we have assumed that $n\neq c$ no eigenvalue is equal to zero. Hence the determinant is non-zero.

See if its so take it in eigenvalues. We know that sum of eigenvalues is the trace of matrix  as $n \neq  c $ is not equal to $0$ so we can easily conclude that $det(A) \neq 0$ you can also do that as $det(A)=π \lambda_i$ and as $n\neq 0$ so the determinant is also non-zero and its done.


Decomposition of a single 4D rotation I have a $4\times 4$ matrix $M$ which represents a general 4-dimensional rotation.
$$
M = \pmatrix{a_{11} &a_{12} &a_{13} &a_{14}\\a_{21} &a_{22} &a_{23} &a_{24}\\a_{31} &a_{32} &a_{33} &a_{34}\\a_{41} &a_{42} &a_{43} &a_{44}}
$$
There are also six 4D basis rotation matrices $M_{XY}$, $M_{YZ}$, $M_{ZX}$, $M_{XW}$, $M_{YW}$, $M_{ZW}$.

How to decompose 4D rotation $M$ into basis rotations (i.e., to find corresponding angles $T_{XY}$, $T_{YZ}$, $T_{ZX}$, $T_{XW}$, $T_{YW}$, $T_{ZW}$)?

Wikipedia says that any 4D rotation given by its matrix can be decomposed into two isoclinic rotations. However, I do not know if it is possible to decompose them further into basis rotations. Also, there may be another (simpler) way.

You should be able to build your basis matrices by doing something similar to the Gram-Schmidt process: by multiplying by suitably-chosen rotation matrices you can set each of the off-diagonal coefficients in turn equal to 0.  For instance, consider applying a $M_{ZW}$ rotation by some angle $\theta$ to the matrix:
$$M' = M_{ZW}(\theta)\ M = \pmatrix{1&0&0&0\\0&1&0&0\\0&0&\cos\theta&-\sin\theta\\0&0&\sin\theta&\cos\theta}\pmatrix{a_{11}&a_{12}&a_{13}&a_{14}\\a_{21}&a_{22}&a_{23}&a_{24}\\a_{31}&a_{32}&a_{33}&a_{34}\\a_{41}&a_{42}&a_{43}&a_{44}}
=\pmatrix{a_{11}&a_{12}&a_{13}&a_{14}\\a_{21}&a_{22}&a_{23}&a_{24}\\*&*&a_{33}\cos\theta-a_{43}\sin\theta&a_{34}\cos\theta-a_{44}\sin\theta\\*&*&a_{33}\sin\theta+a_{43}\cos\theta&a_{34}\sin\theta+a_{44}\cos\theta}$$
(Here the *s are 'don't-cares' - they have a value, but it's essentially unimportant.)  Now, by choosing $\theta$ correctly, you can make $M'_{34} = a_{34}\cos\theta-a_{44}\sin\theta$ equal to zero without affecting any other upper-triangular elements.  Once you've got that matrix, you can continue in a similar fashion by finding a suitable $M_{YW}(\phi)$ to zero the $M_{24}$ element, then an $M_{XW}$ to zero the $M_{14}$ element, etc.; repeating this procedure for each combination of unequal coordinates (in some canonically-defined order) will eventually lead you to an orthogonal (by hypothesis) lower-triangular matrix, which must therefore be the identity.

You can think of this process in terms of applying successive coordinate-plane rotations, each one on the right of what has gone before, so that ultimately the resulting rotation is the identity.

Then multiplying the identity matrix by the inverses of these coordinate-plane rotations, each one also on the right of what has gone before, you will end up with an equation showing how to express the original rotation as a product of coordinate-plane rotations.

This actually works in any dimension. Since the dimension of SO(n) is n-choose-2 = n(n-1)/2, that should be the maximum number of coordinate-plane rotations needed.

In a given dimension, selecting a fixed ordering for performing this procedure defines a continuous map from the n(n-1)/2 dimensional torus (since each angle may be thought of as a point on circle) to SO(n+1):

Tn(n-1)/2  —>  SO(n)

that is in fact surjective (since each rotation can be expressed as a product of coordinate-plane rotations).  The circle factors of Tn(n-1)/2 are indexed by pairs $(i, j)$ with $1 \le i \lt j \le n$ (since each one represents one coordinate plane).

The 2-plane rotations used in this procedure are sometimes called Householder rotations.


Need help with simple proof by mathematical induction $$ \sum_{k=1}^n k^4 = {(6n^5+15n^4+10n^3-n) \over 30} $$

How can this be proven using mathematical induction? My teacher isn't any help, he just tells me to think about it, but I've read the textbook again and again, and there isn't much in it that would help me prove this statement.

*Sorry to ask multiple questions on the site in such a short time, but I'm just quite desperate for help and I don't have anyone to go to for it. My teacher barely speaks English, and with my poor hearing I can hardly follow the class and I need to pass this exam desperately.

I think your teacher is right, but anyway...

Basic step ($n_0=1$):

Left side:

$\sum_{k=1}^1 k^4 = 1$

Right side:

${(6+15+10-1) \over 30} = 1$

Left and right side are equal. Done.

Induction step:

Assuming it works for $n$, try to show it works for $n+1$ aswell.

So we have:

$\sum_{k=1}^{n+1} k^4 = {(6(n+1)^5+15(n+1)^4+10(n+1)^3-(n+1)) \over 30}$

which is the same as:

$(n+1)^4 + \sum_{k=1}^{n} k^4 = {(6(n+1)^5+15(n+1)^4+10(n+1)^3-(n+1)) \over 30}$

Now take this term (the last of the sum) to the other side of the equation:
$\sum_{k=1}^{n} k^4 = {(6(n+1)^5+15(n+1)^4+10(n+1)^3-(n+1)) \over 30} - (n+1)^4$

And after some basic algebra you get back to:

$\sum_{k=1}^n k^4 = {(6n^5+15n^4+10n^3-n) \over 30}$

which is precisely what we want.

Now since we have shown that it works for 1 in the first step, we can argue that it works for 2, and since it works for 2 it works for 3 - and so on :)

General porcedure: Denote the left hand side as $f(n)$, the right hand side as $g(n)$. Show that $f(0)=g(0)$. Then show that for each $n$ you have $f(n)-f(n-1)=g(n)-g(n-1)$. Note that $f(n)-f(n-1)=n^4$.


Proving that Continuous Open Functions are Strictly Monotonic It is a fact from analysis that a continuous and open real-valued function of a real variable is strictly
monotonic. The proof I know runs something like this: Suppose $f$ is an open and continuous map but is not
strictly monotonic. Consequently, there exist three numbers $a < c < b$ such that either
$$
f(a) \geq f(c) \leq f(b) \;\;\;\; (1)
$$
or
$$
f(a) \leq f(c) \geq f(b) \;\;\;\; (2)
$$
If $(1)$ holds then the exteme value theorem guarantees that $f$ attains its infimum on $[a,b]$; but by assumption,
the infimum is at least as small as $f(c)$ so in fact $f$ attains its infimum on $(a,b)$. Also by assumption,
$f$ carries open intervals to open intervals. With this though we have a contradiction since an open
interval cannot contain it's own infimum. A similar argument yields considering suprema yields an
analagous contradiction. Therefore, $f$ is strictly monotonic.

My question is, Is there a more constructive way to prove this that doesn't involve contradiction? Although I think
the proof given is nice, I don't think I could have come up with it own my own because the consequences of $f$ not being strictly monotonic as exhibited in (1) and (2)
would not have occurred to me. So, it would be good to see a direct way of proving this.

f is open map and continuous, let f:R->R
Let the inverse of f is F.so F is continuous from f(R) to R.
Now lets assume there exists distinct x,y in R such that f(x)=f(y).
WLG lets assume f(x)=f(y)=c for some c in R.
F(x) is not continuous at c as x,y are distinct and hence the contradiction.


Notation of the summation of a set of numbers Given a set of numbers $S=\{x_1,\dotsc,x_{|S|}\}$, where $|S|$ is the size of the set, what would be the appropriate notation for the sum of this set of numbers? Is it

$$\sum_{x_i \in S} x_i
\qquad\text{or}\qquad
\sum_{i=1}^{|S|} x_i$$ or something else?

I'd write either $\displaystyle\quad\sum_{i=1}^{|S|} x_i\quad$ or $\displaystyle\quad\sum_{x\in S} x$.

If the second form is used, then the subscript is just clutter.

Some mathematicians (perhaps especially set theorists?) might write $\displaystyle \sum S$.  This would parallel the way set theorists write $\displaystyle \bigcup S$ where others might write $\displaystyle \bigcup_{x\in S} x$.

Say I had a set A, under an operation with the properties of $+$, then $$\sum_{i\in A} x_i$$ is how I write it.


Algebraic proof that a set generated by irrational rotations is dense in $S^1$. This is exercise 1.9 in Lie Groups, Lie Algebras and Representations - Hall.

  Suppose $a$ is an irrational real number.  Show that the set $E_a$ of the numbers of the form $e^{2\pi i n a}$, $n \in \mathbb{Z}$, is dense in the unit circle $S_1$.  Hint: Show that if we divided $S^1$ into $N$ equally sized "bins" of length $2\pi/N$, there is at least one bin that contains infinitely many elements of $E_a$.  Then use the fact that $E_a$ is a subgroup of $S^1$.

My proof of this proposition is as follows.  Since $a$ is irrational, you can determine that the set of rotations $E_a$ is infinite.  Since $S^1$ is compact we can find two $r_1, r_2$ that are within $\epsilon$ of each other.  Then $r_1^{-1}r_2$ is a small rotation of size $\epsilon$.  Now, $r_1^{-1}r_2$ generates rotations that are within $\epsilon$ distance of any point of $S^1$.

That said, I don't believe that the hint suggested in the problem uses that technique.  My knowledge of algebra is not all that strong so I was hoping someone could shed some light on what is being suggested there.

The hint is similar to your proof without directly using compactness. Instead it suggests you use the pigeonhole principle to show that we can choose arbitrarily small arcs that contain at least two elements of the group (in fact infinitely many), then proceed as in your argument.

I think this should work: by Weirstrass' Theorem , every bounded infinite subset of the plane contains a limit point, say $p$. Then there is a sequence {$ka : k \in \mathbb N$}in $ E_a \cap S^1$  that converges to $p$. Now, for any other $q \neq p$, apply translation to construct a sequence that converges to $q$. I mean, scale the sequence in $E_a$ that converges to $p$ by a rotation by $\alpha -\theta$ , where $p=e^{i\theta} , q=e^{i \alpha} $ Then, if $e^{2\pi ka}\rightarrow p$ ,with some changes, $e^{i2\pi ka(\theta -\alpha)} \rightarrow q$. Now, obviously multiplying by $\theta- \alpha$ may push you out of $E_a$, but you can approximate the difference by a rational number.

EDIT: More clearly,  as you pointed out, you can use this (Cauchy) sequence that converges to $p$ to construct an element $e_a$ in $E_a$ that gives you a rotation by any amount. Repeating this , i.e., the sequence {$ne_a$} gives you the dense subset of $S^1$. (you can construct a choice of $e_a$ for any $\epsilon$ you choose.)


Show that $\frac{-1+\sqrt{3}i}{2}$ is a cube root of 1? I know how to directly prove that complex number is one of the cube roots of 1. But the textbook solution also gives another solution as shown below,

However, I don't understand how it jumps to the conclusion that $\frac{-1+\sqrt{3}i}{2}$ is a root of the quadratic equation following the two equations at the top. Thank you!

Well maybe it is using Complex conjugate root theorem & Vieta's formula to say that $\alpha+ \bar{\alpha}=-1$ and $\alpha \bar{\alpha}=1$. So $\alpha$ and $\bar \alpha$ are roots of $x^2-(-1)x+1$ or $x^2+x+1$. Hence $\alpha$ divides this which divides $x^3-1$ hence $\alpha$ is a root of $x^3-1$.

Note:
$\alpha = \frac{-1+\sqrt3i}2,\;\bar\alpha=\frac{-1-\sqrt3i}2$


Cohomology of finite groups with finite coefficients I'm wondering if the group cohomology of a finite group $G$ can be made nontrivial with a nice choice of a finite $G$-module M. In other words, given a finite group $G$ and a number $n$, does there exist a finite $G$-module $M$ such that $H^n(G,M)$ is non-zero?

I would also be interested in the special case that $G$ is a finite $p$-group and n = 2. Can I always get $H^2(G,M) \ne 0$ for some finite $M$?

Thanks for your help.

Yes, for each $n\ge 0$ there is a $G$-module $M$ (depending on $n$) such that $H^n(G,M)\neq 0$ (provided $G\neq 1$ finite). .

Such an $M$ can be constructed by induction:

First note that $H^i(G,F)=0$ for each free $\mathbb{Z}G$-module $F$ and all $i>0$ by Shapiro's lemma and Brown, VIII, 5.2.
Next, show   $H^1(G,I_G)=\mathbb{Z}/|G|$ where $I_G \trianglelefteq \mathbb{Z}G$ is   the augmentation ideal (and $G$-action is given by multiplication in $\mathbb{Z}G$).
Let $n\ge 2$ and suppose $N$ is a $G$-module such that $H^{n-1}(G,N) \neq 0$. Choose a short exact sequence $0 \to M \to F \to N\to 0$ of $G$-modules with $F$ free. Then, by the long exact sequence in cohomology and 1.  we obtain the exact sequence
$$0=H^{n-1}(G,F) \to H^{n-1}(G,N) \to H^n(G,M) \to H^n(G,F)=0,$$
i.e. $H^n(G,M)\cong H^{n-1}(G,N)\neq 0$.

Note: By starting with $I_G$ you can even arrange $H^n(G,M)=\mathbb{Z}/|G|$.

It is well known that, If $G$ and $M$ are finite $p$-groups, $H^2(G,M)=0$ is equivalent to $H^i(G,M)=0$ for all $i \geq 0$.

If one take any $M$ on which $G$ acts trivially, then $H^1(G,M) \cong Hom(G,M)$, which is not trivial unless $M$ or $G$ are trivial.


Quasicomponents and components in compact Hausdorff space Let $X$ be a compact Hausdorff space, $x,y\in X$ and $\mathcal{A}$ a colection of closed subspaces of $X$ such that for every $A\in \mathcal{A}$ then $x$ and $y$ are in the same quasicomponent of $A$. If $\mathcal{B}$ is a subcolection of $\mathcal{A}$, then $x$ and $y$ are in the same quasicomponent of  $$D=\bigcap_{B\in\mathcal{B}} B $$

($x$ and $y$ are in the same quasicomponent of $A$ if and only if there's no two disjoint open sets $U,V$ such that $A=U\cup V$, $x\in U$ and $y\in V$)

This is the first part of a question that wants me to show that if $X$ is a compact Hausdorff space then $x,y$ are in the same quasicomponent if and only if they are in the same component. The second part is to show that $\mathcal{A}$ has a minimal element which I have done, and the last part is to show that the minimal element is connected which I haven't done.

What I've tried so far is assuming theres a separation $U',V'$ for $x$ and $y$ in $D$. Then tried to extend that separation to any $B$ because $U'=B\cap U$ where $U$ is open in $B$. I fail to prove that $U$ is also closed in $B$.

to show that if $X$ is a compact Hausdorff space then $x,y$ are in the same quasicomponent if and only if they are in the same component.

This is from "General topology" by Riszard Engelking.

It seems the following.

  to show that the minimal element is connected

Let $A$ be a minimal element of the family $\mathcal{A}$. Suppose that the set $A$ is not connected. Then $A$ is a union of two its disjoint clopen non-empty subsets . Since $x$ and $y$ are in the same quasicomponent of $A$, the set $\{x,y\}$ is contained in one of these two sets. Denote this set of $B$. Since the set $B$ is a clopen subset of $A$, each quasicomponent of the set $B$ is a quasicomponent of the set $A$. Hence $B\in\mathcal A$. But $B$ is a proper subset of the set $A$, which contradicts to the minimality of the set $A$.

At first prove that family $A$ from closed subset of compact and hausdoorf topological space $X$ which for $x,y$ be given in $X$ are in a quasicomponet of this closed  subsets then $x,y$ are in a quasicoponet  of intersection of elements of $A$. secound if $A$ is chain with icloudtion then intersection of $A$ is conected at last is simple that see "componets of $X$ are quasicomponets of $X$.


Is 'det' a single function? I should mention this up front: this is a question about semantics. It is extremely formal, and a bit pedantic. Most people won't care.

Is det, i.e the determinant function, a single one? Or is it multiple functions, one for each field?

Let $\mbox{Mat}(K)$ denote the set of square matrices over the field $K$ (This is nonstandard notation). Then $\det(x) \in K$ for $x\in \mbox{Mat}(K)$. So we can view $\det$ in this context as a function $\det: \mbox{Mat}(K) \rightarrow K$. If you take this view, then you are forced to acknowledge that 'det' actually refers to infinitely many functions, one for each field $K$, and that which 'det' we are referring to in a particular expression depends entirely on context.

I should note that here, we take the view that a function MUST possess a domain and codomain, that it is IS NOT just a collection of (input, output) pairs with some sort of implicit domain and codomain.

On the other hand, we may view $\det$ as a single function, taking in ANY square matrix, and returning... something. This is fine, but what would the codomain be? It would have to consist of... everything, and that is not a good codomain for a single function object. Ideally, the codomain would be a bit more expressive. However, this, formally, is not a problem.

So, which is it? The former or the latter?

The meaning that mathematicians usually implicitly have in mind is that there are lots of separate functions written "$\det$".  Specifically, for each field $K$ and each $n\in\mathbb{N}$, there is a separate function $\det:M_n(K)\to K$ (which to be completely pedantic perhaps should be written $\det_{n,K}$, but that quickly gets unwieldy so no one does it).  Note here that $M_n(K)$ is the standard notation for the set of $n\times n$ matrices with entries in $K$.

There are also many other functions that are commonly written as "$\det$".  For instance, it is very common to restrict one's attention to invertible matrices and write $\det:GL_n(K)\to K^\times$, where $GL_n(K)$ is the set of invertible $n\times n$ matrices and $K^\times$ is the set of nonzero elements of $K$.  This is of course not literally the same function as the function $\det:M_n(K)\to K$, but no one ever bothers to distinguish it notationally.

I would add that absolutely none of this is special to "$\det$".  There are oodles and oodles of similarly context-dependent notations in math; for instance, the symbol $+$ can refer to a vast variety of different binary operations, with the specific operation in any given instance usually left to be inferred from context.  Mathematical notation is meant to be understood by humans, and humans are very good at resolving ambiguity from context (and bad at processing overly cluttered but completely unambiguous notation).

What about define it as a function with following domain and codomain
$$
\mathrm{det}: \bigcup_{n=1}^{\infty}\ \bigcup_{K:\ K\ \text{is a field}} M_{n\times n}(K)\to \bigcup_{K:\ K\ \text{is a field}} K
$$


Can $x^{x^x}=k$ be solved using the W function? The lambert W function is defined to be the inverse of $f(x)=xe^x$, and the equation $x^x=k$ can be solved fairly easily using the function:
$$x^x=k$$
$$\ln(x^x)=\ln(k)$$
$$x\ln(x)=\ln(k)$$
$$e^{\ln(x)}\ln(x)=\ln(k)$$
$$\ln(x)=W(\ln(k))$$
$$x=e^{W(\ln(k))}$$

My question is, can $ x^{x^x}=k$  be solved using the lambert W function, and if so, can power towers of any height be solved using some iterative process?

can $x^{x^x}=y$ be solved using the lambert W function, and if so,

Using Maurer's notation, the equation ${^n}x=y$ is solvable using Lambert's $W$ function only for $n=2$.

If we loosely define "transcendence degree of an equation" to be the height of the highest exponential tower with $x$ on top, contained in the equation minus one, then $W$ solves ONLY equations with (transcendence) degree at most 1 (that is, with exponentials of total height 2), but NOT all such equations.

The equation $x^{x^x}=y$ is written as ${^3}x=y$, consequently it has (transcendence) degree 2 (tower with 3 x's minus 1)

If memory serves right, the most general equation solvable by $W$ is something like:

$$a\cdot x^m\cdot d^{b\cdot x^n+c}=y$$

which as you can see can be reduced to having degree 1 at some point in the algebraic solution process, by taking $n$-th roots on both sides of the equation.

Elaborating on the second point, note that adding one plain $x$ to the equation above and it fails solvability. That is, the equation

$$a\cdot x^m\cdot d^{b\cdot x^n+c}+x=y$$

for example, is now NOT solvable by $W$, although it is of the same transcendence degree as the one above, so the definition of "transcendence degree" of an equation, needs to be made a liitle more rigorous, by specifying some additional configuration constraints, which vary from case to case.

can power towers of any height be solved using some iterative process?

Sure they can. Omitting details about existence, the following equivalence holds:

$${^n}x=y\Leftrightarrow x=^{\frac{1}{n}}y$$

and this is the $n$-th order super root of $y$, which, after taking care of domains and ranges, can be approximated using backwards numerical iteration.

Such a method, for example, was presented in Maple by Robert Israel, above, long time ago on sci.math.

No, the Lambert W function can't help with $x^{x^x}$.


Shifting Nth Root Algorithm Does anyone have a very simple dumbed-down explanation of the shifting nth root algorithm on paper (finding an nth root with a method similar to long division)?  I know very basic addition, subtraction, multiplication, division, and exponents.  I would like to learn the shifting nth root algorithm (to calculate nth roots) in these very very basic terms, like a kid in school learns long division for the first time.  The only explanation I have found is the wikipedia article, and it is not simple enough for me with my limitiations.  Thank you.

OK, so let us set up an example of calculating $\sqrt[3]{9876.5432}$. We arrange the radicand in blocks of length $3$ so that we have $009:876:543:200:000...$. The first round is slightly different than the next rounds. The first step is to find the maximal $\beta$ such that:
$$
\beta^n\leq\alpha=009
$$
This is $\beta=2$. Then we put $y=2$ and $r=9-2^3=1$.

Now we are ready to proceed to the general scheme. Find the largest $\beta$ such that
$$
(10\cdot 2+\beta)^3-10^3\cdot 2^3\leq 10^3\cdot 1+876
$$
The reason this makes sense is that we are actually looking at  $(20+\beta)^3-8000\leq 1876$ or equivalently $(20+\beta)^3\leq 9876$ and finding the largest digit $\beta$. Trying $(20+1)^3=9261$ and $(20+2)^3=10648$ shows that $\beta=1$ is the largest integer satisfying this. So we proceed with this:
$$
\begin{align}
y'&=10y+\beta\\&=20+1\\&=21\\
r'&=10^3r+\alpha-((y')^3-10^3 y^3)\\
&=10^3\cdot 1+876-(21^3-10^3\cdot 2^3)\\
&=615
\end{align}
$$

Now we proceed in the same manner:
$$
(10\cdot 21+\beta)^3-10^3\cdot 21^3\leq 10^3\cdot 615+543
$$
Since $214^3=9800344$ whereas $215^3=9938375$ we see that the next digit is $\beta=4$. Thus
$$
\begin{align}
y'&=10y+\beta\\&=210+4\\&=214\\
r'&=10^3r+\alpha-((y')^3-10^3 y^3)\\
&=10^3\cdot 615+543-(214^3-10^3\cdot 21^3)\\
&=76199
\end{align}
$$

And so on:
$$
(10\cdot 214+\beta)^3-10^3\cdot 214^3\leq10^3\cdot 76199+200
$$
And we see that $2145^3=9869198625$ whereas $2146^3=9883008136$ so we must have $\beta=5$ and
Now we proceed in the same manner:
$$
(10\cdot 21+\beta)^3-10^3\cdot 21^3\leq 10^3\cdot 615+543
$$
Since $214^3=9800344$ whereas $215^3=9938375$ we see that the next digit is $\beta=4$. Thus
$$
\begin{align}
y'&=10y+\beta\\&=2140+5\\&=2145\\
r'&=10^3r+\alpha-((y')^3-10^3 y^3)\\
&=10^3\cdot 76199+200-(2145^3-10^3\cdot 214^3)\\
&=7344575
\end{align}
$$

So frankly, this method is not much different than simply trying out and fine tuning different numbers raised to the $n$-th power. We have found so far, that $\sqrt[3]{9876.5432}>21.45$ and that $21.45$ is the largest four digit number that does not exeed the goal of $9876.5432$ when raised to the third power.

I hope I provided this in a meaningful way, nonetheless!

I made the algorithm in VBA in Excel. For now it only calculates roots of integers. It is easy to implement the decimals as well.

Just copy and paste the code into an EXCEL module and type the name of the function into some cell, passing the parameters.

Public Function RootShift(ByVal radicand As Double, degree As Long, Optional ByRef remainder As Double = 0) As Double

   Dim fullRadicand As String, partialRadicand As String, missingZeroes As Long, digit As Long

   Dim minimalPotency As Double, minimalRemainder As Double, potency As Double

   radicand = Int(radicand)

   degree = Abs(degree)

   fullRadicand = CStr(radicand)

   missingZeroes = degree - Len(fullRadicand) Mod degree

   If missingZeroes < degree Then

      fullRadicand = String(missingZeroes, "0") + fullRadicand

   End If

   remainder = 0

   RootShift = 0

   Do While fullRadicand <> ""

      partialRadicand = Left(fullRadicand, degree)

      fullRadicand = Mid(fullRadicand, degree + 1)

      minimalPotency = (RootShift * 10) ^ degree

      minimalRemainder = remainder * 10 ^ degree + Val(partialRadicand)

      For digit = 9 To 0 Step -1

          potency = (RootShift * 10 + digit) ^ degree - minimalPotency

          If potency <= minimalRemainder Then

             Exit For

          End If

      Next

      RootShift = RootShift * 10 + digit

      remainder = minimalRemainder - potency

   Loop

End Function


Why do we care about dual spaces? When I first took linear algebra, we never learned about dual spaces. Today in lecture we discussed them and I understand what they are, but I don't really understand why we want to study them within linear algebra.

I was wondering if anyone knew a nice intuitive motivation for the study of dual spaces and whether or not they "show up" as often as other concepts in linear algebra? Is their usefulness something that just becomes more apparent as you learn more math and see them arise in different settings?

Edit

I understand that dual spaces show up in functional analysis and multilinear algebra, but I still don't really understand the intuition/motivation behind their definition in the standard topics covered in a linear algebra course. (Hopefully, this clarifies my question)

Let $V$ be a vector space (over any field, but we can take it to be $\mathbb R$ if you like,
and for concreteness I will take the field to be $\mathbb R$ from now on;
everything is just as interesting in that case).  Certainly one of the interesting concepts
in linear algebra is that of a hyperplane in $V$.
For example, if $V = \mathbb R^n$, then a hyperplane is just the solution set to an equation
of the form
$$a_1 x_1 + \cdots + a_n x_n = b,$$
for some $a_i$ not all zero and some $b$.
Recall that solving such equations (or simultaneous sets of such equations) is one
of the basic motivations for developing linear algebra.
Now remember that when a vector space is not given to you as $\mathbb R^n$,
it doesn't normally have a canonical basis, so we don't have a canonical way
to write its elements down via coordinates, and so we can't describe hyperplanes
by explicit equations like above.  (Or better, we can, but only after choosing
coordinates, and this is not canonical.)
How can we canonically describe hyperplanes in $V$?
For this we need a conceptual interpretation of the above equation.  And here linear
functionals come to the rescue.  More precisely, the map
$$\begin{align*}
\ell: \mathbb{R}^n &\rightarrow \mathbb{R} \\
(x_1,\ldots,x_n) &\mapsto a_1 x_1 + \cdots + a_n x_n
\end{align*}$$
is a linear functional on $\mathbb R^n$, and so the above equation for the
hyperplane can be written as
$$\ell(v) = b,$$
where $v = (x_1,\ldots,x_n).$
More generally, if $V$ is any vector space, and $\ell: V \to \mathbb R$ is any
non-zero linear functional (i.e. non-zero element of the dual space), then
for any $b \in \mathbb R,$ the set
$$\{v \, | \, \ell(v) = b\}$$
is a hyperplane in $V$, and all hyperplanes in $V$ arise this way.
So this gives a reasonable justification for introducing the elements of the dual
space to $V$; they generalize the notion of linear equation in several variables
from the case of $\mathbb R^n$ to the case of an arbitrary vector space.
Now you might ask: why do we make them a vector space themselves?  Why do we want
to add them to one another, or multiply them by scalars?
There are lots of reasons for this; here is one: Remember how important it is,
when you solve systems of linear equations, to add equations together, or
to multiply them by scalars (here I am referring to all the steps you typically
make when performing Gaussian elimination on a collection of simultaneous linear
equations)?  Well, under the dictionary above between linear equations
and linear functionals, these processes correspond precisely to adding together
linear functionals, or multiplying them by scalars.  If you ponder this for a bit,
you can hopefully convince yourself that making the set of linear
functionals a vector space is a pretty natural thing to do.
Summary: just as concrete vectors $(x_1,\ldots,x_n) \in \mathbb R^n$ are naturally
generalized to elements of vector spaces, concrete linear expressions
$a_1 x_1 + \ldots + a_n x_n$ in $x_1,\ldots, x_n$ are naturally generalized to linear functionals.

Here is something to add to all the answers and discussions up to now.

I will explain with an example. Let us think of the space of polynomials
of degree $n$ in the real interval $[0,1]$. We want to consider a basis.
For example the monomials $\{1, x, x^2, \cdots, x^n \}$. Another basis
could be the Bernoulli polynomials $\{B_0(x), B_1(x), \cdots , B_n(x)\}$.

The dual basis $\{B^*(x)\}$ satisfies the property:

\begin{equation}
\langle B^*_i(x) , B_j(x) \rangle = \delta_{ij}.
\end{equation}
Given a polynomial $p(x)$ we would like to find a representation of
it in terms of its bases $\{B_i\}$.  We could do
Gram-Schmidt decomposition but this is hard to do (well, maybe not
so hard, look for Legendre Polynomials). Another
way is to find the dual basis of $\{B_i\}$.

The dual basis of the Bernoulli polynomials is  derived
here

What is the advantage of using the dual basis?

Let us see:

Functional Analysis

Given any polynomial $p(x)$ in the basis $B_j$ we have that
there exists coefficients $c_i$ such that $p(x)$ is spanned
by the Bernoulli polynomials. That is,
\begin{eqnarray*}
  p(x) =  \sum_i c_i B_i(x) = \sum_i \sum_j \delta_{ij} c_j B_i(x) =
    \sum_i \sum_j \left \langle B^*_i(y) \; , \;B_j(y) \right \rangle   c_j B_i(x)
  = \sum_i \langle B_i^*(y)  ,  \sum_j c_j B_j (y)  \rangle   B_i(x),
\end{eqnarray*}
that is,

\begin{eqnarray}
p(x) =   \sum_i c_i B_i(x) =
   \sum_i \left \langle B_i^*(y) \; , \; p(y) \right \rangle  B_i(x),
  \quad \quad (1)
\end{eqnarray}
Since usually inner products are written as integrals we write
\begin{eqnarray*}
  \int_0^1 \sum_i B_i(x)   B_i^*(y)  p(y) dy = p(x)
\end{eqnarray*}
and then we can interpret the sum of functionals $B^*_i$ weighted by
Bernoulli polynomials $B_i$ as
\begin{eqnarray}
  \sum_i B_i(x)   B_i^*(y)  = \delta(x-y).
\end{eqnarray}

The power of dual basis is that we do no have to do anything like a
Gram-Schmidt orthogonalization to have a nice representation of a function in terms of its bases.  We see that the Bernoulli polynomials are not orthonormal on the monomial basis $\{1, x, \cdots x^i, \cdots \}$. However when representing $p(x)=\sum_i c_i B_i(x)$, we see that the Fourier coefficients $c_i$ are represented by

\begin{eqnarray*}
c_i = \left \langle B_i^*(y) \; , \; p(y) \right \rangle
\end{eqnarray*}
due to the fact that the vectors $B_i(x)$ are linearly independent in
equation (1). That is, the coordinates of the vector
$p(x)$ on the bases $\{ B_i(x) \}$ are the projections of the dual base vectors
$\{B^*_i\}$ into the vector $p(x)$.

Tensor Analysis:

The concept of dual is closely related to the concept of covariant tensor.
A contra-variant tensor can be seen as a vector and its covariant counter-part as the dual. If your base is orthonormal the representation is the same in the original basis and the dual bases. Cartesian tensors do not need super-indices. On the other hand, if your basis is not orthonormal, then the dual basis is neither orthogonal and the representation is different. Here you need to distinguish between covariant and contravariant. That is why you need notations such as $\delta_{ij}^{mn}$,
a second rank contravariant, second rank covariant. The inner product of a contravariant with a covariant (the regular inner product) produces an invariant. Without duality these concepts would not even exist.


Polar roses and a symmetric proof Question:
  Can anyone elaborate on my friend's comments below concerning symmetry and how this more or less proves the result(s) concerning the number of leaves?

I know $r=a\sin n\theta$ and $r=a\cos n\theta$ are $n$-leaved if $n$ is odd and $2n$-leaved if $n$ is even, but I don't really know explicitly why: it is clear with graphing and experimenting, but I was hoping for an algebraic proof, more or less.

Talking with a friend, he said there's a symmetry, which is true of course. That $-r$ is the same as adding odd multiples of $\pi$, which I also know why that is true. That cosine and sine shift signs every $\pi$ (presumably meaning $\cos(x)=-\cos(x+\pi)$ and $\sin(x)=-\sin(x+\pi)$). So scaling the argument by an odd number still gives odd $\pi$. So symmetry. Even number converts to even $\pi$'s. So lose the symmetry.

Progress: It looks as though I can write $\sin(n\theta)=-\sin(n(\theta+\pi))$ when $n$ is odd and $\sin(nx)=\sin(n(x+\pi))$ when $n$ is even (and likewise for cosine). My question, then (I suppose), is how to verbalize that the graph produces $2n$ unique leaves when $n$ is even and the graph is on the interval $\theta\in[0,2\pi)$.

I know $=\sin \theta$ and $=\cos \theta$ are $$-leaved if $n$ is odd and $2$-leaved if $$ is even, but I don't really know explicitly why: it is clear with graphing and experimenting, but I was hoping for an algebraic proof, more or less.

Talking with a friend, he said there's a symmetry, which is true of course. That $−$ is the same as adding odd multiples of $\pi$
, which I also know why that is true. That cosine and sine shift signs every $\pi$ (presumably meaning $\cos()=−\cos(+\pi)$
 and $\sin()=−\sin(+\pi)$. So scaling the argument by an odd number still gives odd $\pi$. So symmetry. Even number converts to even $\pi$'s. So lose the symmetry.

Progress: It looks as though I can write $\sin(\theta)=−\sin((\theta+\pi))$ when $n$ is odd and $\sin()=\sin((+\pi))$ when $n$ is even (and likewise for cosine). My question, then (I suppose), is how to verbalize that the graph produces $2$ unique leaves when $$ is even and the graph is on the interval $\theta∈[0,2\pi)$.


If a function has an inverse then it is bijective? I have some trouble finding the answer to this, can someone help me out:

If I have a general function $f$ with domain $X$ and codomain $Y$, I know nothing about the function (injective, surjective). Say I have found and inverse that is, there is a function $g$ such that for $f(x) = y$, $g(y) = x$ (is this enough for an inverse). Then the function is bijective?

If you have $f: X \to Y$, $g : Y \to X$ with $g(f(x)) = x$ for all $x \in X$, it is still possible for $f$ to not be bijective.
However, $f$ will be a bijection onto its image; i.e., $f$ is a bijection from $X$ to $f(X)$. In particular, $f$ is injective.

If you additionally require that $f(g(y)) = y$ for all $y \in Y$ (i.e. $g$ is a "two-sided inverse"), then $f$ is a bijection from $X$ to $Y$.

If you have f:X→Y,f:X→Y, g:Y→X ,g:Y→X with g(f(x))=x   g(f(x))=x for all x∈X  x∈X, it is still possible for ff to not be bijective. However, ff will be a bijection onto its image; i.e., ff is a bijection from XX to f(X)f(X). In particular, ff is injective.

If you additionally require that f(g(y))=yf(g(y))=y for all y∈Yy∈Y (i.e. gg is a "two-sided inverse"), then ff is a bijection from XX to YY.