Convergence of averaged sine function I have stumbled upon those two problems which I got a little stuck on that is show convergence or divergence for the series

$$\sum_{n=1}^{+\infty}\frac{\cos(n)}{n}$$

and

$$\sum_{n=1}^{+\infty}\frac{\sin(n)}{n}.$$

The following answer is not entirely rigorous as written, but you can actually sum these series explicitly. Moreover I will proceed a bit more generically by considering $\{\cos n\phi,\sin n\phi\}$ for some real $\phi\in(0,\pi]$. (The case $\phi=0$ would give the divergent harmonic series, and other real $\phi$ can be determined by symmetry & periodicity.) Recalling Euler's formula $e^{i \phi}=\cos\phi+i \sin \phi$, observe that

$$f(\phi):=\sum_{n=1}^\infty \frac{\cos n\phi}{n}+i \sum_{n=1}^\infty \frac{\sin n\phi}{n}=\sum_{n=1}^\infty \frac{e^{i n\phi}}{n}=\sum_{n=1}^\infty \frac{z^n}{n}$$

where $z:=e^{i \phi}$. We may recognize this last expression in terms of the Taylor series of the logarithm as $f(\phi)=-\ln(1-z)=-\ln(1-e^{i\phi}).$ (To check this, compute $f'(z)$ term by term and sum the resulting geometric series.) This can be simplified by observing that
$$1-e^{i \phi}=e^{i\phi/2}\left(e^{-i\phi/2}-e^{i\phi/2}\right)=-2i e^{i\phi/2}\sin\frac{\phi}{2} = \left(2\sin\frac{\phi}{2}\right) e^{i(\phi-\pi)/2}.$$ and so $f(\phi)=-\ln\left(2\sin\dfrac{\phi}{2}\right)-i\left(\dfrac{\phi-\pi}{2}\right)$. As a check, note that $f(\pi)=\sum\limits_{n=1}^\infty \dfrac{1}{n}(-1)^n=-\ln 2$. With this in hand, we have $f(1)=-\ln\left(2\sin\dfrac{1}{2}\right)+i\left(\dfrac{\pi-1}{2}\right)$; to get the cosine and sine series, simply take the real and imaginary parts respectively. So we have summed the series.

We know that $(-1) \le \cos n$

So $(-1)/n \le \cos n/n$

Summation $(-1)/n$ is divergent. So by Comparison test of first type, summation $\cos n/n$ is divergent.


Prove that if $2^{k_1}l_1=2^{k_2}l_2$, then $k_1=k_2$ and $l_1=l_2$. Suppose $k_1,k_2,l_1,l_2$ are natural numbers such that $l_1$ and $l_1$ are odd. Prove that if $2^{k_1}l_1=2^{k_2}l_2$, then $k_1=k_2$ and $l_1=l_2$.

First, let $l_1 = 2m_1 +1$ and $l_2=2m_2+1$ where $m_1,m_2\in\mathbb{N}$.

$$\begin{align*}
2^{k_1}l_1&=2^{k_2}l_2\\
\frac{l_2}{l_1}&=2^{k_1-k_2}\\
\ln\left(\frac{l_2}{l_1}\right)&=({k_1-k_2})\ln(2)\\
\ln(l_2)-\ln(l_1)&=({k_1-k_2})\ln(2)\\
\ln(2m_2+1)-\ln(2m_1 +1)&=({k_1-k_2})\ln(2)\\
\end{align*}$$

I am stuck at this step because I can't see how I can get RHS or LHS to $0$.

Can anyone give me a hit or suggestion to keep going or start to rewrite a new proof?

Thanks.

Suppose $k_1\neq k_2$. If $k_2 > k_1$, then dividing by $2^{k_1}$ gives
$$ l_1 = 2^{k_2 - k_1} l_2. $$
As $k_2-k_1 > 0$, we deduce that $l_1$ is even. Contradiction.
Similarly, if $k_2 < k_1$, we obtain a contradiction.
So we conclude that $k_1 = k_2$  which clearly implies $l_1 = l_2$.

We note that an odd number divided by an odd number gives an odd number. We also note that $2^n$ for $n \not=0$ is always an even number. So, for $\frac{l_2}{l_1}=2^{k_1-k_2}$ to be true, both sides must be equal to 1, which is equivalent to $l_1=l_2$ and $k_1=k_2$.


Minimal operations to sort a sequence I have a sequence $A = a_1,a_2,...,a_n$. I can choose any element and increase or decrease it by one and a single increase or decrease counts as a single step. I want to make the array sorted in non decreasing in minimal numbers of steps.

I want to prove the following. Consider all unique numbers in the sequence A. Then the optimal answer will be one of the possible combinations of the unique numbers.

e.g If the array $A = 1, 5, 8, 1$. All unique values $S = {1, 5, 8}$. Then the optimal answer will be one of possible sorted combinations of length 4 where elements are taken from S

1 1 1 1 => cost=(1-1)+(5-1)+(8-1)+(1-1)=11

1 5 5 5 => cost=(1-1)+(5-5)+(8-5)+(5-1)=7

5 5 5 8

1 5 8 8

etc.

That is, the optimal answer is the sorted sequence such that all values of the optimal answer are taken from the set S. This implies that we don't need to take intermediate values under consideration.

Let $B = b_1, b_2, \ldots, b_n$ be a minimum-cost sorted sequence derived from $A = a_1, a_2, \ldots, a_n$.  Note that $b_i \geq \min A$; otherwise the cost could be decreased by adding 1 to all $b_i$ that are less than $\min A$.

Suppose $B$ contains an element that is different from all the $a_i$'s.  Let the first such element be $b_j$.

Either $j=1$, or $b_j > b_{j-1}$ because $B$ is nondecreasing and $b_j = b_{j-1}$ for $j>1$ would mean that $b_j$ is not the first element of $B$ not in $A$.

Let $b_j, b_{j+1}, \ldots, b_k$ be the subsequence of $B$ such that all its elements are not in $A$ and either $b_{k+1}$ is in $A$ or $k=n$.  Every $b_i$ for $j \leq i \leq k$ is such that $b_i \neq a_i$.  In addition, $b_k < b_{k+1}$ if $k < n$ and $b_{j-1} < b_j$ if $j>1$.

If $b_i > a_i$ happened more often than $b_i < a_i$ in the subsequence, it would be possible to decrease the cost of $B$ by decreasing all the $b_i$ for $j \leq i \leq k$ by $1$ without violating the monotonicity constraint.  Hence this is not possible.

Likewise, $b_i < a_i$ cannot happen more often than $b_i > a_i$ in the subsequence.  It must be that the $b_i$ such that $b_i > a_i$ are equal in number to those for which $b_i < a_i$.  Therefore moving the segment between $j$ and $k$ down so that $b_j = b_{j-1}$ if $j>1$ or $b_1 = \min{A}$ if $j=1$ does not change the cost of $B$ and makes $b_j$ an element of $A$.

In conclusion, if there is a minimum-cost sorted $B$ with an element not in $A$, then there is also another minimum-cost sorted $B'$ that either has all its elements in $A$, or has its first element outside of $A$ later than $B$.  Since the latter case can only occur finitely many times, there is an optimal sorted $B$ whose elements are all in $A$.


Functional equation involving $f(x^4)+f(x^2)+f(x)$ Find all increasing functions $f$ from positive reals to positive reals satisfying $f(x^4) + f(x^2) + f(x) = x^4 + x^2 + x$.

It's easy to show that $f(1)=1$, and I was also able to show that
$$f(x)-x = f(x^{(8^k)}) - x^{(8^k)}$$
for all integers $k$.

But where to go from here? Any hints would be much appreciated.

Let $g(x) = f(x)-x$. We obtain $g(x^4)+g(x^2)+g(x)=0$. Define $a_n = g(x^{2^n})$ for every $n\in\Bbb Z$ where $x\ne 1,x>0$. It follows that
$$
a_{n+2}+a_{n+1}+a_n=0,
$$ and from this 2nd order linear equation we get for some $A,B$,$$
a_n =\begin{cases}A,\quad n\equiv 0 \;\text{(mod }3)\\ B,\quad n\equiv 1 \;\text{(mod }3)\\-A-B,\quad n\equiv 2 \;\text{(mod }3)\end{cases}.
$$

It says that $(a_n)$ is a $3$-periodic sequence. Now we claim that $a_n = 0$ for every $n$. Since $f(x)=x+g(x)$ is increasing, it holds for each $k=0,1,2$ and all $n\equiv k\text{ (mod} \;3)$, $x>1$ that
$$
x^{2^n}+a_k \le x^{2^{n+1}}+a_{k+1}\le x^{2^{n+2}}+a_{k+2}.
$$ Let $n\equiv k\text{ (mod} \;3)$ tend to $-\infty$ to obtain
$$
a_k\le a_{k+1}\le a_{k+2}
$$ for each $k=0,1,2$. This implies $a_0=a_1=a_2=0$ and thus $a_n = 0,\ \forall n$ follows for $x>1$. The case where $0<x<1$ can be treated similarly by noting that
$$
x^{2^n}+a_k \ge x^{2^{n+1}}+a_{k+1}\ge x^{2^{n+2}}+a_{k+2}.
$$
This proves $f(x)=x+g(x)=x$ for all $x>0$.

Assuming $f(x)$ as a real function, calling $g(x) = f(x)-x$ we have

$$
g(x^4)+g(x^2)+g(x) = 0
$$

now calling $e^y = x$ we have

$$
g(e^{4y})+g(e^{2y})+g(e^y) = 0
$$

or

$$
G(4y)+G(2y)+G(y) = 0
$$

This recurrence equation has as solution

$$
G(y) = C_1 y^{-\frac{2 i \pi }{3 \ln 2}}+C_2 y^{\frac{2 i \pi }{3 \ln 2}}
$$

because

$$
\mathcal{G}(\log_2 (4y))+\mathcal{G}(\log_2(2y))+\mathcal{G}(\log_2 y)=0\\
\mathcal{G}(\log_2 y+2)+\mathcal{G}(\log_2 y +1)+\mathcal{G}(\log_2 y)=0
$$

and making $z = \log_2 y$ we get

$$
\mathcal{G}(z+2)+\mathcal{G}(z+1)+\mathcal{G}(z)=0
$$

which can be characterized as second order linear difference equation.

Then follows

$$
G(y) = C'_1 \cos \left(\frac{2 \pi}{3}  \log_2 y\right)+C'_2\sin \left(\frac{2 \pi}{3}  \log_2 y\right) = g(e^y) = f(e^y)-e^y
$$

and finally

$$
f(x) = C'_1\cos\left(\frac{2\pi}{3}\log_2(\ln x)\right)+C'_2\sin\left(\frac{2\pi}{3}\log_2(\ln x)\right)+x
$$

In case of $f(x)$ increasing then the solution is $f(x) = x$ with $C'_1=C'_2=0$

NOTE

Use this MATHEMATICA script to verify the recurrence equation as well as the final solution.

Clear[G]
G[y_] := Cos[(2 \[Pi] Log[2, y])/3] c1 + Sin[(2 \[Pi] Log[2, y])/3 ] c2
G[4 y] + G[2 y] + G[y] // FullSimplify

f[x_] := Cos[(2 \[Pi] Log[2, Log[x]])/3] c1 + Sin[(2 \[Pi] Log[2, Log[x]])/3 ] c2 + x
Assuming[x > 0, f[x^4] + f[x^2] + f[x] - x^4 - x^2 - x // FullSimplify]


Picking Same Color Probability So i recently came across this question,

Marla has m bottles of marbles. Each bottle has n marbles of n different colours. Marla mixes all the marbles from all the bottles together. Now, she picks up n random marbles and puts them into another empty bottle.

Remember that now there can be more than one marble of a same colour in this bottle.
Marla picks up one marble from this bottle, notes down the colour and puts the marble back into the same bottle. The second time she picks up a marble from the same bottle, if it is of the same colour, then she succeeds.
What is the probability that she succeeds?

This is a sample test case that was provided.

Eg. Sample Input

2 2

Sample Output

0.666666666666

Explanation

The probability of getting both the marbles in the bottle of the same colour is 0.33333333333333.

The probability of then succeeding is 1.

The probability of getting both the marbles in the bottle of different colours is 0.666666666666.

The probability of then succeeding is 0.5.

The total probabilty of succeeding is ( 0.333333333333 x 1 ) + (0.666666666666 x 0.5)= 0.666666666666

How do i go about this?
Probability is not my forte, so i am not really sure how to go about this.

I have made 3 cases (n=m),(n>m) and (n

Let's just consider the probability of getting a specific color of marble twice, and then, due to the symmetry of the problem, we may multiply by $n$ to get the total probability of success for Martha.

First, the probabilities of getting $k$ marbles of that color in your bottle: there are $nm$ marbles, and so there are $nm\choose n$ ways to pick $n$ marbles to put into a bottle.  Of those ways, there are $m\choose k$ ways to pick $k$ marbles from the $m$ marbles of the desired color, leaving ${nm-m}\choose {n-k}$ choices of other-colored marbles to fill up your bottle. So the probability of getting $k$ marbles of this color is:
$$\frac{{m\choose k}\cdot{nm-m\choose n-k}}{nm\choose n} $$
Now, given that you have $k$ marbles in your bottle, then the probability of picking one of those marbles is simply $k/n$. So your probability of getting $k$ marbles of one color and then picking that color of marble twice will be
$$\frac{{m\choose k}\cdot{nm-m\choose n-k}}{nm\choose n}\cdot\frac{k^2}{n^2} $$
So your probability of success, multiplying by $n$ to account for all colors, is the sum of these probabilities for all possibilities of $k$, from 1 to $m$ or $n$, whichever comes first:
$$\Pr(Success)=\sum\limits_{k=1}^{\min(m,n)}\frac{{m\choose k}\cdot{nm-m\choose n-k}}{nm\choose n}\cdot\frac{k^2}{n} $$

...so yeah, as to a nice, closed form solution, I'm not too sure, but just going from counting equally likely events, I believe this would be your answer

Total number of marbles are mn

Out of these mn marbles, m are of same colour. For our event to a success we need atleast 2 marbles of same colour.

We know $$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$

Hence the number of ways for selecting 2 marbles to be of same colour will be -
$$\binom{m}{2} *\binom{mn-m}{n-2} $$

So the probability of getting 2 marbles of the same color is

$$\frac{{m\choose 2}\cdot{mn-m\choose n-2}}{mn\choose n}$$

As we have 2 marbles in our bottle, the probability of picking one of those marbles is simply 2/n. So oour probability of getting 2 marbles of one color and then picking that color of marble twice will be

$$\frac{{m\choose 2}\cdot{mn-m\choose n-2}}{mn\choose n}\cdot\frac{2^2}{n^2}$$

To take into account all colours, we must multiply the result by n.

Hence, the probability for the event to be successfull is --

$$\frac{{m\choose 2}\cdot{mn-m\choose n-2}}{mn\choose n}\cdot\frac{2^2}{n}$$

Please correct me, if i am wrong. Its been a long time i came across probability questions.


Coin flipping probability game ; 7 flips vs 8 flips Your friend flips a coin 7 times and you flip a coin 8 times; the person who got the most tails wins. If you get an equal amount, your friend wins.

There is a 50% chance of you winning the game and a 50% chance of your friend winning.

How can I prove this? The way I see it, you get one more flip than your friend so you have a 50% chance of winning if there is a 50% chance of getting a tails.

I even wrote a little script to confirm this suspicion:

from random import choice

coin = ['H', 'T']

def flipCoin(count, side):
    num = 0
    for i in range(0, count):
        if choice(coin) == side:
            num += 1
    return num

games = 0
wins = 0
plays = 88888

for i in range(0, plays):
    you = flipCoin(8, 'T')
    friend = flipCoin(7, 'T')
    games += 1
    if you > friend:
        wins += 1

print('Games: ' + str(games) + ' Wins: ' + str(wins))
probability = wins/games * 100.0
print('Probability: ' + str(probability) + ' from ' + str(plays) + ' games.')

and as expected,

Games: 88888 Wins: 44603
Probability: 50.17887678876789 from 88888 games.

But how can I prove this?

Well, let there be two players $A$ and $B$. Let them flip $7$ coins each. Whoever gets more tails wins, ties are discounted. It's obvious that both players have an equal probability of winning $p=0.5$.

Now let's extend this. As both players have equal probability of winning the first seven tosses, I think we can discard them and view the 8th toss as a tiebreaker. So let's give player $A$ the 8th toss: if he gets a tail, he wins, otherwise, he loses. So with $p = 0.5$, he will either win or lose this 8th toss. Putting it like this, we can see that the 8th toss for player $A$ is equivalent to giving both players another toss and discarding ties, so both players have winning probabilities of $0.5$.

Based on the answers so far you can get the following induction.
If 2 players A and B play a game with these rules: most number of tails wins.
Each player tosses the coin N times. We have P(Awin-Ntosses)=P(Bwin-Ntosses) = 50%.
You tried to demonstrate that if player B gets an extra coin toss the odds of him winning will  remain 50%.
that would mean that P(Awin-Ntosses)=P(Bwin-N+1 tosses)
If that is indeed true then using induction you can obtain P(Awin-Ntosses)=P(Bwin-N+x tosses). Induction is valid for any N or x. So therefore we can state that the following statement is true: Two players A and B play a game of toss. A tosses the coin 0 times and B tosses the coin
5 million times. The winner is the player that tosses most tails. Both players have the same  50% chance to win the game.

I strongly advise you to evaluate this statement and if you still consider it to be true, then it means you would be willing to play such a toss game for a small stake.


Prove that $ f(x) $ has at least two real roots in $ (0,\pi) $ Let $ f $ be a continuous function defined on $ [0,\pi] $. Suppose that

  $$ \int_{0}^{\pi}f(x)\sin {x} dx=0,   \int_{0}^{\pi}f(x)\cos {x} dx=0 $$

Prove that $ f(x) $ has at least two real roots in $ (0,\pi) $

Here is one root:
Let $F(x) = \int_{0}^x f(t) \sin t dt$. Then $F(0)=0$ and $F(\pi)=\int_{0}^\pi f(t)\sin tdt=0$. So by the intermediate value theorem, there exists $0<c<\pi$ such that
$$
0=F'(c) = f(c)\sin c.
$$
But since $\sin c\neq 0$, we get that $f(c)=0$.

This statement is not true. Actually, we should prove that $$\int_0^\pi f(x)[\cos x-\sin x]dx=0$$Now, define $$f(x)=\begin{cases}a\left(e-e^{4x\over \pi}\right)+1&,\quad 0\le x\le {\pi\over 4}\\1&,\quad {\pi\over 4}<x\le \pi\end{cases}$$where $a\ge 0$. It is clear that $f(x)\ge 1$ over $[0,\pi]$ for all $a\ge 0$ hence it has no root. However, by defining $$I=\int_0^\pi f(x)[\cos x-\sin x]dx$$we immediately conclude that $I=-2<0$ for $a=0$ and $\lim_{a\to \infty } I=\infty$. Based on intermediate value theorem and since $I$ is continuous w.r.t. $a$, there is a value for $a$ for which $I=0$, hence the statement is not true.

Here is a sketch of the two functions using Python:


Can you be 1/12th Cherokee? I was watching an old Daily Show clip and someone self-identified as "one twelfth Cherokee". It sounded peculiar, as people usually state they're "1/16th", or generally $1/2^n, n \in \mathbb{N}$.

Obviously you could also be some summation of same to achieve 3/32nds, etc., but will an irreducible fraction with the numerator 1 always need a power-of-two denominator? More generally does it always require a power-of-two denominator?

Assumptions (as per comments):

Nobody can trace their lineage infinitely
Incest is OK, including transgenerational, but someone can't be their own parent.

This depends on the model. Instead of arguing that we have only $46$ chromosomes and cross-overs or whatever the mechanism is called are not that common, let us assume a continuous model.
That is, a priori, everybody can be $\alpha$ Cherokee for any $\alpha\in[0,1]$ and the rules are as follows

Everybody has exactly two parents.
If the parents have Cherokee coefficients $\alpha_m, \alpha_f$, then the child has $\alpha=\frac{\alpha_m+\alpha_f}2$
In a sufficiently large but finite number of generations ago, people had $\alpha\in\{0,1\}$

It follows by induction, that $\alpha$ can always be expressed as $\alpha=\frac{k}{2^n}$ with $k,n\in\mathbb N_0$ and $0\le k\le 2^n$. Consequently $\alpha=\frac1{12}$ is not possible exactly (though for example $\frac{85}{1024}\approx\frac1{12}$ would be possible).
It doesn't matter if there is any type of inbreeding taking place anywhere in the tree (or then not-tree) of ancestors.
The only way to obtain $\alpha$ not of this form would involve time-travel and genealogical paradoxes: If you travel back in time and paradoxically become your own grandparent and one of the other three grandparents is $\frac14$-Cherokee and the others are $0$-Cherokee, you end up as a solution to $\alpha=\frac{\frac14+\alpha}4$, i.e. $\alpha=\frac1{12}$.

There is more than one way to count genetics. Suppose that 120 years ago all four pairs of your mother's great grand parents (8 in all) were pregnant with their first child, and both pairs of your father's grandparents (4 in all) were pregnant with their first child.

Thus at that moment in time you have 12 ancestors, of mixed generations back, just entering child-bearing years.

If exactly one of those twelve ancestors is full-blooded Cherokee, and the other eleven have no Cherokee blood, then:

Yes, you are clearly 1/12th Cherokee.

Whether you are counting back in even generations or in even years is an arbitrary choice, and it is incorrect to state that only one makes sense. Perhaps there are other valid ways to count ancestry back that would enable other fractions and counting.


How does the average energy from equation (41.15) of Feynman's lectures approach $kT$ as omega goes to zero? Feynman said the following equation should approach $kT$ as $\omega$ goes to zero, or $T$ goes to infinity.
$$\langle E \rangle = \frac{\hbar \omega}{e^{\hbar \omega/kT}-1}$$
Does anybody know how to prove this?

Get use of L'Hopital's rule where ' stands for the derivative
$$\lim_{\omega \to 0} \langle E\rangle= \lim_{\omega \to 0}\frac{(\hbar \omega)'}{(e^{\frac{\hbar \omega}{kT}}-1)'}  = \lim_{\omega \to 0} kT e^{-{\frac{\hbar \omega}{kT}}}=kT$$

Use the Taylor series for the exponential function,
$$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots.$$
In addition to giving you the $\omega\to 0$ limit, it will give you the low-frequency corrections.


Are the graphs I sketched of the derivative of a function and the function correct? I know that when the function is increasing, the derivative is positive. What I am confused about for the first one is that when I am sketching the derivative, I don't know what direction it goes (even though i know it is positive). What I am thinking is that where the graph has a min or max, then the derivative must have an x-intercept there. That is how I came up with my graph for the first two questions.

For the third question, I tried doing the opposite. So the derivative is positive thus the function is increasing and vice versa. Where the derivative has an x-intercept, the function has a min or max but the question is still a little confusing to me.

I think it would be helpful for you to discretize your graph along X axis and calculate the derivative using some finite differences formula.


Linearizing min function Problem How can I linearize $\min(x_1,x_2,x_3)$ in a maximization linear programming problem? Please help me. I've tried many things but I didn't solve.. My LP equations are as follows:

Objective function is:  maximize $z=\min(x_1,x_2,x_3)$

Constraints:

$0.6 x_1 + 0.8 x_2 \leq 4500 \times 20$

$0.2 x_2 \leq 4500 \times 3$

$0.3 x_2 \leq 4500 \times 6$

$0.4 x_1 + 0.6 x_2 \leq 4500 \times 10$

$0.3 x_1 \leq 4500 \times 5$

How you implement min($x_1, x_2, x_3$) in an LP solver depends on what you are trying to do with it.  Since you are maximizing it you can do the following

$$
{\rm maximize } \ z
$$
subject to
$$
z \le x_1 \\
z \le x_2 \\
z \le x_3
$$
$z$ will take the value of min($x_1, x_2, x_3$) in any optimal solution.

assume $y=\min\{x_1,x_2,x_3\}$

Now we need to maximize $y$

since value of $y$ can either be $x1$, $x2$ or $x3$, we can safely add the following constraints:

$$x_1\geq y \qquad x_2 \geq y \qquad x_3\geq y$$

or

$$x_1-y\geq 0 \qquad x_2-y \geq 0 \qquad x_3-y\geq 0$$

now solve the problem.


What is the expectation of a random variable raised to the $n$th power? If $Y=X^n$, with $n$ and the expectation and variance of $X$ known, what is the expectation and variance of $Y$?

To give this an answer, as Stefan Hansen comments:

  Knowing the expectation and variance of X is not enough information to know the moments of all orders.

As an explicit example, let $X$ be normally distributed with mean zero and variance 1, and let $W$ be +1 or -1 with probability 1/2 (the Rademacher distribution).  Then $EX = EW = 0$ and $E X^2 = E W^2 = 1$, but $E X^4 = 3$ while $E W^4 = 1$.

In the special case of the normal distribution, the distribution of a normal random variable are completely determined by its mean and variance, hence so are its higher moments.  You can find various formulas for those moments on Wikipedia.

Here's a partial answer: If $X$ is a Bernoulli variable, and $n\gt0$, then $E[X^n]=E[X]$ and $Var(X^n)=Var(X)$.


Solutions for $x^x=y^y$? How many solutions are there for $x^x=y^y$? I believe they are uncountable, but what is the proof? I have an example stating that $\frac{1}{2}^{\frac{1}{2}}=\frac{1}{4}^{\frac{1}{4}}$. What other examples and proofs of this are there?

The function $x^x$ has a single global minimum at $x = \frac{1}{e}$. You will notice that it is strictly decreasing from $x = 0$ to $x = \frac{1}{e}$ and strictly increasing from $x = \frac{1}{e}$ to infinity.

Moreover it is continuous, so you can use the Intermediate Value Theorem to deduce how many times the function can take on certain values.

The example is one solution.

What about $x = \frac12,$ $y = \frac12$? Is that a solution too? Why or why not?

Can you come up with any other solutions? For how many values of $x$ can you find a solution?


How many possible ways are there to seat down 20 men and 20 women on a bench of length 40, such that no two women are adjacent? How many possible ways are there to seat down 20 men and 20 women on a bench of length 40, such that no two women are adjacent?

At first I was sure the answer was 20!*20!. Because we seat down the women first with spacing between them, and choose their placement. Then we seat down the men and choose their position. Turns out the answer is :

21*20!*20!

I dont understand why ?

Pick one single man (in $20$ ways).

Arrange the women in $20!$ ways and then the men in $19!$, putting them in between the women. Now this man can be sitting to the right of any of the $20$ women, or he can be at the rightmost corner of the bench, hence he has $21$ positions to go, making

$$21\times 20! \times 19! \times 20 = 21\times20!\times20!$$

The reason why the last man only has $21$ choices, is that if he were to be able to sit either $WXMW$ or $WMXW$, where $X$ is the last man, $W$ were the women already seated, and $M$ the man already in between them, then I would be counting repeated positions, because I could have taken $M$ as the last man to be seated and had $X$ in there already.

If you first allocate the women on 20 predestined seats, then you have 20! possibilites to allocate them.

However, having an equidistant spacing of 1 seat between two women is only one in a few options. Let o be a woman, and x be a man:
oxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxox

This is your spacing. However,
oxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxoxxo
also is a possible spacing. Now two men set besides each other, but that's okay.In fact, there are many other such spacings that are okay, which come out to , who would have guessed it, 21.
(The two equidistant spacings, the first where a woman is on the leftmost chair, the second where a man is on the leftmost chair. Then, for the seating with one "xx", you can swap that "xx" block with any other "x" block)

So, if you factor in the number of spacings to your calculation, you get the desired result: 21*20!*20!


Find the coefficient of $x^{29}$ in the given polynomial. The polynomial is :

$$
\left(x-\frac{1}{1\cdot3}\right) \left(x-\frac{2}{1\cdot3\cdot5}\right) \left(x-\frac{3}{1\cdot3\cdot5\cdot7}\right) \cdots \left(x-\frac{30}{1\cdot3\cdot5\cdots61}\right)
$$

What I've done so far : The given polynomial is an expression of degree $30$. Hence, the coefficient of $x^{29}$ will be the negative of the sum of the roots. But the resulting sum is too complicated to handle and I think I'm doing it wrong.


Probability of no repeated pairs Suppose there is a class with 40 students. The students carry out an initial project in randomly assigned pairs. If the students carry out a second project, what is the probability that none of the randomly assigned pairs of the second project is the same as in the first project? In other words, there can be no repeated pairs in the two projects.

Can anyone provide an answer to this question by using a simulation in R and then by providing a formula?

This is a straightforward application of the principle of inclusion and exclusion.  There are $$\frac{40!}{2^{20}20!}$$ ways to choose $20$ pairs from $40$ students.  The second time around we must exclude any pair that was chosen the first time.  There are $20$ such pairs, and $$\frac{38!}{2^{19}19!}$$ ways to make $19$ other pairs, which gives  $$\frac{40!}{2^{20}20!}-20\frac{38!}{2^{19}19!}$$  But now any pairing that repeats two pairs has been subtracted out twice, so we must add those back in, and so on.  This gives $$\sum_{j=0}^{20}(-1)^j\binom{20}{j}\frac{(40-2j)!}{2^{20-j}(20-j)!}$$ admissible pairings. Doing the arithmetic, there are $191549525877429961604096$ admissible pairings out of $319830986772877770815625$ possible ones, which gives a probability of $$\frac{191549525877429961604096}{319830986772877770815625}\approx0.5989085917227133$$

If we generalize the problem to $n$ pairings out of $2n$ students, the same reasoning applies, and we find that as $n\to\infty$ the probability that there are no repeated pairs approaches $e^{-1/2}\approx0.6065306597126334$, so that we are already pretty close to the limit when $n=40.$

Probability of person being in same pair second time round is 1/19
So probability of being in different pair is 18/19

So probability of everyone being in different pair is (18/19)^20


Prove that if $\lim _{x\to \infty } f(x)$,then $\lim_{x\to \infty} f(x)=0$ Let $f:\Bbb R\to \Bbb R$ be a continuous function such that $\int _0^\infty f(x)\text{dx}$ exists.

Prove that if

$\lim _{x\to \infty }  f(x)$,then $\lim_{x\to \infty} f(x)=0$

If $f$ is non-negative then $\lim _{x\to \infty }  f(x)$ must exist and  $\lim_{x\to \infty} f(x)=0$

My try
To prove that $\lim_{x\to \infty} f(x)=0$ we should show that $\exists G>0$ such that $x>G\implies |f(x)|<\epsilon $ for any $\epsilon>0$
But I can't find out how to show this.
Please help.

This is pretty standard. Let $F(x) = \int_{0}^{x}f(t)\,dt$ then we are given that $\lim_{x \to \infty}F(x) = L$ exists. Now by continuity of $f$ and fundamental theorem of calculus we have $F'(x) = f(x)$ and we are given that $\lim_{x \to \infty}f(x) = \lim_{x \to \infty}F'(x) = M$ also exists. We can see via mean value theorem that $$F(x + 1) - F(x) = F'(\xi) = f(\xi)$$ where $x < \xi < x + 1$. Letting $x \to \infty$ in the above equation and noting that $\xi \to \infty$ we get $$L - L = M$$ or $M = 0$ so that $f(x) \to 0$ as $x \to \infty$.

Second statement is false and counterexample is in the following figure (taken from the masterpiece A Course of Pure Mathematics by G. H. Hardy):

For all $n\in \mathbb{N}$ larger than 1,

Let
$$
f(x)=
\begin{cases}
0 \ for \ x\leq n
\\
n^4x-n^5 \ for \ n<x\leq n+1/n^3
\\
-n^4+2n+n^5 \ for \ n+1/n^3<x\leq n+2/n^3
\end{cases}$$
By some calculation
$\int_n^{n+1}f(x)dx=1/n^2$. Then $\int_0^\infty f(x) = \sum_{n=2}^\infty 1/n^2$.

But $lim_{x \to \infty} f(x)$ does not exist.


Let R be a ring such that $a^2 = a$ , $\forall a$ $\in R$ . Prove that R is commutative. Im not sure what to do for this one, any help would be appreciated.

$$(a+b)^2 = (a+b) \Rightarrow a^2+ab+ba+b^2 = a+b$$ Since $a^2 = a$ and $b^2 = b$ we have that $ab = -ba$ by cancellation. Now $$(a+a) = (a+a)^2 = a^2+2a^2+a^2= a+a+a+a \Rightarrow a+a = 0 \text{ for all } a\in \mathbb{R}.$$ This says that each element is its own (additive) inverse and so $-ba = ba$ which finally gives $$ab = -ba = ba$$

First, we show that every element of R is its own additive inverse. First, note that 2x = (2x)^2 = 4(x^2) = 2(x^2). This tells you 4x = 2x, and thus 2x = 0, for all x in R. Now, x + y = (x+y)^2 = x^2 + xy + yx + y^2 = x + y + xy + yx. From this we get xy + yx = 0, xy = -yx, which coupled with our lemma and the fact that additive inverses are unique gives xy = yx.


Find perimeter and angle of triangle using three 3d vectors . Given the following, three vectors:
$$\vec{a} = 3\mathrm{i} - 2\mathrm{j} + 5\mathrm{k}\\\vec{b} = \mathrm{i} - 6\mathrm{j} + 6\mathrm{k}\\\vec{c} = 2\mathrm{i} + 3\mathrm{j} - \mathrm{k},\\$$
find the perimeter and angles of the triangle using three 3D vectors.

Studying for exam and this question is giving me difficulty. My attempt at finding the perimeter is finding length of vectors and then adding them together. E.g $|A|^2=9+4+25=38$ and so on, correct ? How would I go about finding the perimeter and angles of this triangle.

The side vectors of the triangle are given by the differences of the position vectors of the vertices. For example

$$\vec a -\vec b = 2i+4j-k$$

is one of the sides whose length is $\sqrt{4+16+1}=\sqrt{21}.$

Can you do the same with the other two edges?

The following figure may help to better understand the solution:

Regarding the angles: Let's calculate another side vector:

$$\vec a -\vec c=i-5j+6k.$$

The length of this vector is $\sqrt{62}.$

So,

$$\vec x=\frac{\vec a- \vec b}{\sqrt{21}}$$
and

$$\vec y=\frac{\vec a- \vec c}{\sqrt{62}}$$

are two unit vectors. The scalar product of these two vectors will give you the cosine of the angle of the two corresponding edges, the angle at $\vec a$.

Can you do that with the other two pairs of edges?

Perimeter is the sum of all sides.. Add all three sides and find the magnitude
i.e |A|+|B|+|C|=SOme number then again find the magnitude of this number you will find the answer and for angle use law of cosine


Continuity for topological spaces After reading the definition of a continuous map on general topological spaces, my question is the following:
Suppose $f$ is continuous from $\mathbb R$ to $\mathbb C$ given by $x \mapsto e^{ix}$.  Now this makes since the circle $S_1$ is closed in the ambient space $\mathbb C$ and so is $\mathbb R$. But $f$ is supposed to be continuous on any subset of $\mathbb R$ as well. Now if I restrict $f$ to $(-4\pi, 4\pi)\to \mathbb C$ given by the same mapping, I get the same closed circle, but its preimage is now not closed.   Or am I supposed to be viewing this interval with a subspace topology and considering it as the whole space, in which case the interval is both open and closed. Finally, if this is the case, am I correct in stating that a continuous function may map an open set to a closed one if the mentioned open set is also closed?

You need to distinguish between the two different functions $f:\mathbb R \to \mathbb C$ given by $f(x)=e^{ix}$ and its restriction $g=f|_{(-4\pi, 4\pi)}$. The definition of continuity is sensitive to the domain and its topology. So, both of these functions are continuous since the inverse image of an open in the codomain $\mathbb C$ is open in the topology of the domain. The domain for $f$ is $\mathbb R$, but the domain for $g$ is $(-4\pi,4\pi)$, and indeed $(-4\pi, 4\pi)$ is closed in this space.

Indeed, restricting the domain formally gives you a function $\tilde f: (-4\pi,4\pi) \to \Bbb C$. Then indeed $\tilde f$ will be continuous.

As to your other point, we have no control on what a continuous mapping maps things to. We only know that the preimage of any open is again open.

But even in the present case, a part of the image $S^1$ in $\Bbb C$ (say the image of an open set) will neither be open nor closed in $\Bbb C$, because it does not contain all its limit points (hence not closed), but has empty interior (hence cannot be open).


How to draw a intersection of two cones graphically What is the best software we can use to draw the intersections of two cones such as ellipses and hyperbolas mathematically and clearly.

Please send some drawaings and software you have used?

Best software is "Grapher" on Macintosh in the utilities folder.


Why did my friend lose all his money? Not sure if this is a question for math.se or stats.se, but here we go:

Our MUD (Multi-User-Dungeon, a sort of textbased world of warcraft) has a casino where players can play a simple roulette.

My friend has devised this algorithm, which he himself calls genius:

Bet 1 gold
If you win, bet 1 gold again
If you lose, bet double what you bet before. Continue doubling until you win.

He claimed you will always win exactly 1 gold using this system, since even if you lose say 8 times, you lost 1+2+4+8+16+32+64+128 gold, but then won 256 gold, which still makes you win 1 gold.

He programmed this algorithm in his favorite MUD client, let it run for the night. When he woke up the morning, he was broke.

Why did he lose? What is the fault in his reasoning?

Suppose, for simplicity, that the probability of winning one round of this game is $\frac{1}{2}$, and the probability of losing is also $\frac{1}{2}$. (Roulette in real life is not such a game, unfortunately.) Let $X_0$ be the initial wealth of the player, and write $X_t$ for the wealth of the player at time $t$. Assuming that the outcome of each round of the game is independent and identically distributed, $(X_0, X_1, X_2, \ldots)$ forms what is known as a martingale in probability theory. Indeed, using the bet-doubling strategy outlined, at any time $t$, the expected wealth of the player at time $t + 1$ is
$$\mathbb{E} \left[ X_{t+1} \middle| X_0, X_1, \ldots, X_t \right] = X_t$$
because the player wins or loses an equal amount with probability $\frac{1}{2}$ in each case, and
$$\mathbb{E} \left[ \left| X_t \right| \right] < \infty$$
because there are only finitely many different outcomes at each stage.

Now, let $T$ be the first time the player either wins or goes bankrupt. This is a random variable depending on the complete history of the game, but we can say a few things about it. For instance,
$$X_T = \begin{cases} 0 & \text{ if the player goes bankrupt before winning once} \\
 X_0 + 1 & \text{ if the player wins at least once} \end{cases}$$
so by linearity of expectation,
$$\mathbb{E} \left[ X_T \right] = (X_0 + 1) \mathbb{P} \left[ \text{the player wins at least once} \right]$$
and therefore we may compute the probability of winning as follows:
$$\mathbb{P} \left[ \text{the player wins at least once} \right] = \frac{\mathbb{E} \left[ X_T \right]}{X_0 + 1}$$
But how do we compute $\mathbb{E} \left[ X_T \right]$? For this, we need to know that $T$ is almost surely finite. This is clear by case analysis: if the player wins at least once, then $T$ is finite; but the player cannot have an infinite losing streak before going bankrupt either. Thus we may apply the optional stopping theorem to conclude:
$$\mathbb{E} \left[ X_T \right] = X_0$$
$$\mathbb{P} \left[ \text{the player wins at least once} \right] = \frac{X_0}{X_0 + 1}$$
In other words, the probability of this betting strategy turning a profit is positively correlated with the amount $X_0$ of starting capital – no surprises there!

Now let's do this repeatedly. The remarkable thing is that we get another martingale! Indeed, if $Y_n$ is the player's wealth after playing $n$ series of this game, then
$$\mathbb{E} \left[ Y_{n+1} \middle| Y_0, Y_1, \ldots, Y_n \right] = 0 \cdot \frac{1}{Y_n + 1} + (Y_n + 1) \cdot \frac{Y_n}{Y_n + 1} = Y_n$$
by linearity of expectation, and obviously
$$\mathbb{E} \left[ \left| Y_n \right| \right] \le Y_0 + n < \infty$$
because $Y_n$ is either $0$ or $Y_{n-1} + 1$.

Let $T_k$ be the first time the player either earns a profit of $k$ or goes bankrupt. So,
$$Y_{T_k} = \begin{cases} 0 && \text{ if the player goes bankrupt} \\
Y_0 + k && \text{ if the player earns a profit of } k \end{cases}$$
and again we can apply the same analysis to determine that
$$\mathbb{P} \left[ \text{the player earns a profit of $k$ before going bankrupt }  \right] = \frac{Y_0}{Y_0 + k}$$
which is not too surprising – if the player is greedy and wants to earn a larger profit, then the player has to play more series of games, thereby increasing his chances of going bankrupt.

But what we really want to compute is the probability of going bankrupt at all. I claim this happens with probability $1$. Indeed, if the player loses even once, then he is already bankrupt, so the only way the player could avoid going bankrupt is if he has an infinite winning streak; the probability of this happening is
$$\frac{Y_0}{Y_0 + 1} \cdot \frac{Y_0 + 1}{Y_0 + 2} \cdot \frac{Y_0 + 2}{Y_0 + 3} \cdot \cdots = \lim_{n \to \infty} \frac{Y_0}{Y_0 + n} = 0$$
as claimed. So this strategy almost surely leads to ruin.

I suppose he was betting on a red/black or even/odd system. Now, there is a "0" which is green/not even or odd, so if you hit that zero, you lose.

There are 36 numbers you can hit, plus that 0 . Even if you chose red or black, even or odd, there are bigger chances to lose than win.


Prove that one of the lines represented by $ax^2+2hxy+by^2=0$ will bisect the angle between the coordinate axes if $(a+b)^2=4h^2$. Prove that one of the lines represented by $ax^2+2hxy+by^2=0$ will bisect the angle between the coordinate axes if $(a+b)^2=4h^2$.

Solution
I calculated the two lines represented by $ax^2+2hxy+by^2=0$ as follows;
here.
$$ax^2+2hxy+by^2=0$$
Multiplying by $a$ on both sides and adding $h^2y^2$ to both sides :
$$ax+hy=\pm y\sqrt{h^2-ab}.$$

What should I do next?

$$(a+b)^2=4h^2 \Rightarrow a+b=\pm 2h$$
Let $$ax^2+2hxy+by^2=0$$
be factorized as $$(y-m_1x)(y-m_2x)=0$$
Then,
$$m_1+m_2=-\frac{2h}{a}$$
$$m_1m_2=\frac{b}{a}$$
We have,
$$1+\frac{b}{a}=\pm \frac{2h}{a}$$
So,
$$1+m_1m_2=\pm (m_1+m_2)$$
If either of $m_1$ or $m_2$ is zero, the other must be $\pm 1$, which means we are done.

If none of them are zero, let $m_1=km_2$.
$$1+km_2^2=\pm m_2(1+k)$$
$$km_2^2 \pm m_2(1+k) +1=0$$
which gives $m_2=\pm 1, \pm \frac{1}{k}$
which means $m_1=\pm \frac{1}{k},\pm 1$ correspondingly.

Thus at least one of them must bisect the coordinate axes.

If one of the line of
$$ax^2 +2hxy+by^2=0 \tag{1}$$
bisects the coordinate axes then the eqn. of that line might be $y=x$ OR, $y =-x$.
Combining these eqns we get $y=\pm x$ and substituting this value of $y$ in eqn. (1) we get
\begin{eqnarray}
 ax^2+2hx(\pm x)+by^2=0 \\
 ax^2±2hx^2+by^2=0 \\
 x^2(a±2h+b)=0 \\
 a±2h+b=0 \\
 a+b=±2h
\end{eqnarray}

Also, $(a+b)^2=4h^2$


Difficult Derivative? I'm in a single-variable calculus course, in which we recently covered logarithmic differentiation.  The professor proved it that works when $f(x)>0$, and when $f(x)<0$.  I've been trying to find a way to derive that kind of function when $f(x)=0$, but I'm not sure if it's possible, or what.  I've thought of this example, that resists all my efforts to differentiate, but seems to be differentiable, (and even appears to have a value of zero).

Find

  $$f'\left(\frac{3\pi}{2}\right)\quad \rm where \quad f(x)=(\sin{x} + 1)^x .$$

Is there any way I can find this derivative (if it exists), beyond numerically computing the limit from the definition of the derivative? Or, vice versa, how can I prove that this derivative doesn't exist?

Thanks,
  Reggie

Oddly enough, this is a case where you actually do well to revert to the definition of the derivative:

$$f'\left({3\pi\over2}\right)=\lim_{h\rightarrow0}{f\left({3\pi\over2}+h\right)-f\left({3\pi\over2}\right)\over h}$$

You don't have to do any numerical computation.  Note that $\sin({3\pi\over2}+h)=-\cos h$, so for $f(x)=(1+\sin x)^x$ the definition becomes

$$f'\left({3\pi\over2}\right)=\lim_{h\rightarrow0}{(1-\cos h)^{{3\pi\over2}+h}\over h}$$

since $f({3\pi\over2})$ is clearly $0$.  But for (small) nonzero $h$, we have $$0\lt|1-\cos h|^{{3\pi\over2}+h}\lt|1-\cos h|$$since $|1-\cos h|\lt1$ and $3\pi/2\gt1$, from which, using the definition of the derivative for the cosine function and the fact that $(\cos x)'=-\sin x$, it follows that

$$0\le \lim_{h\rightarrow0}\left|{(1-\cos h)^{{3\pi\over2}+h}\over h}\right|\le\left|\lim_{h\rightarrow0}{1-\cos h\over h}\right|=|\sin0|=0$$

hence

$$ f'\left({3\pi\over2}\right)=0$$

You can find the first derivative algebraically using the product and chain rules.

I rewrote the function as $y=e$ to the power of $\ln \sin x +1$. Then I rewrote this as e to the power of u. I used the chain rule to find $\frac{dy}{du}$, which was the original function (as the derivative of e to the x is just e to the x). The derivite of the inner function $\ln \sin x+1$ required use of the product and chain rules. We can find the derivitive of the overall function to be $\ln \sin x+1$ times $\cos x \csc x +2 \cos x$. Then we can substitute in for $x$. Notice how $e$ to the power of $ln 0$ is just $0$.


Condition on coefficients of polynomial to guarantee special zero-distribution Consider polynomial below:

$P(s) = a_4s^4 + a_{3}s^{3} + a_2s^2 + a_1s + a_0 $

The question is: Under which conditions on coefficient of above polynomial, we claim that the zeros of polynomial constitute such a special distribution like this :

$ z_1  = α $

$z_2= -β $

$z_{3,4} = -ζ∓jγ$

where $  |z_1 |<|z_2 |$  and    $  (α,β,γ,ζ>0)$ .

Any Ideas would be appreciated.

First note that $a_4 \ne 0$, otherwise there couldn't be four zeros.  To simplify, divide the polynomial by $a_4$, so we can assume $a_4 = 1$, i.e. it's a monic polynomial.

The set of roots is symmetric under complex conjugation, so your monic polynomial is as well: the $a_i$ are real.  Since $0$ is not a root, $a_0 \ne 0$.

Now you want one root on the positive real axis, one on the negative real axis,
and the other two in the left half plane off the real axis.  Let $p_0 = P$, $p_1 = p'$, $p_2$, $p_3$, $p_4$ be the Sturm sequence for $P$ (in degenerate cases there might be fewer than these).  The number of roots in the positive real axis
and in the negative real axis can be obtained by Sturm's theorem in terms of the number of of sign changes in the Sturm sequence at $x = +\infty$, at $x=0$ and as $x = -\infty$: there should be one more sign change at $-\infty$ than at $0$, and one more at $0$ than at $+\infty$.

For there to be three roots in the left half plane and one in the right half plane, take the generalized Sturm sequence using the Euclidean algorithm starting with $R_0(z) = -a_2 z^2 + a_0$ and $R_1(z) = -z^3 + a_1 z$: there should be two more sign changes at $+\infty$ than at $-\infty$.


Having difficulty with statistics problem (Stem-leaf) diagram I never used this website before, but if anyone could help me with this then that will be great. Also, if I tagged my question in the wrong topic, please point me in the right direction as well.

What numbers represents following Stem-and-Leaf Diagram? Stems: 2, 3, 4 Leaf: 0,3,5,6: 1, 4, 6: 7,9

A. 2, 3, 4, 0, 1, 7, 3, 4, 9, 5, 6, 6

B. 0, 3, 5, 6, 1, 4, 6, 7, 9

C. 20, 22, 24, 25, 31, 33, 34, 39, 41, 42

D. 20, 23, 25, 26, 31, 34, 36, 47, 49

If your diagram looks like the following:

\begin{align}
2 &\mid 0\ 3\ 5\ 6 \\
3 &\mid 1\ 4\ 6 \\
4 &\mid 7\ 9
\end{align}

Then the stems are the tens digits and the leaves that appear to the right of each stem are the units digit for that stem.

So, from the top stem, we get $20, 23, 25, 26$. From the middle stem, we get $31,34,36$. From the bottom stem, we get $47, 49$.

When making a stem plot by hand, you should always specify the position of the decimal point
at the head of a stem plot. Here are examples using R (first) and Minitab to show how this may be handled in software output.

 x = c(20, 23, 25, 26, 31, 34, 36, 47, 49)
 stem(x)

   The decimal point is 1 digit(s) to the right of the |

   2 | 0356
   3 | 146
   4 | 79

 y = x/10
 y
 ## 2.0 2.3 2.5 2.6 3.1 3.4 3.6 4.7 4.9
 stem(y)

## The decimal point is at the |

   2 | 0356
   3 | 146
   4 | 79

Sometimes, the stem is 'split'. Then leaves 0, 1, 2, 3, 4 appear
opposite the first mention of the stem number, and 5, 6, 7, 8, 9
appear opposite the second.  Each software package has
its own rules about splitting, which can be overridden if you like.

In Minitab output, the left column accumulates
the number of observations from the top or bottom, and the
number in parentheses signifies the number of observations
on the line that contains the median.

 MTB > set c10
 DATA> 20, 23, 25, 26, 31, 34, 36, 47, 49
 DATA> end
 MTB > stem c10

 Stem-and-leaf of C10  N  = 9
 Leaf Unit = 1.0

  2   2  03
  4   2  56                # Notice 4 + 2 + 3
 (2)  3  14                # gives the total
  3   3  6                 # sample size n = 9
  2   4
  2   4  79

 MTB > stem c10;
 SUBC> incr 10.  # subcommand overrides default splitting

 Stem-and-Leaf Display: C10

 Stem-and-leaf of C10  N  = 9
 Leaf Unit = 1.0

  4   2  0356
 (3)  3  146
  2   4  79

Suppose that, on average, a huge international construction company experiences 35
industrial accidents a day worldwide. Here is a simulation in R of the number
of accidents a day (assuming a Poisson distribution) over 100 days, with results shown in a stem plot. Notice that R 'splits the stem' here to
keep lines of leaves from getting too long. Numbers of accidents
were 21, 24, 25, ..., 45, 46, 46.

 w = rpois(100, 35)   # simulates 'fake' data
 stem(w)

 ## The decimal point is 1 digit(s) to the right of the |

 2 | 14
 2 | 5555566677777888888899
 3 | 001222222223334444444
 3 | 55555566666666777778888888899999
 4 | 00000111112222233334
 4 | 566

 length(w); mean(w);  var(w)
 ## 100           # n
 ## 34.71         # sample mean
 ## 33.86455      # sample variance

A typical result with Poisson data is for the sample mean
and variance to be roughly the same number.


Question in Complex Analysis Let $f$ be a analytic in Domain $D$ which is an open disk of radius 1 centered at $z=0$ , it is given that in Domain $D$ , $|f(z)|\leq 1-|z|$ then prove that $f(z)=0$ in Domain $D$

It is asked that we have to prove $f(z) = 0$ using Cauchy's Integral Formula

I have attached my solution here My solution Please verify this!!

Your solution is correct. A few notes: the second page could be substituted with a reductio ad absurdum: if $f(0)\neq 0$ and $f$ is constant, there is a $z:1-|z|<|f(0)|=|f(z)\le 1-|z|$.

In general, there's a faster approach to this kind of problems, by means of the cauchy inequality (which is an easy corollary of Cauchy integral formula) : given a function $f$ which is holomorphic in $B(z_0,R)$, we have that, $\forall r<R$:

$$|f^{(n)}(z_0)|\le n! r^{-n}\underset{|z-z_0|=r}{\sup}|f|$$

By Cauchy inequality and the hypotesis,

$|f^{(n)}(0)|\le n!r^{-n}\underset{|z|=r}{\sup}|f|\le n!r^{-n}(1-r)$

Letting $r\to 1^-$, we obtain

$$f(0)=0\\
\forall_{n>0}f^{(n)}(0)=0$$

Thus we have proved that $f$ is zero on the open unit disc

Let $z\in D$ and $d$ be the distance from $C_r$ to $z$.

By CIF, $|f(z)|=|1/(2\pi i)\oint_{C_r}f(\zeta)/(\zeta-z)\DeclareMathOperator{\d}{d}\d\zeta|\le\oint_{C_r}|f(\zeta)/(\zeta-z)|\d\zeta\le\oint_{C_r}(1-|\zeta|)/|\zeta-z|\d\zeta\le(1-r)/d\to0$ as $r\to1$.

This is essentially what you did.


Prove that if $3\mid a^2+b^2$ then $3\mid a$ and $3\mid b$. For elements $a$ and $b$ in the ring $\Bbb{Z}$ prove that if $3\mid a^2+b^2$ then $3\mid a$ and $3\mid b$.

I tried proving it but I just don't manage to. Maybe I am missing some basic claims in the number Theory?

I would appreciate hints and, generally, your help.

I am sorry, there was another data I was given, I was fixed and resent to me.

The original question was to prove that $c\mid a^2+b^2$ implies $c\mid a$ and $c\mid b$, which as many answers show isn't true.

But this is true if you take the assumption that there isn't a square root of $-1$ mod $c$. Take mod $c$ of the equation to get that

$$a^2 = - b^2 \mod c$$

which can only have a solution $a \neq 0$ or $b \neq 0$ if $-1$ has a square root mod $c$. I.e., there is a number $x$ such that $x^2 = -1 \mod c$.

Indeed, assuming $a$ and $b$ are both relatively prime to $c$, then we can take $b^{-1}$ to get

$$(b^{-1}a)^2 = -1 \mod c.$$

But if no solution exists, our assumption of co-primality cannot hold. If we take the prime factors of $c$ and use modular arithmetic, then we can show $\gcd(a,p) \geq p$ for each prime factor. (Note that $x^2=-1 \mod p$ has a solution if it does mod $c$.) Now this only means $\prod p_i$ divides $a$ and $b$, but if we cancel out these common factors we should be able to repeat the procedure to get full divisibility.

Seeing your edit, the solution follows from my answer by showing mod $3$ that no solution $x^2=-1$ exists. It's a simpler case since $3$ is prime. Sorry I didn't make the above argument tighter, but you should be able to work out the specific (and easier) case from the above.

It's not hard to see the general claim is if and only if as well.

Here, we see that $(a^2 + b^2, 2ab) = ((a + b)^2, 2ab)$ .
By the elementary definition of the gcd of two numbers, we can safely assume that $gcd = 3$ or $3$ is a common divisor. In doing so, we get $3$ divides $ab$ and also divides $(a + b)^2$. Any square number divisible by $3$ must be of the form $3q$. So any case of "$a$" being even or "$b$" being even automatically gets removed and there we have the final answer


Mismatch between distances on a great circle? I was trying to derive by myself the distance on a great circle.

To confirm my calculations are correct I was comparing against the formula (4) in this link, which is something like

$$
\cos(\delta_1)\cos(\delta_2) \cos(\lambda_1 - \lambda_2) + \sin(\delta_1) \sin(\delta_2)
$$

According to wikipedia instead the formula looks more:

$$
hav(\Theta) = hav(\varphi_2 - \varphi_1) + \cos(\varphi_1) \cos(\varphi_2) hav(\lambda_2 - \lambda_1)
$$

Where

$$
hav(\theta) = \sin^2\left( \frac{\theta}{2}\right) = \frac{1 - \cos(\theta)}{2}
$$

They don't seem equivalent to me, I've tried to use some trig to prove they're an identity, but unless I'm making mistakes it doesn't seem the case.

My derivation leads exactly to the first formula I've shown. The question is... which one is correct, are they the same?

Note that
$$
\operatorname{hav}\Theta = \operatorname{hav}(\varphi_1-\varphi_2)+\cos(\varphi_1)\cos(\varphi_2)\operatorname{hav}(\lambda_1-\lambda_2)
$$
iff
$$
1-\cos\Theta = 1-\cos(\varphi_1-\varphi_2)+\cos(\varphi_1)\cos(\varphi_2)[1-\cos(\lambda_1-\lambda_2)]
$$
iff
$$
\cos\Theta = \cos(\varphi_1-\varphi_2)+\cos(\varphi_1)\cos(\varphi_2)[\cos(\lambda_1-\lambda_2)-1],
$$
and recall $\cos(\varphi_1-\varphi_2)=\cos\varphi_1\cos\varphi_2-\sin\varphi_1\sin\varphi_2$.


Homogeneity does not suffice for a map between vector spaces to be linear The following problem is taken from Sheldon Axler's book Linear Algebra Done Right, more precisely Exercise 1. from Chapter 3:

Problem: Give an example of a function $f : \mathbb{R}^2 \to \mathbb{R}$ such that $f(av) = af(v)$ for all $a \in \mathbb{R}$ and all $v \in \mathbb{R}^2$ but $f$ is not linear.

I tried to include things like absolute values and square roots in order to handle the homogeneity, but I did not had any success in constructing such an example yet by doing so.

Do you know such an example? Thank you in advance!

consider the map: $$f:\mathbb{R}^2 \to \mathbb{R}\\ (x,y) \mapsto \begin{cases}
 x \text{ if } y=0 \\
0 \text{ else }\end{cases} $$
then this is homogenous $f(a(x,y))(f(ax,ay))=\begin{cases}ax \text{ if } y=0 \\ a0\text{ else }\end{cases}$, but not linear, since $$f(1,1)+f(1,-1)=0 +0 = 0 \neq f(2,0) =f((1,1)+(1,-1)).$$


Compute probabilities of balls not being drawn from urns Independent events $A_1$ through $A_n$, where $A_n$ is event that nth ball is not drawn in a random draw from urn full of $n^2$ balls labeled $1, 2, \dots, n^2$

What is $\lim_{n \to \infty} P(A_n)$?

What is the $P(\bigcap_{n=2}^{\infty} A_n)$?

What does continuity have to do with this?

Anyway

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

Thus $\lim_{n \to \infty} P(A_n) = \lim_{n \to \infty} \frac{n^2-1}{n^2}$

$P(\bigcap_{n=1}^{\infty} A_n) = \prod_{n=1}^{\infty} P(A_n)$ by independence

$\to \prod_{n=1}^{\infty} P(A_n) = P(A_1)P(A_2)P(A_3) \dots = 0P(A_2)P(A_3) \dots = 0$

$P(\bigcap_{n=2}^{\infty} A_n) = \prod_{n=2}^{\infty} P(A_n) = \prod_{n=2}^{\infty} \frac{n^2-1}{n^2}$

Can you take it from here?


Probability of getting 3 cards in the same suit from a deck When three cards are randomly selected at a time from a standard deck
  of 52 playing  cards, what is the probability that all of these three
  cards are in the same suit (heart,  diamond, spade, or club)?

I'm mortified to ask for help again about probability math. If you look at my profile you probably can see that I've asked several questions about probability. When I thought I can wrap my head around this kind of math, then I had this question and I knew I did not.

Can you please give me some hints on this question and advice me some useful tips for learning to solve this kind of math?

P/s : I already had the answer for this, but I don't know how to solve it!!!

Three cards are selected from a standard deck of $52$ cards. Disregarding the order in which they are drawn, the possible outcomes are $\binom{52}{3}$. Out of these, how many include all cards of the same suit (say hearts)? There are $\binom{13}{3}$ ways in which you can get all 13 heart cards.

Since there are 4 suits, there are $4\binom{13}{3}$ ways in which all cards drawn are of the same suit. Thus the probability is:

$$\frac{4\binom{13}{3}}{\binom{52}{3}}\approx 5.18\%$$

$$4~\frac{^{13}C_3}{^{52}C_3}=0.0518 $$
$4$ suits per deck

$3$ cards required in a suit


an outline of "intuitive mathematics"? This question is related to the third answer in this post.

There seems to be a difference between the intuitive idea of a thing (such as a function) and "models" of that thing in mathematics (such as the way we view functions as being a kind of set of ordered pairs). Another instance of this is the natural numbers and the set-theoretic construction of the natural numbers. If our set theory does not have an axiom of infinity which allows us to define $\mathbb{N}$, then are we disallowed the use of induction? Or is induction something "we just believe" independent of any model? A concrete question that comes to mind is: if there was no axiom of infinity, could we prove the statement "If $X_1,\dots,X_n$ are sets, then there exists a set $\{X_1,\dots,X_n\}$"? (this is a statement that I believe requires induction and the axiom of infinity). Could "$X_1,\dots,X_n$" even be stated without the formal notion of $\mathbb{N}$?

My questions are this:

1) Do the formal models, such as set theory, actually justify "intuitive mathematics"? My feeling is that they don't. I take the numbers $0,1,2,\dots$ and induction for granted (conceptually prior to set theory).

2) If not, then why is the formality of set theory often used as justification for "intuitive mathematics"? For example, you might see authors explaining induction by mentioning the inductive set, which is kind of irrelevant if induction is something we just take for granted.

3) Are there textbooks or documents which attempt to outline what constitutes "intuitive mathematics" without appealing to models or the formality of set theory?

I guess I'm interested in understanding "what comes first". What should we all take for granted? What is the metatheory (is this the right word?)?

There is no way to "prove" or "justify" one's intuition. What you can do, is view properties that you would expect something to have, and check to see that the "set theoretic" construction agrees with them (with proof.) Similarly, I think that you can take a construction that you find "intuitive" and prove that it is equivalent to a more "formal," or "different" approach." In some sense, you could make your intuition precise, and frame it in mathematical terms, and then prove its equivalence to a more formal definition.

I think that this is a good example.

Alternatively, one might feel that the one of the three constructions of $\mathbb{R}$ are intuitive, and prove them equivalent to the other two:

The completion of all Cauchy sequences in $\mathbb{Q}$.
Construction by Dedekind Cuts
Define $\mathbb{R}$ to be an Ordered Field with the Least Upper Bound Property.

Poincare has some nice views on intuition/logic in mathematics: Poincare on intuition.

Here is a brief quotation:

And then," say the philosophers, "it still remains to show that the object which corresponds to this definition is indeed the same made known to you by intuition; or else that some real and concrete object whose conformity with your intuitive idea you believe you immediately recognize corresponds to your new definition. Only then could you affirm that it has the property in question. You have only displaced the difficulty."

That is not exactly so; the difficulty has not been displaced, it has been divided. The proposition to be established was in reality composed of two different truths, at first not distinguished. The first was a mathematical truth, and it is now rigorously established. The second was an experimental verity. Experience alone can teach us that some real and concrete object corresponds or does not correspond to some abstract definition. This second verity is not mathematically demonstrated, but neither can it be, no more than can the empirical laws of the physical and natural sciences. It would be unreasonable to ask more."

Perhaps you would also be interested in intuitionsim, as developed by Brouwer.

Induction still works in the absence of the axiom of infinity.

Without AxInf, you still have $0 = \emptyset, 1 = \{0\}$, $2 = \{0, 1\}$, etc.

For a set $X$, if we define the the successor of $X$ to be $X \cup \{X\}$, then every set has a successor. In short, we have a collection whose members are exactly $0, 1, 2, 3, \dots$ This collection is not a set, but we can still do induction. The collection of natural numbers is still well-ordered, and they satisfy the Peano axioms.

See Andrés E. Caicedo's answer to this question. What are the consequences if Axiom of Infinity is negated?. In particular, "ZFC with infinity replaced by its negation is biinterpretable with PA, Peano Arithmetic."


Combining two distributions to get a single one? Consider the following experiment:

Let $X$ be a random variable which has a beta distribution with some fixed parameters.

Let $Y$ be a Bernoulli random variable with parameter $X$.

In other words, we have a two step process:
1) get a number $x \in [0,1]$ from a beta distribution
2) get a number $y \in \{0,1\}$ from a coin flip with heads probability $x$.

Is it possible to say how $y$ will be distributed (in terms of the beta distribution parameters, I suppose)?

Yes. This is a very common application of Bayesian inference. In fact the Wikipedia article for prior probability has your exact example in the lede. The beta distribution is described as a prior distribution, and the Bernoulli distribution is described as a sampling distribution.

Here is the answer to your question: with a $X \sim \operatorname{Beta}(a,b)$ prior and a $\operatorname{Bernoulli}(X)$ likelihood, the posterior distribution is $\operatorname{Beta}(a',b')$ where $a'$ is the number of successes observed and $b'$ is the number of failures observed.

On the first observation no data has been collected yet, so $a'=a$ and $b' = b$ and we just have $\operatorname{Beta}(a,b)$.  This is also because the expected value of Bernoulli distribution is just $p$.


10 fish are caught. (Probability Question) Question says: Suppose that 10 fish are caught at a lake that contains 5 distinct types of fish.

a) how many different outcomes are possible, where an outcome specifies the numbers of caught fish of each of the 5 types?

(okay so I am not 100% clearly understood after reading the question. So suppose there are like 2 A, 2 B, 2 C, 2 D, 2 E fish. The outcome is 2 of each 5 types of fish. So is it like 5x5x5x5x5x5x5x5x5x5, 5^10? but then again what if some types are not caught?? It sounds so simple but it's confusing me a lot. Any tips would be helpful.)

b) how many outcomes are possible when 3 of the 10 fish caught are trout?
(so three of the 10 fish have to be one type. That means 4x4x4x4x4x4x4x(3x2x1)? is this correct?)

Thank you!

a) One possible outcome is (2,2,2,2,2), which indicates 2 of type A, 2 of type B, etc.  Another possible outcome is (9,0,1,0,0), which indicates 9 of type A, 1 of type C.  Hence you have an ordered list of five nonnegative integers, whose sum is 10.  Count these.

b) Now we know that the first element of the list is 3, so we have four remaining digits, that sum to 7.  For example (3,7,0,0,0) or (3,2,2,2,1).  Hence you can count ordered lists of four nonnegative integers, whose sum is 7.  Count these.

I am unsure whether my answer is correct or not. Someone please comment. But first let me put the full question as it is from its source.

Suppose that 10 fish are caught at a lake that contains 5 distinct types of fish.
(a) How many different outcomes are possible, where an outcome specifies the numbers of caught fish of each of the 5 types
(b) How may outcomes are possible when 3 of the 10 fish caught are trout?
(c) How many when at least 2 of the 10 are trout?

Let me add one more to above (please comment if the solution I wrote for this one is correct or not)

(d) What will be the solution to (c) when 5 different types of fishes are caught?

Solution
(a) This is star and bar theorem 2 problem. Let us assume there are 10 stars (representing fishes) put in a row. Now we separate them into 5 groups (representing fish types) by putting 4 bars among them. (Notice that if we put bar before first star on extreme left, it will denote that the first group is empty. Same for bar put after last star and two bars put together without any star between them) So we have total 14 positions (10 stars and 4 bars) out of which we have to select 4 for bars which can be done in $\binom{10+5-1}{5-1}=\binom{14}{4}=1001$ ways.
(b) When 3 out of 10 are trout, there can be maximum 4 different types of fishes in remaining 7. With same above logic, this can be done in $\binom{7+4-1}{4-1}=\binom{10}{3}=120$ ways.
(c) When at least 2 out of 10 are trout, there can be maximum 5 different types of fishes in remaining 8. This can be done in $\binom{8+5-1}{5-1}=\binom{12}{4}=495$ ways
(d) This becomes star and bar theorem 1 problem. When there is exactly 2 trouts, there should be 4 types of fishes in remaining 8. So not we order 8 stars in a row. They will have 7 gaps separating them. We put 3 bars in those gaps to separate them into 4 groups. This can be done in $\binom{8-1}{4-1}=\binom{7}{3}=35$ ways. If there are more than 2 trouts, then the remaining 8 fishes must contain 5 types of fishes. This can be done in $\binom{8-1}{5-1}=\binom{8-1}{5-1}=\binom{7}{4}=35$ ways. By product rule we get $35\times 35=1225$


If $a, b \in G$ are such that $a^4=e$ and $a^2*b=b*a$ show that $a=e$. Let $(G,*)$ be a group. If $a, b \in G$ are such that $a^4=e$ and $a^2*b=b*a$ show that $a=e$.
$a^2*b=b*a \implies ab=a^{-1}*b*a$.
I can't proceed further.

You just have to combine your two equations!
If $a^4 = e$ and $a^2b = ba$, then
$$
b = a^4b = a^2(a^2b) = a^2(ba) = (a^2b)a = (ba)a = ba^2
$$
and therefore $a^2 = e$. But if $a^2 = e$, then $b = a^2b = ba$, so $a=e$.
Hope that helps,

Proof:
As a⁴=e. Then
b=a⁴b
b=(a²)²b
since a²b=ba
b=ba²
Since G is a group
b=a²b
Since
a²b=ba
b=ba=>a=e.


Exactly 5 Platonic solids: Where in the proof do we need convexity and regularity? The famous statement that

    Only five convex regular polyhedra exist.

is usually proven as follows:

Let $P$ be a convex regular polyhedron with

V vertices,
E edges and
F faces.

Moreoever, let

n be the number of edges of each face and
c be the number of edges which meet at a vertex.

Then $F=\frac{2E}{n}$ and $V=\frac{2E}{c}$ and $c\geq 3, n\geq 3$.

Substituting this into Euler's polyhedron formula
$$
V-E+F=2,
$$

and doing some easy calculations, one gets that only

$$
(V,E,F)\in\{(4,6,4),(8,12,6),(20,30,12),(6,12,8),(12,30,20)\}
$$
are possible combinations.

So far, so good.

    But where do we need that $P$ is convex and regular?

(I think the regularity is at least used for $n\geq 3, c\geq 3$, isn't it?)

As already said in the comments, regularity means being composed of equal faces, thus enabling to connect the numbers $V$, $E$ and $F$ by some algebraic relations. This is the left part of Euler's identity
$$V-E+F=2$$
Now, convexivity is, in fact, the right-hand side.

In general, for a surface $S$, the formula reads
$$V-E+F=\chi(S)$$
where $\chi$ is the Euler characteristic (defined by the equation above or, alternatively, by the alternating sum of dimensions of homology groups).

If a polyhedron is convex, it can be proven that it's boundary is homeomorphic (topologically equivalent) to a sphere $\mathbb{S}^2$, and $\chi(\mathbb{S}^2)=2$, providing the right part of Euler's equation.

So, convex is just a simplification; the classification really works for all polyhedra homeomorphic to a ball. For some other topology, a different classification may arise.

There are five Platonic Solids because their definition restricts them to polyhedra.

  A Platonic solid is a regular, convex polyhedron. It is constructed by congruent regular polygonal faces with the same number of faces meeting at each vertex.

If all geometric structures are included the result is the twelve FFELLONIC FORMS, a complete series ranging from a triangle to the densest possible space filling honeycomb.


Is $\tau = \inf \{n :X_1+\cdots+X_{n-1}\leq 1\leq X_1+\cdots+X_{n-1}+X_n \}$ a stopping time? Let $X_1, X_2, ...$ be a sequence of i.i.d. non-negative random variables. Let $$\tau = \inf \{n :X_1+\cdots+X_{n-1}\leq 1\leq X_1+\cdots+X_{n-1}+X_n \}.$$
  Is $\tau$ a stopping time for the sequence $X_1, X_2, ...$?

Using the definition of a stopping time, I should verify whether the event $\tau = n$ may or may not be independent of $X_{n+1}, X_{n+2}, ... $.

The expression for $\tau$ makes it difficult for me. Any hints how to solve this problem?

This what I come up with so far (if I'm wrong please tell me!):

by definition: Let $F_n$ be the $\sigma$-algebra generated by $X_1, X_2, ... X_n$.  So $F_n$ is a filtration.

Let $Y_{n-1} = \sum_{i = 1}^{n-1}X_i$ and $Z_n = \sum_{i = 1}^{n}X_i$. So both $Y_n$ and $Z_n$ are adapted to $F_n$.

The definition of stopping time is: $\{\tau = n\} \in F_n$.

Edit: Thanks to Did, I needed to clarify this:

Let $\{Y_k \leq 1\}$ be the set of all $X_k$ such that their sum is less or equal to 1, and $\{Z_k \geq 1\}$ be the set of all $X_k$ such that their sum is greater or equal to 1 $\forall k = 1, 2, 3... n$. Then:

$\tau = \{Y_{1} \leq 1\} \cap \{Y_{2} \leq 1\} ... \cap \{Y_{n-1} \leq 1\} \cap \{Z_{1} \geq 1\} \cap \{Z_{2} \geq 1\} ... \cap \{Z_{n} \geq 1\} \in F_n$ because $F_n$ is closed under intersections.

So $\tau$ is a stopping time. In fact it is a "First hitting time".

You will need $X_1, X_2... X_n$ independence to prove if $Y_n$ and $Z_n$ are a martingale.


$\lim_{n\to\infty}a_nb_n=? $ given that $\lim_{n\to\infty}a_n=0$ and $\lim_{n\to\infty}b_n=\infty$ A. $\lim_{n\to\infty}a_nb_n=?$ given that $\lim_{n\to\infty}a_n=0$ and $ \lim_{n\to\infty}b_n=\infty$
B. $\lim_{n\to\infty}{a_n \over b_n}=?$ given that $\lim_{n\to\infty}a_n=0$ and $\lim_{n\to\infty}b_n=0$
C. $\lim_{n\to\infty}{a_n \over b_n}=?$ given that $\lim_{n\to\infty}a_n=\infty$ and $\lim_{n\to\infty}b_n= \infty$

These seem to be difficult situations I come across when I have to evaluate limits.

Some examples,

$\lim_{n\to\infty}n^a\sin(1/n^b)$ with $a,b>1$
$\lim_{n\to\infty}{n^a\over \sin(1/n^b)}$ with $a,b>1$
$\lim_{n\to\infty}{1/n^a\over \sin(1/n^b)}$ with $a,b>1$
$\lim_{n\to\infty} nf(x_n)$ [where we only know that $\lim \ f(x_n) = 0]$

My first question is : Are the two situations  A and B  equivalent ? I feel that they are equivalent as $\lim_{n\to\infty}b_n=\infty \iff \lim_{n\to\infty}{1\over b_n}=0$, The problem is that I'm not sure if we can apply the product rule for limits to take the limit inside.

My second question is : What are some ways to solve such problems ? I only know that we can apply stolz- cesaro in some cases.

These are usually called indeterminate forms, not because we can't compute the limit, but because no general theorem exists.

Just to make a couple of easy examples for your fourth question.

Consider $b_n=\frac{1}{n}$. Then
$$
\lim_{n\to\infty} b_n=0
$$
and
$$
\lim_{n\to\infty} nb_n=\lim_{n\to\infty}n\frac{1}{n}=1
$$
If we take instead $c_n=\frac{1}{n^2}$ we have
$$
\lim_{n\to\infty} c_n=0,\qquad
\lim_{n\to\infty} nc_n=\lim_{n\to\infty}\frac{1}{n}=0
$$
With $d_n=\frac{1}{\sqrt{n}}$ we have
$$
\lim_{n\to\infty} d_n=0,\qquad
\lim_{n\to\infty} nd_n=\lim_{n\to\infty}\sqrt{n}=\infty
$$

So no general rule can be given.

You may want to check that
$$
\lim_{n\to\infty} n^a\sin\frac{1}{n^b}=
\begin{cases}
\infty & \text{if $a<b$}\\
1 & \text{if $a=b$}\\
0 & \text{if $a>b$}
\end{cases}
$$

For example, let's look at your question B, the ratio of sequences.

What is the answer when we set $a_n=\frac{1}{x}$, $b_n=\frac{1}{x}$?

What about when $b_n=\frac{1}{x^2}$?

When $a_n=\frac{1}{x^2}$ and $b_n=\frac{1}{x}$?


Confirmation on probability identity I need a confirmation regarding a probability formula:

$ P(A\cup B)=P(A\cap B)+P({ A }^{ c }\cap B)+P(A\cap { B }^{ c }) $

I am asked if this identity is true.

For me, it is true.
Can you please confirm?

Thanks in advance.

Hint Draw a Venn diagram of the sets $A$ and $B$. Color the three sets $A \cap B$, $A^C \cap B$ and $A \cap B^C$. What do you get?

[sorry, didn't see the other answer, which says the same as mine]

if you have the decomposition of $A \cup B$ into three disjoint sets, then the result follows from the additivity of the probability measure. so can you prove that:
$$
A \cup B = A \cap B + A \cap B^c + A^c \cap B
$$
where the $+$ means disjoint union?


Hypothesis testing for sample means within a normal distrubution The context of the question is that a bakery bakes cakes and the mass of cake is demoted by $X$ such that $X \sim N(300, 40^2)$. A sample of 12 cakes is taken and the mean of the sample is 292g. The question wants me to find the $p$-value and test to see if the mean has changed at 10% significance.

So I know how to carry out the test as $\overline{X_{12}} \sim N(300,\frac{40^2}{12})$, But what would the p-value I'm trying to calculate be? I know the p-value is 0.244.

The hypothesis testing:
$$H_0: \mu =300\\
H_1:\mu \ne 300 \\
z=\frac{\bar{x}-300}{40/\sqrt{12}}=-0.6928\\
p\text{-value}=P(z<-0.6928)=0.244 \ \\
\text{Reject $H_0$ if $p<\frac{\alpha}{2}$}: \ 0.244\not < 0.05 \Rightarrow \text{Fail to Reject} \ H_0.$$
Note: $p$-value calculation:

1) In MS Excel: $=NORM.S.DIST(-0.6928;1)$.

2) WolframAlpha.

3) Z table.

$\newcommand{\P}{\mathbb{P}}$It appears that you have $X\sim N(\mu, 40^2)$ (known variance), and your null hypothesis is that $\mu = 300$, with alternative hypothesis $\mu\ne 300$. For this, we will be using a two-tailed test.

Under the null hypothesis, we have $\overline{X}_{12}\sim N\left(300, \frac{40^2}{12}\right)$, or $T := \frac{\overline{X}_{12} - 300}{40/\sqrt{12}}\sim N(0,1)$. The $p$-value is the probability of getting a "more extreme" result for the test-statistic $T$ than observed under the null hypothesis (note the observed value is $\color{blue}{\frac{292-300}{40/\sqrt{12}}}$), which for our two-tailed test means that the $p$-value is

$$\P\left(|T| > \left| \frac{292-300}{40/\sqrt{12}}\right| \right) \quad \text{where }T \sim N(0,1).$$

If you compute this probability, it seems you get double the answer you wrote. So the answers probably used a one-tailed test, which would be the case if our alternative hypothesis was $\mu < 300$, in which case the $p$-value would be

$$\P\left(T < \frac{292-300}{40/\sqrt{12}} \right) \quad \text{where }T \sim N(0,1).$$

This will get you the reported answer. The reason I used a two-tailed test is because I interpreted the alternative hypothesis as being with a $\ne$ sign rather than $<$, because the wording of the question was "mean has changed".


Problem related to positive definite , symmetric , skew symmetric matrix Pick out the true statement(s):

(a) There exist $n \times n$ matrices with real entries such that $(I-(AB-BA))^n=0$.
(b) $A $ is a symmetric and positive definite $n\times n$ matrix then $(\operatorname{tr}(A))^n\ge n^n \det(A)$.
(c) Let $A$ be a $5 \times 5$ skew -symmetric matrix with real entries. Then $A$ is
singular.

(a) If $C=AB-BA$ we see that $\operatorname{tr} C = 0$. If $(I-C)^n=0$, then $I-C$ is nilpotent and all eigenvalues are zero. Hence $\operatorname{tr} (I-C) = 0$, which implies that $\operatorname{tr}C = n$, a contradiction. Hence no such matrix can exist.

(b) The arithmetic and geometric means are related by
${x_1+...+x_n \over n} \ge \sqrt[n]{x_1\cdots x_n}$, which is equivalent to
$(x_1+...+x_n)^n \ge n^n x_1\cdots x_n$. Since $\operatorname{tr} A = \lambda_1+\cdots + \lambda_n$ and $\det A = \lambda_1 \cdots \lambda_n$, we obtain the desired result.

(c) All eigenvalues of a skew symmetric matrix are imaginary. Since it is real and has odd dimension, it must have one real eigenvalue. The only number that is both real and imaginary is zero. Hence $A$ is singular.

a) never true

Note that $AB - BA$ can never have $1$ as an eigenvalue. Hence
$$
\text{If } \lambda \text{ is an eigenvalue of } I -(AB-BA) \text { then } \lambda \ne 0$$

c) If $\lambda$ is an eigenvalue, then so is $\lambda$. Any odd dimensional matrix must have a real eigenvalue. So zero is an eigenvalue. What can you conclude?

b) is easy


how to find the number of solutions for $\ln(x) =x^2/8$ using simple math principles? how to find the number of solutions for  $\ln(x) = x^2/8$ with high school level math (basic logarithm rules or something of that nature)( and without graphing $\ln(x)$ and $x^2/8$). (sorry for the poor formatting, feel free to edit)
I tried to change the equation to someting familiar to use a logarithm rule:
$$\displaylines{
x^2/8 = \ln(x) \cr
x/8^{0.5} = lan(x)^{0.5} \cr
x = 8^{0.5}*lan(x)^{0.5}
}$$
Changing the representation of the equation using the basic logarithm rule
$$\displaylines{
x^2/8 = \ln(x) \cr
x = e^{x^2/8}
}$$
But my tries got me nowhere. Any suggestions?

Let $y:=x^2$ (as $x>0$, no solution is introduced). The equation can be written
$$y=e^{y/4}.$$
The exponential is a convex function, so it has at most two intersections with a straight line.
Now as $0<e^{0/4},4>e^{4/4}$ and $12<e^{12/4}$, we have these two intersections.

try visualizing the graphs of ln(x) and x^2 / 8. The first is a log-curve, and the second in a parabola. They will  intersect at at most two points. so you will have at most two solutions.
Since the log-curve starts (or ends) at the origin, and the parabola will "circumnavigate" the origin going from -infinity to +infinity, it's safe to say that they will intersect at exactly one point.


Get a point by a given point, degree and distance I have a point $(x_1, y_1)$, an angle $A$ and a distance $D$

How do I get the point $(x, y)$ which is $D$ unit away from $x_1, y_1$ with angle $A$.

In the right image, the answer will be $(x_1 - D\sin A, y_1 + D\sin A)$

And in the left one $(x_1 + D\sin (180-A), y1 + D\sin(180-A))$

Is there some kind of formula that could cover all cases $(A<90, 90<A<180, 180<A<270, 270<A<360$?

The solution is $(x_1-D\cos A, y_1+D\sin A)$ for all angles $A$


$\frac {\operatorname d\!y}{\operatorname d\!x}$ for $\sqrt{xy}=1$ To find $\displaystyle \dfrac {\operatorname d\!y}{\operatorname d\!x}$ for $\sqrt{xy}= 1$:

$$\dfrac{\sqrt y}{2\sqrt x}+\dfrac{y'\sqrt x}{2\sqrt y}=0\\
\dfrac{y'\sqrt x}{2\sqrt y}=\dfrac{-\sqrt y}{2\sqrt x}\\
y'=\dfrac{-y}x$$

Is this correct? and can you show me how to get the same answer solving using $\ln$?

$\sqrt{xy}=1\implies xy=1 \implies xy'+y=0 \implies y'=-\dfrac yx$.
So, yeah, this is true.

$$\ln \sqrt {xy} = \ln 1$$
$$\frac{\ln{x}+\ln{y}}{2}=0$$
$$\ln y=-\ln x$$
$$\dfrac{\operatorname d}{\operatorname d\!x} \ln y= - \dfrac{\operatorname d}{\operatorname d\!x}\ln x $$
$$\dfrac{\operatorname d \ln y }{\operatorname d\!y} \dfrac{\operatorname dy}{\operatorname d\!x}  =-\frac{1}{x}$$
$$\dfrac{\operatorname dy}{\operatorname d\!x}  =-\frac{y}{x}$$


Euclid: The Game - Level 23: Construct an external common tangent to two given circles in five moves? In Euclid: The Game - Level 23, we are given two non-concentric circles:

The goal is to create one of the tangents shown in the following image:

Note: it suffices to create a line segment on one of these $2$ tangents. This segment need not actually touch both circles, it just needs to be in the right direction.

Part of the challenge in Euclid: The Game is to do this with a minimal number of "moves", each move standing for the usage of one of the tools given. Creating a point or naming an intersection do not count as a move; everything else does.

The "minimum number of moves" to solve this is said to be $6$. Multiple solutions exist (see below). However, the page mentions that "there are rumours that it is possible to do this level in 5 non-primitive moves".

So the question, is: can someone come up with a solution in $5$ (non-primitive) moves?

SPOILERS BELOW!

I know of two different solutions in $6$ moves:

Solution 1:

 Define a random point $M$ on the circle $c_B$.
 Translate the line $BM$ to $A$, creating a segment $AN$. (move 1)
 Define the intersection $O = c_A \cap AN$.
 Compass the length $ON$ centered at $B$, this creates a circle $c_B'$. (move 2)
 Create the midpoint $P$ of $AB$. (move 3)
 Create the circle $c_P$ centered at $P$ with radius $PB$. (move 4)
 Define  $Q = c_P \cap c_B'$.
 Create the ray $BQ$. (move 5)
 Translate $AQ$ to $R$, creating a segment $RS$. (move 6)

Solution 2:

 Define a random point $M$ on the circle $c_B$.
 Translate the line $BM$ to $A$, creating a segment $AN$. (move 1)
 Define the intersection $O = c_A \cap AN$.
 Create the ray $BA$. (move 2)
 Create the ray $MO$. (move 3)
 Define $P = BA \cap MO$.
 Create the midpoint $Q$ of $AP$. (move 4)
 Create the circle $c_Q$ centered at $Q$ with radius $PQ$. (move 5)
 Define  $R = c_Q \cap c_A$.
 Create the ray $PR$. (move 6)

Remarks:

Both $6$-moves-solutions start similarly by picking a random point $M$. Perhaps picking a not-so-random point (for example on the intersection of the circles, see next item) to start with can reduce the number of points needed?
You can move the centers of the circles, for example creating intersecting circles! In some of the levels, this is necessary to get the minimum number of moves, so perhaps it can be of use here too?

Here is a simple one in 5 moves -

Draw two parallel lines through centers of the circles - 2 moves
Draw a line through two intersections (on same side) - 1 move
Draw the perpendiculars to the chords through the centers, in the two circles - 2 moves

The intersections of these two perpendicular lines with the perspective circles are defining the segment as required in the question - those are actually the tangent points on the two circles.

I do not think a drawing is needed.


Ant cannot travel to opposite side of cube in even amount of steps Say an ant starts at vertex 1 of a cube and wants to travel to vertex 8. The ant cannot do this in 4 steps, nor any even amount of steps in general. How can I express this? It seems that after every first move of one, the number is guaranteed to be odd because there will be an even amount of moves made after the additional first move.

The cube graph is an example of a bipartite graph, where the vertices can be  divided into two group such that the edges always connect a vertex from one group to a vertex in the other group, never to another vertex in the same group.

The group can be seen in your diagram by moving vertex $1$ in amongst the other odd-numbered vertices and vertex  $8$ likewise into the even-numbered group. Then you will always traverse an odd number of edges to reach a vertex in the other group from the starting vertex, plus all cycles are even length, etc.

As you can see you have bipartite graph here and one of the things of bipartite graph is that for any u,v vertices all the paths from u to v has to odd or even (can't be 1 path odd while there is other even) so if we have bipartite graph and we show 1 odd path then all the paths are odd.


Computing the Picard group of $\mathbb{Z}[\sqrt{-19}]$ I want to compute the Picard group of $\mathbb{Z}[\sqrt{-19}]$, which is not a Dedekind domain. The problem is that I don't even know where to begin.

Any ideas would be helpful.

Thanks

Edit- Since there's some confusion by what I mean by the "Picard group".

$\textit{Definition}$

Let A be an integral domain. Let K denote it's field of fractions. Consider the set of all invertible fractional ideals F(A). Then F(A) forms a group. Consider the subgroup of non-zero principal fractional ideals P(A). So any element in P(A) looks like xA where x $\in$ K$^*$. Then the Picard group of A is defined as F(A)/P(A).

When A is a Dedekind domain, the Picard group coincides with the class group.

Edit 2 - So here's why I'm having a hard time with this problem.

Had $\mathbb{Z}\sqrt{-19}$ been integrally closed, we could proceed using the Minkowski bound and Kummer-Dedekind theory. Unfortunately, $\mathbb{Z}[\sqrt{-19}]$ is $\textit{not}$ integrally closed.

The integral closure in it's field of fractions $\mathbb{Q}(\sqrt{-19})$ is actually strictly bigger, given by $\mathbb{Z}[\frac{1 + \sqrt{-19}}{2}]$.

And thus, I can't use routine methods to try and understand this group.

In case of imaginary quadratic field, there is a nice correspondence between the Picard group of orders with discriminant $D$ and reduced primitive quadratic form with the same discriminant. More precisely, the form $ax^2+bxy+cy^2$ corresponds to the ideal class of $(a,(-b+\sqrt{D})/2)$.

For more information and proof, consult chapter 7 of the well-known Primes of the Form $x^2+ny^2$ by David Cox.

The order $\mathbb{Z}[\sqrt{-19}]$ has discriminant $-56$, and there are three primitive reduced forms:
$$x^2+19y^2 \qquad 4x^2-2xy+5y^2 \qquad 4x^2+2xy+5y^2$$
they respectively translate to the trivial ideal, $(4,1+\sqrt{-19})$ and $(4,-1+\sqrt{-19})$.

Hence the Picard group of $\mathbb{Z}[\sqrt{-19}]$ is cyclic of order $3$ generated by $(4,1+\sqrt{-19})$.

You want to use the Kummer-Dedekind Theorem to factor the prime ideals of norm at most the Minkowski bound.

I recommend you follow 2.6.1 of https://jrsijsling.eu/notes/ant-notes.pdf


Find $(f^{-1})' (a) $ for $f(x) = x - \frac {2}{x}$, $x < 0, a = 1$ The textbook answer is $\frac {1}{3}$, I went through all the steps, but couldn't interpret it.
Below were my steps.
Find $(f^{-1})' (a) $ for $f(x) = x - \frac {2}{x}$, x < 0, a = 1
First I tried to find the inverse, rearrange the equation to $$x = y - \frac{2}{y} \implies x = \frac{y^{2}-2}{y}$$
Rearrange, $xy=y^{2}-2 \implies y^{2}-xy-2 = 0$
Then, I tried to use the equation $\frac{-b \pm \sqrt{b^{2}-4ac}}{2a}$, and I got:
$y = \frac{x \pm \sqrt{x^{2}+8}}{2}=f^{-1}$
Next, I tried to find $(f^{-1}){'}$, for both $+$ and $-$
$\frac{1}{2} \cdot (x' + ((x^{2}+8)^{\frac{1}{2}})')$
Apply chain rule:
$((x^{2}+8)^{\frac{1}{2}})'$ = $\frac{1}{2}(x^{2}+8)^{\frac{-1}{2}} \cdot2x$
Then do the same for -
And I got $(f^{-1})'$ = $\begin{cases} \frac{1}{2}\cdot(1+\frac{1}{2}(x^{2}+8)^{\frac{-1}{2}} \cdot2x) \\ \frac{1}{2}\cdot(1-\frac{1}{2}(x^{2}+8)^{\frac{-1}{2}} \cdot2x) \end{cases}$
When a = 1,
$(f^{-1})'=\begin{cases} \frac{\sqrt{9}+1}{2\sqrt{9}} = \frac{2}{3} \\ \frac{\sqrt{9}-1}{2\sqrt{9}} = \frac{1}{3} \end{cases}$
I need some help. It says when $x < 0$, but both of my answers are positive. How do I interpret it?

The domain of $f$ must be the range of $f^{-1}$, i.e. $f^{-1}$ must only output negative values for any value of $x$. So, you should reject the solution $y=\frac{x+\sqrt{x^2+8}}{2}$ and notice $$\frac{x-\sqrt{x^2+8}}{2} \lt \frac{x-\sqrt{x^2}}{2} \le 0 $$ which satisfies our requirement and the corresponding answer is $\frac 13$.

The domain of $f$ is the interval $(-\infty,0)$.
Thus, $g:=f^{-1}$ should be such that its range is $(-\infty,0)$.
This way, one can see that
$$
g(s) = \frac{s-\sqrt{s^2+8}}{2}
$$
is the proper solution for the inverse of $f$,
(because we always have $g(s)<0$), but
$g(s) = \frac{s+\sqrt{s^2+8}}{2}$ would not be, because its values
can be positive (for example we have $g(0)>0$).
Now that the proper inverse function is known, we can see from your calculations
that $\frac13$ is the correct answer and $\frac23$ is wrong.

It says when $x < 0$, but both of my answers are positive. How do I interpret it?

While the output of the function $g=f^{-1}$ should be negative,
its derivative $g'=(f^{-1})'$ can be positive, because these are different objects.


Integral of $\tan(\ln(x+1))/x$ How can I solve this integral?

$$\int{\frac{\tan(\ln(x+1))}{x}dx } $$

I tried substitution: $x+1 = e^t $,

$$\int{ \frac{e^t \cdot \tan(t)}{e^t -1 }dt} $$

and U-substitution: $u = \tan(t), \ v=\ln(e^t-1) $, but it doesn't work.

Thank you!

Hint: Use a substitution $u=\ln(x+1)$ so that $du=\frac{1}{x}\,dx$. Can you solve the rest of it from here?

Edit: Follow the other dude :)


Absolute convergence of series $\sum\limits_{n=2}^\infty{(-1)^{n-1}\over (n+(-1)^n\sqrt{n})^{2\over 3}}$ Exam absolute convergence of series $$\sum\limits_{n=2}^\infty{(-1)^{n-1}\over (n+(-1)^n\sqrt{n})^{2\over 3}}$$
WolframAlpha says it diverges, but I don't know how to show it. As it's not monotonic $\Rightarrow$ I can't estimate it from below. I tried to do Сauchy criteria, but I got mess with radicals and got nothing

$$|u_n|=\frac {1}{n^\frac 23 (1+\frac {(-1)^n}{\sqrt {n}})^\frac 23}$$

$$\sim \frac {1}{n^\frac 23} $$

thus it is not   absolutely convergent.

$$u_n=$$
$$\frac {(-1)^{n-1}}{n^\frac 23}(1-\frac 23\frac {(-1)^n}{\sqrt {n}}(1+\epsilon (n)) )$$

the first series is alternate and the second is absolutely convergent thus $\sum u_n $ is conditionally convergent.

"Amusing" downvote to an otherwise perfectly valid answer.

Consider the function $u(x)=x^a$, then $u'(x)=ax^{a-1}$ hence, if $x_n\sim n\sim y_n$, $$\frac1{u(x_n)}-\frac1{u(y_n)}=\frac{u(y_n)-u(x_n)}{u(x_n)u(y_n)}\sim(y_n-x_n)\frac{u'(n)}{u(n)^2}=(y_n-x_n)\frac{a}{n^{a+1}}$$ The series in the question corresponds to $x_n=n-\sqrt n$, $y_n=n+1+\sqrt{n+1}$ and $a=\frac23$, then $y_n-x_n\sim2\sqrt n$ hence $$\frac1{u(x_n)}-\frac1{u(y_n)}\sim\frac{4/3}{n^{2/3+1/2}}$$ and $\frac23+\frac12>1$ hence the series converges.


Solve $z^3=\overline{z}$ in $\mathbb{C}$ Solve an equation in $\mathbb{C}$: $$z^3=\overline{z}$$

Now clearly $$|z|^3= |z^3|=| \overline{z}| =|z|\implies |z|=0 \;\;\vee\;\;|z|=1 $$

So if $|z|=0$ we get $z_1=0$ and if $|z|=1$ we get $$ z^4 = z \cdot  \overline{z} =1$$
and so we get $$0=z^4-1 = (z-1)(z+1)(z+i)(z-i)$$
so $z_{2,3}=\pm1$ and $z_{4,5}=\pm i$. So this equation has 5 solutions.

Second approach: we have $$z^3=\overline{z}\implies \overline{z}^3=z \implies (z^3)^3=z$$
so we get the equation $z^9-z=0$ which has $9$ (different) solution!?

What is going wrong here?

In that way from $z^8=1$ we obtain four extra solutions which doesn’t satisfy the original equation, notably in the derived form $z^4=1$, as in the real case when we square equations in order to eliminate square roots.

if $|z|=1$ then let $z=\cos{\alpha}+i\sin{\alpha}=>z^3=\cos{3\alpha}+i\sin{3\alpha}=\overline{z}=\cos\alpha-i\sin\alpha=>\cos3\alpha=cos\alpha$ and $\sin3\alpha=-sin\alpha\cdots$


2x2 Matrix Inverse Question Let $a, b, c, d\in \mathbb{R}$. I'm asked to determine the products of the ($2x2$) matrices

$$
\begin{bmatrix}
a&b\\
c&d\\
\end{bmatrix} and
\begin{bmatrix}
d&-b\\
-c&a\\
\end{bmatrix}
$$

and show that the first matrix is invertible if and only if $ad \neq bc$.

Calculating the product is pretty straightforward:

$$
\begin{bmatrix}
a&b\\
c&d\\
\end{bmatrix}
\begin{bmatrix}
d&-b\\
-c&a\\
\end{bmatrix}
= \begin{bmatrix}
ad-bc&-ab+ba\\
cd-dc&-bc+ad\\
\end{bmatrix}
$$

And I know that a matrix A is invertible if I can find a matrix B such that $AB=I$ or $BA = I$. Any hints on how to show that the first matrix is invertible if and only if $ad \neq bc$?

From your step we have

$$\begin{bmatrix}
ad-bc&-ab+ba\\
cd-dc&-bc+ad\\
\end{bmatrix}= \begin{bmatrix}
ad-bc&0\\
0&ad-bc\\
\end{bmatrix}$$

and we need $ad-bc\ne 0$.

Notably form here we can see that

$$\begin{bmatrix}
a&b\\
c&d\\
\end{bmatrix}^{-1}=\frac1{ad-bc}\begin{bmatrix}
d&-b\\
-c&a\\
\end{bmatrix}$$

You can use the fact that a square matrix is invertible if and only if it has a non-zero determinant. If ad = bc then it is easy to show matrix A will have determinant 0.


explicit upper bound of TREE(3) TREE(3) is the famously absurdly large number that is the length of a longest list of rooted, 3-colored trees whose $i$th element has at most $i$ vertices, and for which no tree's vertices can be mapped to the vertices of a subsequent tree preserving color and inf relationships.

Some lower bounds of TREE(3) have been proven. Among them are $A^{A(187196)}(1)$, where the superscript denotes function iteration and $A(x) = 2\uparrow^{x-1}x$ (using Knuth up-arrows) is a version of the Ackermann function. More bounds appear in this Wikia article and here. These bounds are rather unsatisfying because there is no indication of how tight they are.

The only upper bound of TREE(3) that I have seen (other than, trivially, TREE($n$) for $n > 3$) has a similar derivation as a longest sequence of a more general type of graph; I can't find a reference at the moment. This upper bound is also unsatisfying because it is non-constructive.

Does anyone know of an explicit expression that is an upper bound of TREE(3)? Or is it so large that there is no hope to construct one?

$\ TREE(3)\ $ is far above the $\Gamma_0$-level of the fast growing hierarchy.

See here how insane large it is : https://sites.google.com/site/largenumbers/home/appendix/a/numbers3

I am not sure whether an upper bound is known. Nested Ackermann-functions do not come near to $\ TREE(3)\ $ at all. Conway chains and even Bowers's $3D$-arrays are still MUCH MUCH MUCH too small.

TREE(3) is known to be finite, but few finite upper bounds have been proven that do not rely on constructions such as TREE(n) and n>3.

As for lower bounds, saying $g_{64}$ (Graham's number) is a lower bound for TREE(3) is like saying 1 is a lower bound for Graham's number. Even $g_{187,196}$ has proven to be an EXTREMELY WEAK lower bound for TREE(3).

See
https://en.wikipedia.org/wiki/Kruskal%27s_tree_theorem
and
https://www.youtube.com/watch?v=3P6DWAwwViU.


Strict subadditivity of the norm in uniformly convex Banach Spaces I would like to prove the following:

Let $(X,\|\cdot\|)$ be a uniformly convex Banach Space. Then the norm is strictly subadditive, i.e., $\forall x,y\in X\backslash\{0\}:\|x+y\|<\|x\|+\|y\|\,$.

I started my proof by considering two elements satisfying $\|x+y\|=\|x\|+\|y\|$. By using all properties of a unifomly convex space it should follow that $x=y=0$, but I always failed.

Can someone give me a hint how to proceed?

If $x$ is not a multiple of $y$, $\Vert y\Vert\ge\Vert x\Vert>0$, and $\Vert x+y\Vert=\Vert x\Vert+\Vert y\Vert$, then
$$
\biggl\Vert  {x\over\Vert x\Vert}+ {y\over\Vert y\Vert} \biggr\Vert
\ge
\biggl\Vert  {x\over\Vert x\Vert}+ {y\over\Vert x\Vert} \biggr\Vert
-
\biggl\Vert  {y\over\Vert x\Vert}- {y\over\Vert y\Vert} \biggr\Vert
={{\Vert x\Vert +\Vert y\Vert}\over \Vert x\Vert}
-\Vert y\Vert\Bigl({1\over\Vert x\Vert}-{1\over \Vert y\Vert}\Bigr)=2.
$$

So $X$ is strictly convex if and only if for distinct norm one vectors $x$, $y$, $\Vert x+y\Vert<2$.

$X$ is uniformly convex if and only if for each $\epsilon>0$, there is a $\delta>0$ so that for any $x$, $y$ in the closed unit ball of $X$ with $\Vert x-y\Vert\ge\epsilon$ one has $$\Bigl\Vert{ x+y\over 2}\Bigr\Vert\le 1-\delta.$$

It should be easy to see from the above that a uniformly convex normed space is strictly convex.

I think this property is not verified in uniformly convex Banach spaces in general, in fact it is not even true in the uniformly convex Banach space $(\mathbb{R},|.|)$:  $|1+2|=|1|+|2|$


The solution set of the equation $|2x - 3| = - (2x - 3)$ The solution set of the equation $\left | 2x-3 \right | = -(2x-3)$ is

$A)$ {$0$ , $\frac{3}{2}$}

$B)$ The empty set

$C)$ (-$\infty$ , $\frac{3}{2}$]

$D)$ [$\frac{3}{2}$, $\infty$ )

$E)$ All real numbers

The correct answer is $C$

my solution:
$\  2x-3   = -(2x-3)$ when $2x-3$ $\geqslant$  $0$ $\Rightarrow$ $x$ = $\frac{3}{2}$
$-(2x-3) = -(2x-3)$ when $2x-3$ $<$  $0$ $\Rightarrow$ $0$ = $0$
I can't get how the answer is  presented in interval notation (-$\infty$ , $\frac{3}{2}$].

Any help is appreciated.

To have $|y| = -y$, you need to have $y \leq 0$; now set $y = 2x-3$, so you want $2x-3 \leq 0 \Leftrightarrow x \leq {3 \over 2} \Leftrightarrow x \in (-\infty, {3 \over 2}]$.

case -  when $|2x-3| = -2x+3$ so than there will be $-2x+3=-2x+3$ --- from what will result $0=0$ so for this case the answer will be E).

2.case - when $|2x-3| = 2x-3$ so than there will be $2x-3=-2x+3$ --- 4x=6 --- $x=6/4 =3/2$ --- so in this case $x=3/2$

Note that because in your exercise , in this equation there is sign of equality and not is inequality for this you not can getting interval solutions


Intersection of a function having increasing derivative with tangent line I came across the following exercise in Spivak:
Suppose that $f(x)$ is a differentiable function and suppose that $f'(x)$ is increasing. Show that every tangent line of $f(x)$ intersects the graph only once.

My intuition is as follows:
Consider the tangent line $t(x)$ for any point $a$. Clearly, we have $t'(a) = f'(a)$. Since $f'(x)$ is an increasing function whereas $t'(x)$ is constant, the two graphs can never intersect again for $x\geq a$ because $f(x)$ "grows faster" than $t(x)$.

How do make this argument precise?

Consider the tangent $l$ to $f(x)$ at $x=a$. This has gradient $f'(a)$. Let $c>a$. Then by the mean value theorem, there is some $a<b<c$ such that $f'(b)(c-a)=f(c)-f(a)$. This means $f(c)=f(a)+f'(b)(c-a)>f(a)+f'(a)(c-a)$ which is the value of $l$ at $c$. Do the same for the case $c<a$.

Since $f'$ is increasing, the function $f$ is strictly convex, which means that its graph always lies "below the line segment joining any two points of the graph".

Now assume that a tangent line $t$ at $f$ in the point $(x_1,f(x_1))$ intersects the graph of $f$ twice, at $(x_1,f(x_1))$ and $(x_2,f(x_2))$. Since the graph of $f$ has to lie below the two points $(x_1,f(x_1))$ and $(x_2,f(x_2))$, the value of $(f-t)'$ cannot be zero at $x$, which is a contradiction, since $t$ should be a tangent at $f$ in $x_1$, i.e. $f'(x_1) = t'(x_1)$.


Prove or disprove that the set of even integers forms a group under addition possible duplicate but want work checked and verified

let $l =$ any even number and $m =$ any even number

$l+m = w$ where $w$ is any element in the even integers thus it is closed under addition

$2k+0=2k$ and $0-2k=2k$ so there is an identity both left and right

every even number can have its negative added to get 0 so there is an inverse for every element in the even integers i.e $2+(-2)=0$

Its associative

Your proof is okay, but it can be improved.

Closed: For any $l,m \in \mathbb{E}$ (where $\mathbb{E}$ is the set of even integers) $l = 2a$ and $m = 2b$ for some $a,b \in \mathbb{Z}$. Thus, $l + m = 2a + 2b = 2(a+b) \in \mathbb{E}$.

Identity: $0 \in \mathbb{E}$ and $0 + m = m + 0 = m$ for all $m \in \mathbb{E}$.

Inverse: For any $2k \in \mathbb{E}$, the number $2(-k) \in \mathbb{E}$ and $2k + 2(-k) = 2k - 2k = -2k + 2k = 0$.

Associative: The integers under addition are associative, so $\mathbb{E} \subset \mathbb{Z}$ inherits that property.

It's $2\Bbb Z$.  It's an infinite abelian group, and  is cyclic, so isomorphic to $\Bbb Z$.

$\phi:\Bbb Z\to 2\Bbb Z$ by $\phi(k)=2k$ is an isomorphism.