What is the probability of multiple independent events occurring in a given amount of time? I thought about this problem the other day and my limited knowledge in basic probability theory was not enough to help me figure this one out. Say you have 3 probabilities:

You have a 50% chance of dying tomorrow due to ant abduction(P(A) = 0.5).
You have a 40% chance of dying tomorrow by being beaten by a baboon (P(B) = 0.4).
You have a 30% chance of dying tomorrow by being cannibalized by Canadians (P(C) = 0.3).

Then, what is the probability that you will die tomorrow, assuming the previous three reasons are the only ways you could die?

I thought of doing something like P(A or B or C), and since A,B,C are disjoint, then P(A or B or C) = P(A) + P(B) + P(C). This is clearly wrong, since the probabilities can't exceed 1. What would be an appropriate way of thinking about this?

The probability that you will not die tomorrow, is the probability all events will not happen:
$$\mathbb{P}(\text{you will not die})=(1-0.5)(1-0.4)(1-0.3)=\frac{21}{100}$$
So the probability that you will die tomorrow is:
$$\mathbb{P}(\text{you will die})=1-\mathbb{P}(\text{you will not die})=1-\frac{21}{100}=\frac{79}{100}$$

This problem may be easily solved using the common formula.
From the given situation it is obvious that any one of the happenings from three you have described is enough for your death.
You tried in good way. Just few more calculations was needed. The formula you used was almost correct. The correct formula is as follows:
$$
\begin{split}
P(A\cup B\cup C)&=P(A)+P(B)+P(C)-P(A \cap B)-P(B \cap C)-P(C \cap A) + P(A \cap B \cap C)\\
&= (0.5) + (0.4) + (0.3) - (0.5\cdot 0.4) + (0.4\cdot 0.3) + (0.3\cdot 0.5) + (0.5\cdot0.4\cdot0.3)\\
&= 0.79
\end{split}
$$
See more here: https://en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle


Show that if $f(x) > 0$ for all $x \in [a,b]$, then $\int_{a}^b f(x) dx > 0$ Assume $f$ is Riemann integrable and nonnegative over $[a,b]$. Show that if $f(x) > 0$ for all $x \in [a,b]$, then $\int_{a}^b f(x) dx > 0$.

This seems very obvious to me. One thing I would do is if the function is not continuous, break up the integrals it is continuous on into infinitely many small integrals which all must a approach a positive number. I still am trying to see how to make this argument rigorous.

One can show without much difficulty that a Riemann integrable function on a bounded interval has at least a point of continuity. (There exists infinitely many of them actually)Let c be a point of continuity of $f(x)$ on the interval $ [a,b]$ . Note that since $f(c)>0$ there exists a neighborhood about $x=c$ such that $f(x) > \frac{f(c)}{2}$ on that neighbourhood. Thus we can easily conclude that the integral is strictly positive.

Hmm, going through the Lebesgue integral actually seems to be the simplest path here.

$f$, being Riemann integrable on a bounded interval, is actually Lebesgue integrable, and its Riemann integral equals its Lebesgue integral.

Now the preimages $$f^{-1}\bigl[(1/n,\infty)\bigr]$$ must have measure zero for every $n$, because if one of them has positive measure $m$, then the integral would be at least $\frac mn$, that is, positive.

But the union of these (countably many) preimages is all of $[a,b]$ if $f$ is positive everywhere; on the other hand a countable union of sets of measure zero will itself have measure zero. This contradicts the (unspoken) assumption that the interval $[a,b]$ is nontrivial.


Solution Verification: Maximum number of edges, given 8 vertices Suppose a simple graph G has 8 vertices. What is the maximum number of
  edges that the graph G can have?

The formula for this I believe is

  n(n-1) / 2

where n = number of vertices.

8(8-1) / 2 = 28.  Therefore a simple graph with 8 vertices can have a maximum of 28 edges.  Is this correct?

Yes, it is. The maximum number of edges is simply the number of pairs of distinct vertices; if there are $n$ vertices, this is

$$\binom{n}2=\frac{n!}{2!(n-2)!}=\frac{n(n-1)}2\;.$$

n(n-1)/2

every vertex connected to every other vertex

ex {a,b,c,d,e}

a connected to b
a connected to c  = 4
a connected to d
a connected to e

b connected to c
b connected to d  = 3
b connected to e

c connected to d  =2
c connected to e

d connected to e   =1

This is an arithmetic progression with common difference=1 and base term as 1

there is a nice formula to calculate its sum as n(n-1)/2

hence apply the formula


Why do we use the Euclidean metric on $\mathbb{R}^2$? On the train home, I thought I would try to prove $\pi$ is irrational. I needed a definition, so I used:

$\pi$ is the area of the unit circle.

But what is a circle?

A circle is the set of tuples $(x,y)$ satisfying "$x^2 + y^2 \leq 1$".

That seems a little awkward - is there a more natural way to say it?

The standard distance metric on $\mathbb{R}^2$ is "$\sqrt{(x1-x2)^2 + (y1-y2)^2}$". A circle is the unit ball of this metric, and $\pi$ is its area.

But why do we use this metric in the first place? Why not use the Taxicab metric - it's linear so it'd seem to satisfy even more nice mathematical properties.

Because of the Pythagorean theorem.

And what property of $\mathbb{R}^2$ lets us prove the Pythagorean theorem?

This one took a little more thought. I think it comes about because we want to associate lengths to lines that are equivalent under translations and rotations. Translations seem pretty natural - but I'm having trouble defining rotations without being circular:

If you multiply a coordinate tuple by the appropriate matrix, you get a rotation - but the matrix's entries involve $\sin$ and $\cos$, which I would define as the coordinates on the unit circle(which is circular).
I can define the set of all infinite lines that go through the origin. Except for the vertical line, you might assume they all have the form $L_m = \{(x,y)|y = mx\}$ for some $m \in \mathbb{R}$. Given an $m \in \mathbb{R}$, I can't find an obvious way to get the point on the unit circle corresponding to it.
The complex number route just gives $\sin$ and $\cos$. Circular, again.
We can determine whether two vectors are orthogonal using the dot product. That reduces defining the entire unit circle to defining just a single quadrant(if we also assume the length of $-1 * x$ is the same as $x$).
A quick Google search shows circles were a primitive notion in Euclid's Elements.
There is another question/answer given here, but I was still left confused as to what a rotation actually is in $\mathbb{R}^2$:
Why is the Euclidean metric the natural choice?

So my questions are:

Out of all the possible metrics, why choose the Euclidean metric as the natural one? I think it's rotation that's the root cause, but that might end up being a red herring.
How do we define a natural equivalence of points up to rotation from first principles?
This is related to #1. If we define the set of infinite lines through the origin, you can choose a representative element from each one and form a perimeter curve(assuming a continuous choice function). What properties does the circle have that would make it a natural choice?

The Euclidean metric is special because it comes from what is called an inner product, and up to scaling it is the only metric that does so. This allows you to talk about angles between vectors in a sensible way, which you cannot do with other metrics.

So really we don't choose to use the Euclidean metric, so much as we choose to use the dot product (the only inner product on $\mathbb R^2$, up to scaling) and we get the Euclidean metric as a result.

We use 2-norm simply because it works i.e. it satisfies properties of Euclidean space.

No dimension is more important than other (e.g. Chebyshev distance doesn't qualify).
$\mathbb{R}^2$ is a real coordinate space. So integers have to work too.
For positive integers $x, y, z$ $ x^n+y^n=z^n$ doesn't work for $n > 2$. [1]
There is no reference point. E.g. distance using 1-norm depends of which vector you choose as an axis
When you rotate distance doesn't change, hence the unit circle.

[1] https://en.wikipedia.org/wiki/Fermat%27s_Last_Theorem


consistency of estimator Let $X_1,\ldots,X_n$ be an independent sample from random variable with pdf
$$ f(x)= \frac 1 \theta e^{-x/\theta}, \qquad x ≥ 0 $$
How to check that the estimator $\theta(X) = \dfrac{X_1 + X_2} 2$ is consistent?

Note that you estimator does not depend on the sample size $n$, hence

$$
Var(\theta(X))  =\frac 1 4(\theta^2 + \theta^2)=\frac{\theta^2}{2}.
$$
The variance (MSE, as this estimator unbiased) does not converges to $0$. Generally it is not sufficient condition to deduce that it is not consistent, but it gives the intuitive understanding that the probability mass does not concentrate at some point (parameter).    More formally, for all $\epsilon > 0$,

$$\lim_{n\to \infty} P(| \theta(X) - \theta|\ge \epsilon) = \lim_{n\to\infty}\left( P(X_1+X_2\ge2(\theta+\epsilon))+ P(X_1+X_2\le 2(\theta - \epsilon))\right)$$

$$=P(X_1+X_2 \notin[2(\theta \pm \epsilon)]) = 1-\int_{2(\theta - \epsilon)}^{2(\theta + \epsilon)}f_{X_1+X_2}(x)dx \approx1.$$
In the last step you can use the fact $X_1+X_2 \sim \Gamma(2, 1/\theta)$, thus for small enaugh $\epsilon$, this integral will equal almost $0$.

Use definition of consistent estimator along with the fact that sum of exponentially distributed r.v.s has Gamma distribution.

In other words, you want to prove that $\mathbb{P}[\sum_{i=1}^{n}X_{i}\geq n\cdot\epsilon]\rightarrow0$ as $n\rightarrow\infty$. Since $\sum_{i=1}^{n}X_{i}$ has Gamma distribution with parameters $n$ and $\theta$, you want to show that $\Gamma(n,\frac{n\epsilon}{\theta})/\Gamma(n)$ tends to $1$ for $\epsilon>0$ and $\theta>0$.


Why does $(3\sqrt3)^2 = 27$? How does $(3\sqrt3)^2 = 27$?

I've tried to solve this using binomial expansion and using the FOIL method from which I obtain $9 + 3\sqrt3 +3\sqrt3 + 3$.
it has been a while since I've done this kind of thing so it may be something obvious that I can't see.


How can I show that the Lucas numbers are given by the sum of $F_{k-1}+F_{k+1}$ I was wondering how I can show that the Lucas numbers are also given by the formula $F_{k-1}+F_{k+1}$ in terms of Fibonacci numbers. I got the formula for the Lucas numbers to be $L_k=[\frac{1}{2}(1+\sqrt{5})]^k+[\frac{1}{2}(1-\sqrt{5})]^k$ and I know the recurrence relation is just the sum of the two previous numbers (i.e. $L_k=L_{k-1}+L_{k-2}$), but I cannot seem to get how to use this information to show that $L_k=F_{k-1}+F_{k+1}$.

Sure, we could use the closed forms for the sequences:
\begin{align}
L_n &= \left(\frac{1+\sqrt{5}}{2}\right)^n + \left(\frac{1-\sqrt{5}}{2}\right)^n\\
F_n &= \tfrac{1}{\sqrt{5}} \left[ \left(\frac{1+\sqrt{5}}{2}\right)^n - \left(\frac{1-\sqrt{5}}{2}\right)^n \right]
\end{align}

So a bit of algebra:
\begin{align}
F_{n-1}+F_{n+1}
&= \tfrac{1}{\sqrt{5}} \left[ \left(\frac{1+\sqrt{5}}{2}\right)^{n-1} - \left(\frac{1-\sqrt{5}}{2}\right)^{n-1} \right] + \tfrac{1}{\sqrt{5}} \left[ \left(\frac{1+\sqrt{5}}{2}\right)^{n+1} - \left(\frac{1-\sqrt{5}}{2}\right)^{n+1} \right]\\
&=\tfrac{1}{\sqrt{5}}\left[ \left(1+\left(\frac{1+\sqrt{5}}{2}\right)^2\right) \left(\frac{1+\sqrt{5}}{2}\right)^{n-1} - \left(1+\left(\frac{1-\sqrt{5}}{2}\right)^2\right)\left(\frac{1-\sqrt{5}}{2}\right)^{n-1}  \right]\\
&=\tfrac{1}{\sqrt{5}}\left[ \left(\sqrt{5}\frac{1+\sqrt{5}}{2}\right) \left(\frac{1+\sqrt{5}}{2}\right)^{n-1} - \left(\sqrt{5}\frac{\sqrt{5}-1}{2}\right)\left(\frac{1-\sqrt{5}}{2}\right)^{n-1}  \right]\\
&=\left(\frac{1+\sqrt{5}}{2}\right)^n + \left(\frac{1-\sqrt{5}}{2}\right)^n
= L_n
\end{align}


Let $\{X_n\}$ be i.i.d $N(0,1)$ random variables. Show that $\limsup_{n\rightarrow\infty} \frac{|X_n|}{\sqrt{\log n}}=\sqrt2$ a.s. Let $\{X_n\}_{n\ge1}$ be independent $N(0,1)$ random variables. Show that
  $$\limsup\limits_{n\to\infty} \frac{\left|X_n\right|}{\sqrt{\log(n)}}=\sqrt{2} \qquad \text{a.s.}$$

I aim to prove this using the fact that

$$\limsup\limits_{n\to\infty} X_n = b \quad \iff \quad \text{for all } \varepsilon>0 \ : \ \Biggl\{ \begin{array}{l}
\mathbb{P}(X_n \le b+\varepsilon \text{ eventually})=1, \text{ and} \\
\mathbb{P}(X_n > b-\varepsilon\text{ i.o.})=1.
\end{array}$$

I show the first of these two conditions as follows:

\begin{align*}
&\hspace{-2em}\mathbb{P}\left(\frac{\left|X_n\right|}{\sqrt{\log(n)}} > \sqrt{2} + \varepsilon\right)\\
&= \mathbb{P}\bigl(|X_1|>(\sqrt{2}+\varepsilon)\sqrt{\smash[b]{\log(n)}}\bigr) & \text{as $X_n$'s are identically distributed}\\
&=\int_{(\sqrt{2}+\varepsilon)\sqrt{\smash[b]{\log(n)}}}^{\infty} |x|\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} \, dx\\
&=\int_{(\sqrt{2}+\varepsilon)\sqrt{\smash[b]{\log(n)}}}^{\infty} x\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}} \, dx\\
&=\frac{1}{\sqrt{2\pi}}\int_{(\sqrt{2}+\varepsilon)^2\log(n)}^{\infty} e^{-u}du & \text{by making the substitution $u=\frac{x^2}{2}$}\\
&=-\frac{1}{\sqrt{2\pi}}\bigl[e^{-\infty}-e^{-\log(n)(\sqrt{2}+\varepsilon)^2}\bigr]\\
&=\frac{1}{\sqrt{2\pi}}n^{-(\sqrt{2}+\varepsilon)^2}
\end{align*}

Thus, we have:
\begin{align*}
\sum_{n=1}^\infty \mathbb{P}\Biggl(\frac{\left|X_n\right|}{\sqrt{\log(n)}} > \sqrt{2} + \varepsilon\Biggr)&=\sum_{n=1}^{\infty}\frac{1}{\sqrt{2\pi}}n^{-(\sqrt{2} + \varepsilon)^2}\\
&<\infty \qquad \text{ since $\sqrt{2}+\varepsilon>1$}
\end{align*}

So, by the Borel–Cantelli Lemmas:

\begin{align}
&\mathbb{P}\Biggl(\frac{\left|X_n\right|}{\sqrt{\log(n)}} > \sqrt{2} + \varepsilon\text{ i.o.}\Biggr)=0\\
&\implies \mathbb{P}\Biggl(\frac{\left|X_n\right|}{\sqrt{\log(n)}} \le \sqrt{2} + \varepsilon\text{ eventually}\Biggr)=1
\end{align}

It remains then to show that

$$\mathbb{P}\Biggl(\frac{\left|X_n\right|}{\sqrt{\log(n)}} > \sqrt{2} - \varepsilon \text{ i.o.}\Biggr)=1.$$

To do this, I would like to run a symmetric argument to the one above but here we need the final series to diverge which will happen if and only if $\sqrt{2}-\varepsilon \le1$, which happens if and only if $\varepsilon\ge \sqrt{2}-1=0.414\ldots$ and so $\varepsilon$ is getting away from zero, which we can't have. Unless I am making a stupid mistake or missing something obvious, I don't see a way around this issue. Is this approach doomed to fail or is there a way to fix it up? Or is there just a better approach in general for such a problem. Thanks in advance.

The problem is #1 (a) from this exam

I can only see one mistake in your proof: you have
$$
P(|X_1|>(\sqrt2+\epsilon)\sqrt{\log n})=\int_{(\sqrt2+\epsilon)\sqrt{\log n}}^\infty\color{red}{|x|}\varphi(x)\,dx
$$
where $\varphi(x)$ is the pdf of $X_1$. This is not correct; the right hand side of what your wrote is equal to $E\left[|X_1|;X_1>(\sqrt2+\epsilon)\sqrt{\log n}\right]$. To correct this, you need to get rid of the $\color{red}{|x|}$. Remember, to find a probability, you integrate the pdf, and to find the expected value of a function, you integrate the function times that pdf.

The integral can no longer be written in terms of elementary functions. However, following the hint, you can get an asymptotic equivalent for the integral, which is sufficient to do the Borel-Cantelli tests.

Notice that your reasoning does not change if you substitute  $\sqrt 2$ with $1$. It means, at least, that
$$
\limsup_{n\to\infty} \frac{|X_n|}{\sqrt{\log n}}\le 1 \quad \text{a.s.}
$$

My suspicion is that it goes to zero.


Longest path in a graph. How do I prove that the longest path in a graph that starts from the vertex $\ v_1 $, includes all the adjacent vertices of $\ v_1 $?

You claim:

  For every vertex $v_1$, the longest path that starts in $v_1$ includes all adjacent vertices of $v_1$

This is false. If you have a graph with four vertices, $a,b,cd$, and three edges, $(a,b)$, $(b,c)$, $(c,d)$ then the longest path that starts at $b$ is the path $b\to c\to d$ and it does not include $a$.

Let $P$ be a longest path in the graph that stars at vertex $v_1$. We will prove that $P$ contains all neighbours of $v_1$. Suppose not. Then, there is a vertex $v_2$ adjacent to $v_1$ such that $P$ does not contain $v_2$. But that implies that we can join path $P$ with path $v_1v_2$, obtaining a path longer than $P$, a contradiction to the fact that $P$ is a longest path.


Finding an equation for a circle given its center and a point through which it passes No idea how to do this, I used to have these conic shapes committed to memory but I forget them already.

I am supposed to find an equation for the circle that has center $(-1, 4)$ and passes through $(3, -2)$.

I tried graphing it, but it didn't help since the point was not in a straight line with the center and I have no idea what the diameter is.

Recall the:

 Distance Formula:

The distance between the points $(x_1,y_1)$ and $(x_2,y_2)$ is
$$
D=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}
$$

Example:

The radius of your circle is the distance between the points  $(-1,4)$ and $(3,-2)$. Using the Distance Formula:
$$
D= \sqrt{  \bigl(3-(-1)\bigr)^2 + (-2-4)^2  }
= \sqrt{  4^2 + (-6)^2  }
= \sqrt{ 16+36 }
=  \sqrt{ 52 } .
$$

What is the equation of the circle?

It is important to realize   the "equation of the circle" is: a point $(x,y)$ is on the circle if and only if the coordinates of the point $x$ and $y$ satisfy
the equation. So, how to get the equation? What is the relationship between the $x$ and $y$ coordinates of a point on the circle?

Well, let $(x,y)$ be a point on the circle. The big idea is:

$$
\text{The distance from the point }(x,y)\text{  to the center }(-1,4)\text{ is }\sqrt{52}.$$

So, using the distance formula (with $(x_2,y_2)=(x,y)$ and $(x_1,y_1)=(-1,4)$) , it follows that
$$
\sqrt{52}=\sqrt{\bigl(x-(-1)\bigr)^2 +(y-4)^2}.
$$

Or
$$
52=(x+1)^2 +(y-4)^2.
$$

The shortcut would be to just use the following formula (But it's important to realize why you'd use it and where it comes from):

Equation of a Circle

The equation of the circle with center located at $(a,b)$ and with radius $r$ is
$$
r^2=(x-a)^2 +(y-b)^2
$$
Note that the radius squared is on the left-hand side of the equation.

 The following may help:

1 find equation of the circle passing through (1,0) and (0,1) and having the smallest possible radius .
2  show that cot 15/2 degree = √2+√3+√4+√6 .


Why is it that $\omega + \omega_1 = \omega_1$? I have seen some other questions about the proofs around ordinals and $\omega$ and $\omega_1$, but some of the answers have confused me.

An ordinal $\alpha$ is countable if $\alpha < \omega_1$.

$\omega_1$ is the first uncountable ordinal and it is assumed that for $\alpha, \beta < \omega_1$, $\alpha + \beta < \omega_1$.

Now, my initial thought, similar to the answers in the 'Prove $\omega + \omega_1 = \omega_1$' style of questions is that we substitute $\omega$ and $\omega_1$ as $\alpha$ and $\beta$ respectively. Then I would obtain $\omega + \omega_1 < \omega_1$.

Similarly, I could substitute again into $\alpha + \beta \geq max \{\alpha, \beta \}$ to get $\omega + \omega_1 \geq max \{\omega, \omega_1 \} = \omega_1$. So we have an upper and lower bound and can claim $\omega + \omega_1 = \omega_1$.

But my question is, how does this work since $\omega_1$ is uncountable (and so we cannot say $\omega_1 < \omega_1$)?

HINT: Use the definition of ordinal addition, and the fact that adding two countable ordinals is countable.

Every ordinal number can be expressed as a multiple of $\omega^2$ plus an ordinal number less than $\omega^2$. If you take any multiple of $\omega^2$ and left add $\omega$, you get back the exact same ordinal number. If you take any countable multiple of $\omega^2$ and right add any ordinal number less than $\omega^2$, you still get a countable ordinal. Thus, $\omega_1$ is a multiple of $\omega^2$ and so $\omega + \omega_1 = \omega_1$.


How to evaluate $\lim_{x\to\infty}\left(\frac{x+n}{\ln(x+n)}-\frac x {\ln x}\right)$? The question is somewhat related to the discussions I had about this other post. It seems for every integer $n>1$ the limit
$$\lim_{x\to\infty}\left(\frac{x+n}{\ln(x+n)}-\frac x {\ln x}\right)$$
does exist. But I wasn't able to obtain a closed form for it (I admit that I didn't try enough). So, is there a closed form for it?

Hint. By the Mean Value Theorem,
$$\frac{x+n}{\ln(x+n)}-\frac{x}{\ln x}=((x+n)-x)\cdot\frac{d}{dx}\left(\frac{x}{\ln x}\right)_{x=t}=n\left(\frac{1}{\ln t}-\frac{1}{\ln^2 t}\right)$$
with $x<t<x+n$. Now $t$ goes to infinity as $x\to\infty$.

see my answer
$$\lim_{x\to\infty} \frac{x+n}{\ln(x+n)}-\frac{x}{\ln x}$$
$$=\lim_{x\to\infty} \frac{x\left(1+\frac nx\right)}{\ln x\left(1+\frac{n}{x}\right)}-\frac{x}{\ln x}$$
$$=\lim_{x\to\infty} \frac{x\left(1+\frac nx\right)}{\ln x+\ln \left(1+\frac{n}{x}\right)}-\frac{x}{\ln x}$$
$$=\frac{x\left(1+0\right)}{\ln x+\ln \left(1+0\right)}-\frac{x}{\ln x}$$
$$=\frac{x}{\ln x}-\frac{x}{\ln x}=0$$
the limit exists for all real n


Inequality for Distribution of points in space Consider the space $\mathbb{R}^2.$ We are given $n$ mutually different points $x_1,..,x_n \in \mathbb R^2.$

We can then introduce expressions

$$f_i(x_1,...,x_n) =  n^2 \vert x_i \vert^2 + \sum_{j \neq i } \frac{1}{\vert x_j-x_i \vert^2 }.$$

I would like to know if there exists an explicit constant $c>0$, independent of the points, (as large as possible) such that
$$ n^2+ \sum_{i=1}^n f_i(x_1,...,x_n) \ge c n^3.$$

I will do two cases by hand to give you a feeling:

$n=1$: This case is clear, as

$$1+ f_1(x_1) \ge 1 =c1^3$$ with $c=1.$

$n=2$: This case is already more tricky, however

$$4 + f_1(x_1,x_2) + f_2(x_1,x_2) =4+ 4(\vert x_1 \vert^2+\vert x_2 \vert^2) + \frac{2}{\vert x_1-x_2 \vert^2}$$
and thus using the parallelogram identity we find
$$ 4 + f_1(x_1,x_2) + f_2(x_1,x_2)=4+ 2 (\vert x_1-x_2 \vert^2 + 1/\vert x_1-x_2 \vert^2) + 2 \vert x_1+x_2 \vert^2.$$

Now, we may use that $t+1/t  \ge 2$ for $t > 0$ to infer that

$$ 4 + f_1(x_1,x_2) + f_2(x_1,x_2)\ge 4+ 4 = 2^3.$$

So somehow one could conjecture that $c=1$ is possible, but I don't know whether this is true in general.

No, this is not true, here is the sketch of a sequence of distributions where your sum asymptotically grows slower than $n^3$. In the following I am using the symbol $\asymp$ for expressions whose ratio is bounded above and below by some positive constants independent of $n$, and similarly $\lesssim$ for inequalities which hold up to a positive constant.

Start with the following construction: Let $R>1$ be an integer, and let $x_{k,l} = (k,l)$ be the points with integer coordinates $k,l \in \{-R,\ldots,R\}$. These are $n=(2R+1)^2 \asymp R^2$ points with
$$
\sum_{k,l=-R}^R \|x_{k,l}\|^2 = \sum_{k,l=-R}^R (k^2+l^2) \asymp R^4 \asymp n^2.
$$
Furthermore, for any fixed $k_0,l_0 \in \{-R,\ldots,R\}$ we have
$$
\sum_{{k,l=-R}\atop{(k,l) \ne (k_0,l_0)}}^R \frac{1}{\|x_{k,l} - x_{k_0,l_0}\|^2} \le
\sum_{{k,l=-2R}\atop{(k,l) \ne (0,0)}}^{2R} \frac{1}{\|x_{k,l}\|^2} \le
\sum_{m=1}^{2R} \,\, \sum_{\max(|k|,|l|)=m} \frac{1}{\|x_{k,l}\|^2},
$$
which is rearranging the sum into sums over boundaries of concentric squares of radius $m$. The number of integer points on such a boundary (i.e., the number of terms of the inner sum, given the radius $m$) is $\asymp m$, and each term in that sum is $\frac{1}{\| x_{k,l} \|^2} \asymp \frac{1}{m^2}$, so we get that the total sum above is
$$
\asymp \sum_{m=1}^{2R} \frac{1}{m} \asymp \ln R \asymp \ln n.
$$
Summing this up over all lattice points $(k_0,l_0)$, we get that
$$
\sum_{{k,l,k_0,l_0=-R}\atop{(k,l) \ne (k_0,l_0)}}^R \frac{1}{\|x_{k,l} - x_{k_0,l_0}\|^2} \lesssim n \ln n.
$$
For your functions $f_i$, where $i$ now runs through all pairs $(k,l)$ as above, we then get
$$
\sum_{k,l=-R}^R f_{k,l}(x_{-R,-R},\ldots,x_{R,R}) \lesssim n^4 + n \ln n,
$$
which obviously is not quite a counterexample yet. However, from the homogeneity of the terms it is immediate that scaling by a factor $\lambda > 0$ leads to
$$
\sum_{k,l=-R}^R f_{k,l}(\lambda x_{-R,-R},\ldots,\lambda x_{R,R}) \lesssim \lambda^2 n^4 + \lambda^{-2} n \ln n.
$$
A choice of $\lambda = n^{-3/4} (\ln n)^{1/4}$ (so that both terms have the same asymptotic behavior) leads to
$$
\sum_{k,l=-R}^R f_{k,l}(\lambda x_{-R,-R},\ldots,\lambda x_{R,R}) \lesssim n^{2.5} \sqrt{\ln n}.
$$
Somehow I suspect that this is actually the lower bound for the asymptotic behavior of your sum for general $(x_i)$, but I don't have a proof of that.

It is true and follows from expanding the following expression:
Consider a vector $x=(x_1,...,x_N)$, and define vectors $X,Y$ as

$X_i:=\frac{2}{N} x_i$ and $Y_i:=-\frac{2}{N^2} \sum_{j \neq i } \frac{x_i-x_j}{\vert x_i- x_j \vert^2}.$

It follows then that

$$ \vert X+Y \vert^2= \frac{4}{N^2} \vert x \vert^2 + \frac{4}{N^4} \sum_{i=1}^N \vert \sum_{j \neq i } \frac{x_i-x_j }{\vert x_i-x_j \vert^2} \vert^2+ -\frac{4}{N^3} \sum_{i \neq j } \left( x_i-x_j \right) \frac{\left(x_i-x_j\right)}{\vert x_i-x_j \vert^2}$$

Thus, $$ \vert X+Y \vert^2= \frac{4}{N^2} \vert x \vert^2 + \frac{4}{N^4}  \sum_{j \neq i } \frac{1}{\vert x_i-x_j \vert^2}  -\frac{4(N-1)}{N^2}.$$


Calculating area of a parallelogram using length of two diagonals Is there a formula for calculating area of a parallelogram using only the length of diagonals? if so what is it?

No.  Imagine two fixed diagonals that are attached and can rotate in the center.  As they pivot, different parallelograms are induced; if they are almost parallel the area will be close to zero.

If the diagonals are given in vector form, then as @RecklessReckoner said, one diagonal is vector sum $a+b$ and the other is vector sum $a-b$. So you have two equations and two variables, you can find out vectors $a$ and $b$, then simply find the magnitude of their cross product.

If the diagonals are vectors $p$ and $q$, the area can be represented as the magnitude of the cross product: $\frac 14 (p+q) \times (p-q)$.


Linear algebra problem from dummite & foote Let $V$ be a finite dimensional vector space over $\mathbb{Q}$ and suppose $T$ is a nonsingular linear transformation of $V$ such that $T^{-1} = T^2 + T$. Prove that the dimension of $V$ is divisible by $3$. If the dimension of $V$ is precisely $3$, prove that all such transformations $T$ are similar.

So applying $T$ to both sides of the given equation gives us $T^3 + T^2 - I = 0$, hence the minimal polynomial $m(x)$ of $T$ divides $x^3 + x^2 + 1$. But this polynomial is irreducible over $\mathbb{Q}$ by the rational root test, hence $m(x) = x^3 + x^2 + 1$. The structure theorem for finitely-generated modules over PIDs tells that
$$(V,T) \cong \bigoplus_{i=1}^{t}\frac{\mathbb{Q}[x]}{(a_i(x))},$$
where $t \geq 1$ and $a_i(x) \mid a_{i+1}(x)$ and $a_t(x) = m(x)$. But since $m(x)$ is irreducible, we must have each $a_i(x) = m(x)$, therefore
$$(V,T) \cong \left(\frac{\mathbb{Q}[x]}{(f(x))}\right)^t.$$
Now the dimension of $V$ equals $3t$ and is thus divisible by $3$. But doesn't it then also follow that given a dimension $3t$, any two such transformations $T$ make $V$ isomorphic as a $\mathbb{Q}[x]$-module to the above, hence are similar? It seems true in general, not just for $\dim(V) = 3$. Thanks!

Since the polynomial is irreducible, it has distinct roots. Let $a,b,c$ be these distinct roots. Then consider the following two matrices.

\begin{pmatrix}
  a & 0 & 0 & 0 & 0 & 0 \\
  0 & a & 0 & 0 & 0 & 0 \\
  0 & 0 & b & 0 & 0 & 0 \\
  0 & 0 & 0 & b & 0 & 0 \\
  0 & 0 & 0 & 0 & c & 0 \\
  0 & 0 & 0 & 0 & 0 & c
\end{pmatrix}

\begin{pmatrix}
  a & 0 & 0 & 0 & 0 & 0 \\
  0 & a & 0 & 0 & 0 & 0 \\
  0 & 0 & a & 0 & 0 & 0 \\
  0 & 0 & 0 & a & 0 & 0 \\
  0 & 0 & 0 & 0 & b & 0 \\
  0 & 0 & 0 & 0 & 0 & c
\end{pmatrix}

Obviously, these two matrices have the same minimal polynomial $p = x^3 + x^2 - 1$, however they are not similar.


A finite set always has a maximum and a minimum. I am pretty confident that this statement is true. However, I am not sure how to prove it. Any hints/ideas/answers would be appreciated.

Let $S = \{s_1, \ldots,s_n\}$ be a nonempty finite set of size $n > 0$. We will show by induction on $n \in \mathbb N$ that there exist some $m,M \in S$ such that for all $s \in S$, we have that $m \leq s \leq M$.

Base Case: For $n=1$, we have $S = \{s_1\}$, so taking $m = s_1$ and $M=s_1$ trivially satisfies the required condition.

Induction Hypothesis: Assume that the claim holds for $n=k$, where $k \geq 1$.

It remains to prove that the claim holds true for $n = k+1$. To this end, choose any set $S$ with $k+1$ elements, say $S = \{s_1 ,\ldots,s_k,s_{k+1}\}$. Now by the induction hypothesis, the subset:
$$
S' = S \setminus \{s_{k+1}\} = \{s_1 ,\ldots,s_k\}
$$
has a minimum element and a maximum element. That is, we know that there exists some $m',M' \in S'$ such that for all $s' \in S'$, we have that $m' \leq s' \leq M'$. Now observe that $s_{k+1}$ must fall under $1$ of $3$ cases:

Case 1: Suppose that $s_{k+1} < m'$. Then take $m = s_{k+1}$ and $M=M'$. To see why this works, observe that any element in $S$ is either $s_{k+1}$ or some $s' \in S'$, and:
$$
m = s_{k+1} < m' \leq s' \leq M' =M
$$

Case 2: Suppose that $m' \leq s_{k+1} \leq M'$. Then take $m = m'$ and $M=M'$. To see why this works, observe that any element in $S$ is either $s_{k+1}$ or some $s' \in S'$, and:
$$
m = m' \leq s_{k+1} \leq M' = M
$$
$$
m = m' \leq s' \leq M' = M
$$

Case 3: Suppose that $s_{k+1} > M'$. Then take $m =m'$ and $M=s_{k+1}$. To see why this works, observe that any element in $S$ is either $s_{k+1}$ or some $s' \in S'$, and:
$$
m = m' \leq s' \leq M' < s_{k+1} = M
$$

Hence, we have shown that $S$ has a minimum and maximum element, as desired.

I suppose you mean a set of numbers, $S=\{s_1,s_2,\dots,s_n\}$ and $n$ is the finite size, just let

$$
m=\min(s_1,s_2,\dots,s_n)
$$
$$
M=\max(s_1,s_2,\dots,s_n)
$$
Q.E.D.


Stability of periodic solution I am stuck on this question:

  $$x'(t) = x(x-p(t))(x-1),$$ where $p$ is $1$-periodic and $0< p(t) <1$ for all $t$. We want to show the existence of a STABLE periodic solution $\mu$ such that $0< \mu(t) <1$ for all $t$.

If we require $0< x_0 <1$, where we impose the initial condition that $x(0) = x_0$, then it is clear that there exists a solution $\mu$ that satisfies the ODE with this initial condition. Standard result also shows the existence of such periodic solutions. My question is how to show the stability of such solutions by the definition of stable periodic solutions.

Well, let me explain my idea in full extent. First of all, instead of considering first order ODE, I consider an equivalent system of ODEs:

$$ \dot{x} = x(x - p(\tau))(x - 1), \;\; \dot{\tau} = 1 . $$

It's easy to verify that integral curves of original ODE are the same as integral curves of this system.
Also, from this form we get that vector field that is defined by this system is 1-periodic in $\tau$: if $(\tilde{x}(t), t)$ is a solution then $(\tilde{x}(t), t + k)$ is also a solution for any $k \in \mathbb{Z}$. This symmetry naturally allows us to consider behaviour of trajectories only in strip $(t, x) \in \lbrack 0, 1 \rbrack \times (-\infty, +\infty)$ because behaviour of any full trajectory can be "glued" using the "data" from the strip.

From here I'll assume that $p(\tau)$ is a continuous function.
This allows us to use two theorems immediately:
1. Existence and uniqueness theorem. The uniqueness part simply means that integral curves either coincide or have empty intersection.
2. Denote by $w(t; w_0) = (\tau(t), x(t))$ the solution that satisfies IVP $w(0; w_0) = w_0 = (\tau_0, x_0)$. Then function $F(t, u) = w(t; u)$ is continuous w.r.t. $t$ and $u$.

The purpose of the following part is two-fold: it proves that there is a 1-periodic solution and it shows that it is asymptotically Lyapunov stable. I don't know what kind of theorem you used for showing that there is a periodic motion, but I believe that my explanation will be quite simple. Here's an illustration:

As I've said earlier, choose $a$ and $b$ in such way that $0 < a < p_{min} < p_{max} < b < 1$ holds. You can simply check that vector field points along boundary in the same way as on the figure. Such configuration allows us to tell that all trajectories that start at $(\tau, x) \in \lbrace 0 \rbrace \times \lbrack a, b \rbrack$ can leave $(\tau, x) \in \lbrack 0, 1 \rbrack \times \lbrack a, b \rbrack$ only through $(\tau, x) \in \lbrace 1 \rbrace \times \lbrack a, b \rbrack$. To be absolutely precious, you might check vector field at points $(1, a)$ and $(1, b)$, see that trajectories that start at them leave
$(\tau, x) \in \lbrack 0, 1 \rbrack \times \lbrack a, b \rbrack$ in backward time and conclude that all trajectories escape strip only through
$(\tau, x) \in \lbrace 1 \rbrace \times ( a, b )$.

It's time to define "time-1 mapping". We define it by formula

$$ f(x) = F(1, (0, x)).$$

This is a continuous mapping of segment $\lbrack a, b \rbrack$ into itself.
If we were interested only in proving existence of periodic solution (i.e.,
which has $f(x) = x$), we could use a Brouwer fixed-point theorem (which
states that such fixed point exists).
But we need to investigate the stability and we can use simpler arguments. Note that $f(a) > a$ and $f(b) < b$, so $f(a) - a > 0$ and $f(b) - b < 0$; because $f$ is continuous there exists a point $\hat{x}$ where $f(\hat{x}) - \hat{x} = 0$. Therefore, we proved that $f(x)$ has fixed point and our system of ODEs has a 1-periodic solution.

But we can prove more! Recall that I've said that in this case $f(x)$ is monotonically increasing function. It's very simple to prove. Pick any point $(0, x^\ast)$ and consider the integral curve that is passing through this point. As we already know, it will intersect line $\tau = 1$ at point $(1, f(x^\ast)$. Now consider the closed domain $G$ which is bounded by lines $\gamma$,
$\lbrace 0 \rbrace \times \lbrack a, x^\ast \rbrack$, $\lbrack 0, 1 \rbrack \times \lbrace a \rbrace$ and $\lbrace 1 \rbrace \times \lbrack a, f(x^\ast) \rbrack$. By construction there is only one way to escape $G$ -- only through the $\lbrace 1 \rbrace \times \lbrack a, f(x^\ast) \rbrack$ (because escaping through $\gamma$ will violate uniqueness theorem). But this simply means that for any $x \in (a, x^\ast)$ follows $f(x) < f(x^\ast)$. And it holds for any $x^\ast \in (a, b \rbrack$, which simply means that $f(x)$ is increasing function.

But what does that all mean for stability of periodic motion?
 Here is another illustration:

It's a cobweb diagram for our mapping $f$. It's easy to prove that iterations of any point from segment $\lbrack a, b \rbrack$ converge to fixed point. It means that after each period any other solution "starts" at cross-section closer and closer to periodic solution. Using some $\epsilon$-$\delta$ machinery you can prove that periodic solution is asymptotic Lyapunov stable, but we already have the key fact that fixed point on cross-section is attractive. After that the proof should be quite simple.

EDIT

As @Michael pointed out, my illustration is not the only possibility for cobweb diagram. There could be other fixed points as well: at some of them $y = f(x)$ might touch $y=x$ and at some of them they might be intersecting transversely.
The key observation is that there should exist a fixed point $\hat{x}$ such that $f(x) > x$ for $x \in \lbrack \hat{x}-\varepsilon, \hat{x} )$ and
$f(x) < x$ for $x \in ( \hat{x}, \hat{x} +\varepsilon \rbrack$. Why such fixed point exists? It's due to these facts:
1) $y = f(x)$ starts in the $y > x$ part of square $\lbrack a, b \rbrack \times \lbrack a, b \rbrack$ and ends in the
$y < x$ part;
2) we definitely have fixed points for $y = f(x)$.

By reductio ad absurdum we could show that if we have no such fixed points, then curve $y = f(x)$ stays only in $y \geqslant x$ part of square which contradicts the first fact.

The fixed point with such properties would be attractive and will correspond to the stable periodic solution.

Okay, so here is a proof that $f$ is continuous (which takes care of a detail in Evgeny’s proof of existence of a fixed point of $f$).
Setup:
Recall the ODE is:
$$ x’ = x(x-p)(x-1) $$
Assume $p$ is integrable and there are values $p_{min}, p_{max}$ such that for all $t$ we have
$$ 0<p_{min}\leq p(t) \leq p_{max} < 1 $$
Further impose any additional mild assumptions so that solutions with a given initial condition exist, are unique, and do not cross (perhaps we need $p$ continuous for that?)
Let $a,b$ be values such that $0<a<p_{min}<p_{max}<b<1$.
Define Evgeny’s function $f:[a,b]\rightarrow\mathbb{R}$ such that for each $z \in [a,b]$, the value $f(z)$ is defined as  $x(1)$ for a solution $x(t)$ to the ODE with initial condition $x(0)=z$.  We want to show $f$ is continuous over $[a,b]$.
Proof:
Fix $z \in [a,b]$.  Fix a small value $\delta \neq 0$ such that $z+\delta \in [a,b]$.  We want to show that $|f(z+\delta)-f(z)| \leq |\delta|D$ for some positive constant $D$.   Define $x(t)$ as the ODE solution with $x(0)=z$.  Define $y(t)$ as the ODE solution with $y(0)=z+\delta$.  Define $g(t) = y(t) - x(t)$, so that $g(0) = \delta$.  Since $y=x+g$ we get:
$$ y’ = x’ + g’  = x(x-p)(x-1) + g’ $$
On the other hand, since $y$ satisfies the ODE we get:
$$ y’ = y(y-p)(y-1) = (x+g)(x+g-p)(x+g-1) $$
Equating the above two expressions gives:
$$ x(x-p)(x-1) + g’ = (x+g)(x+g-p)(x+g-1) $$
and so:
$$ g’ = y’ - x’ = g\underbrace{[g^2 - gp + 3gx - g - 2px + p + 3x^2 -2x]}_{R(t)} $$
That is, for all $t$ we have
$ g’(t) = g(t)R(t) $.
Now, we already know (from above comments) that any solution to the ODE with initial condition in $[a,b]$ must stay in $[a,b]$.  Thus, $x,y, g$ are bounded  for all time, and there is some positive constant $C$ such that $|R(t)| \leq C$ for all $t$.
For all $t$ we have  $g(t) \neq 0$  (since $\delta\neq 0$ and solutions cannot cross). Thus:
$$ \frac{g’(t)}{g(t)} = R(t)$$
So for all $t \geq 0$:
$$ \log[|g(t)|/|g(0)|] = \int_0^t R(\tau)d\tau $$
So
$$ |g(t)| = |g(0)|e^{\int_0^t R(\tau)d\tau} = |\delta|  e^{\int_0^tR(\tau)d\tau} \leq |\delta| e^{Ct} $$
Thus:
$$ |f(z+\delta) - f(z)| = |y(1) - x(1)| = |g(1)| \leq |\delta| e^{C} $$
$\Box$


Are there any interesting sets of real functions whose cardinality is unknown? Consider the set, let's call it $R$, of all functions $f:\mathbb{R} \to \mathbb{R}$? Are there any interesting subsets $S \subset R$ whose cardinality is (currently) unknown? By interesting, I mean something "natural" (e.g., continuous real functions, the set of all real functions with a countable number of discontinuities, Darboux functions) which analysts might care about.

The motivation for this question is honestly just curiosity. I find this quite interesting, because there's so many types of real functions we can consider. Surely there's some interesting, nontrivial examples worth discussing.

How about the space of all real-valued functions which are not continuous? You can actually show that there are some input-output pairs of real numbers which satisfy the definition of continuity. In other words, you can show continuous functions exist without ever writing one down!

Proof
$(1)$ Let $F(\mathbb{R},\mathbb{R})$ denote the set of all possible single variable, real valued functions.

$(2)$ Now let $S\in F(\mathbb{R},\mathbb{R}) $ denote the set of all functions where if the difference between output values is less than $\epsilon$, then we can find a corresponding $\delta$ which will always be greater than the difference between two arbitrary input values. This is possible, we can just make up a rule where the difference between input values $x_1$ and $x_2$ is twice the difference between $f(x_1),f(x_2)$. Play around with some numbers to get the intuition.

$(3)$ Let $f\in S$. Consider two points of the function $f(x_{1})$ and $f(x_{2})$. Remember since $f\in S$, the difference $|f(x_{1})-f(x_{2})|<\epsilon$ and $|x_1-x_2|<2\delta$ by definition. This fits the definition of continuity, and so continuous functions do exist. You can modify to show its possible to have differentiable functions, monotonic etc.

What is the cardinality of $F(\mathbb{R},\mathbb{R})$? What about the set of monotone functions? What about the set of functions which are differentiable, but the derivative is not differentiable anywhere?

To conclude, you could reasonably conjecture (based on cardinality of power sets) that there are probably a $\aleph_{2}$ number of possible subsets of $F(\mathbb{R},\mathbb{R})$ we could ask ourselves what the cardinality of each of those sets are. Consider a set $X$ where each function is bounded by some constant: $|f(x)|<M$. We could choose any $M\in (1,\infty)$ to be the bound, and that means there are $\aleph_1$ such functions in $X$ alone.


Largest base $b$ such that sum of digits of base $b$ of $11^4$ do not add up to $16$ When $11^4$ is written out in base $10$, the sum of its digits is $16=2^4$. What is the largest base $b$ such that the base-$b$ digits of $11^4$ do not add up to $2^4$?
Is there an elegant solution? I tried plugging in bases, but this obviously does not work. I am thinking Binomial Theorem, but I am not sure how that will translate to other bases.

Assuming you are asking about $11_b^4$ in base $b$, the largest is base $6$, where
$11_6^4=7^4_{10}=15041_6$ with a digit sum of $11_{10}$ or $15_6$
If we don't worry about carries, we have $11^4=14641$ in any base.  This comes from $(b+1)^4=b^4+4b^3+6b^2+4b+1$.  It is only when the $6$ carries that the sum will not be $16_{10}$

Answer appears to be any base >14626.
$$
11^4_{10} = 14641_{10}
$$
$$
11^4_{14626} = 1F_{14626}
$$
Sum of digits in base 14626 appears to be 16. All bases higher than that have a different sum for the digits.


Monotone continuous function Could some one please to verify if this converse statement is true?

The original statement: "Suppose $\,f: D \rightarrow R$ is monotone. If $\,f(D)$ is an interval then $\,f$ is continuous". Which is proved in the text book.

The converse statement: "Suppose $\,f: D \rightarrow R$ is monotone. If $\,f$ is continuous then $\,f(D)$ is an interval"

If $D$ itself is an interval, this is true (even without requiring monotonicity of $f$): If $a<b<c$ and $a, c\in f(D)$, then by the intermediate value theorem, you also have $b\in f(D)$.

If $D$ is any subset of the real numbers, this is wrong in general: $f: [0, 1]\cup [2,3] \rightarrow \mathbb{R}, x \mapsto x$ is obviously strictly monotone and continuous, but $f(D)=[0,1]\cup [2,3]$ is not an interval.


How many unique combinations to wear seven different socks? My friend has seven unique but similar socks.

How many ways does he have to wear them, not double counting same combos?

$\mathbf {Attempt}$

I think $7! - 5!$ , but then you forget you can put the first foot sock on the other foot as well so $7!-5!+6$ is the correct answer?

You are choosing $2$ socks from $7$ socks. That is ${7\choose 2}=\frac{7!}{2!5!}=21$, where you can first choose from $7$ socks, then you can choose from the $6$ remaining socks (so $7\cdot 6=\frac{7!}{5!}$) and due to counting each pair double, you divide by $2!=2$.

I think it's pretty well explained here, also for the general case $\frac{n!}{k!(n-k)!}$:
https://betterexplained.com/articles/easy-permutations-and-combinations/

You have $7$ choices for your left root. Then you have $6$ choices for the right. This gives $42$ choices. However, this counts $AB$ and $BA$ as different, while the problem says to not double count the same pair. So we have half that many choices, so $21$ is the answer.


Probability Distribution Of Random Variable Defined By Gaussian Variables I don't have answers to this question
Please let me know if there are mistakes

Problem:

  $X$, $Y$, and $Z$ are independent Gaussian random variables with:
  $E(X) = 2$$E(Y) = −1$$E(Z) = 4$$Var(X) = 2$ $Var(Y) = 1$  $Var(Z) = 2$
   If the random variable $W$ is given by $W = 3X + 3Y − Z$

  i) Find the mean and variance of $W$ ii) Write down the probability
  distribution for $pW(w)$ iii) What is the probability that a sample
  of $W$ will be between 0 and 5?

Attempt:

i)

\begin{align}
E(W) &= E(3X + 3Y - Z)\\
\ &= E(3X) + E(3Y) - E(Z) \\
\ &= 3E(X) + 3E(Y) - E(Z) \\
\ &= 3(2) + 3(-1) - 4\\
\ &= -1 \\
\end{align}

\begin{align}
V(W) &= V(3X + 3Y - Z)\\
\ &= V(3X) + V(3Y) - V(Z) \\
\ &= 9V(X) + 9V(Y) - V(Z) \\
\ &= 9(2) + 9(1) - 2\\
\ &= 25 \\
\end{align}

ii)

I am not sure how to find $pW(w)$ knowing $W$, $E(W)$ and $V(W)$.

iii)

Given $pW(w)$ from part ii I assume the answer would be:
$P(0 ≤ W ≤ 5) = \int_{0}^5 pW(w) dW$

Thank you

In part a, you mistakenly subtract the third variance rather than adding it. You know how when you multiply a variable by a constant, the variance gets multiplied by the constant squared? Well, the same goes for $(-1).$ So the variance should be $29,$ not $25$

For part $b$ any sum of independent Gaussians is a Gaussian and a Gaussian is fully characterized by its mean and variance. So we have $W\sim N(-1,29)$ and it has PDF $$f_W(w) = \frac{1}{\sqrt{58\pi}}e^{-\frac{(w+1)^2}{58}}.$$

Part c looks good too, and now you know what to plug in. It might be easier to write it in terms of the $N(0,1)$ quantity $\frac{W+1}{\sqrt{29}}$. We have $$0<W<5 \iff \frac{1}{\sqrt{29}}< \frac{W+1}{\sqrt{29}} <\frac{6}{\sqrt{29}} $$ so the answer should be $$P(1/\sqrt{29} < N(0,1) < 6/\sqrt{29}) = \Phi(6/\sqrt{29})-\Phi(1/\sqrt{29})$$ where $\Phi$ is the standardard normal CDF.

Note A previous version mistakenly gave OP an 'all clear' on part (a).


Diffeomorphism from level set onto $S^2$ I'm given a map $\phi : R^4 \rightarrow R^2$ defined by $\phi(x,y,s,t) = (x^2 + y, x^2+y^2+s^2+t^2+y)$.

It's easy to show that the level set $C = \phi^{-1}(0,1)$ is a smooth submanifold of $R^4$ with dimension $2$ by showing that $(0,1)$ is a regular value.

I'm trying to show that this level set is diffeomorphic to $S^2$, the unit sphere in $R^3$. The only thing I can think of is to construct a smooth embedding from $S^2$ into $R^4$ whose image is $C$, since smooth embeddings are diffeomorphisms onto their images.

It's easy enough to get that $C$ is diffeomorphic to $\{(x,y,z) \mid x^4 + y^2 + z^2 = 1\}$, but from there I'm stuck. The map $(x, y, z) \rightarrow (\text{sign}(x) x^2, y, z)$ is a differentiable homeomorphism from $S^2$ onto $C$, but it's not even $C^2$ let alone smooth.

Once you're working with the surface $C=\{ (x,y,z) : x^4+y^2+z^2=1 \}$, how about the map
$$(x,y,z) \mapsto \frac{(x,y,z)}{ \sqrt{x^2+y^2+z^2}} ?$$

$C$ is a ball-type surface centered at the origin, so why not just project to the sphere in this simple way?

The level set is traced out by the intersection of x^2+y=0 and x^2 + y^2 +s^2 +t^2 + y =1. Combine the equations and get y^2 + s^2 + t^2 = 1.


Can limits of be computed via Fourier transforms Suppose, we want to find a   limit of $f(t)$ or $t^n f(t)$ as $t \to \infty$.   Can these limits be computed via fourier tranform of $f(t)$?

A function $f$ is absolutely continuous on $[a,b]$ iff there is a function $g$ that is absolutely integrable on $[a,b]$ and satisfies
$$
                  f(y)-f(x) = \int_{x}^{y} g(t)dt,\;\;\; x,y \in [a,b].
$$
Equivalently, a function $f$ is absolutely continuous on $[a,b]$ if it is continuous on $[a,b]$, has a derivative a.e. on $[a,b]$, the derivative is integrable on $[a,b]$ and $f(y)-f(x)=\int_{x}^{y}f'(t)dt$ for all $x,y\in [a,b]$.

The Fourier transform is perfectly tied to the differentiation operator as follows: $f \in L^2(\mathbb{R})$ is absolutely continuous on every finite interval, with $f' \in L^2(\mathbb{R})$ iff
$$
               \int_{-\infty}^{\infty}|\hat{f}(\xi)|^2+|\xi\hat{f}(\xi)|^2d\xi < \infty.
$$
And $f$ is n-times absolutely continuous with $f,f',\cdots,f^{(n)}\in L^2$ iff
$$
          \int_{-\infty}^{\infty}|\hat{f}(\xi)|^2+|\xi\hat{f}(\xi)|^2+\cdots+|\xi^{n}\hat{f}(\xi)|^{2}d\xi < \infty,\\
    \mbox{ or } \int_{-\infty}^{\infty}(1+|\xi|^{2n})|\hat{f}(\xi)|^2d\xi < \infty.
$$
That's about it when it comes to iff conditions.


Efficient way to find pairs of multiplicative inverses Consider the space $\mathbb{Z}_n$, where $n$ is a reasonably large prime, say, for example, 53. How can I quickly and efficiently write out the multiplicative inverses for each element of $\mathbb{Z}_{53}$? I don't want to use the Extended Euclidean algorithm 26 times (for the 52 total terms in this set). Is there a way to do this that's not just "brute force", so to speak?
To be clear, I want to do this by hand, not using a computer.

Let $[a]$ be a residue class modulo the prime $p$ and let $[b] = [a]^{-1}$ be its multiplicative inverse.  You can, for instance, find this inverse using the extended Euclidean algorithm.  Then,
\begin{align*}
[a^2]^{-1} &= [b^2]  \text{,}  \\
[a^3]^{-1} &= [b^3]  \text{,}  \\
&\vdots
\end{align*}
So use the slow method to get one result, then go through successive powers to fill in more entries in your table of inverses.  At some point, the powers cycle back, perhaps not filling in the entire table.  Now pick one of the unfilled entries and start this process again.  When you run out of empty entries, hooray!, you're done.
One interesting thing that can happen is that you can in one step work out the inverses of the powers of $[c^2]$.  At a later step, you find you are working out the powers of $[c]$ because every even power is already in your table and every (or possibly most) odd powers are new entries.  (Some entries may already be filled in as members of other cycles.)
As an example, modulo $17$, $[2]^{-1} \cong [9]$, so
\begin{align*}
[2^2]^{-1} \cong [4]^{-1} &\cong [9^2] \cong [13]  \\
[2^3]^{-1} \cong [8]^{-1} &\cong [9\cdot 13] \cong [15]  \\
[2^4]^{-1} \cong [16]^{-1} &\cong [9\cdot 15] \cong [16]  \\
[2^5]^{-1} \cong [15]^{-1} &\cong [9 \cdot 16] \cong [8]  \\
[2^6]^{-1} \cong [13]^{-1} &\cong [9 \cdot 8] \cong [4]  \\
[2^7]^{-1} \cong [9]^{-1} &\cong [9 \cdot 4] \cong [2]  \\
[2^8]^{-1} \cong [1]^{-1} &\cong [9 \cdot 2] \cong [1]  \\
[2^9]^{-1} \cong [2]^{-1} &\text{and we've cycled.}
\end{align*}
Note that $2^8 \cong 1$ was a slightly earlier signal that we were about to cycle.
Notice we got half of the entries for the table.  Now try it with $[3]$ and (I think) get the other half (or go around again with another new starting point).
There is a theoretical idea floating around in the above: primitive roots.  Pulling a primitive root out of thin air is hard, so here, we just hope that we either picked a primitive root or picked a small power of a primitive root so we can fill in many table entries.  Even if you don't, you get some table entries.
We could probably estimate the expected number of cycles to complete the table, which probably has a leading term of the approximate size $\frac{\varphi(p)}{\varphi(\varphi(p))}$, where $\varphi$ is the Euler totient function, to capture the number of residue classes that are a small power of one of the primitive roots modulo $p$.

EDIT: I missed that you wanted every element in $\mathbb{Z}_p$; I'm sure there's a faster way than this for that application.
I  will expand on my comment; Euler's theorem states that
$$
a^{\phi(m)} \equiv 1 \mod m
$$
where a is coprime to m and $\phi$ is the Euler totient function. Suppose m is the prime you speak of, then notice that $\phi(m) = m-1$. So then we find that:
$$
a^{m-1}  \equiv 1 \mod{m} \implies a^{m-2} \equiv a^{-1} \mod{m}
$$
Therefore, if we take $m = 53$, as in your example, computing the multiplicative inverse is as simple as computing $a^{51}$, which can vary from being very easy (say if $a = 2$), to very bad (say if $a = 51$).


Square root and principal square root confusion A few months ago I asked a question about the $\pm$ symbol because I was confused about it... I still carry the same confusion (which really bugs me) but I think the real confusion has to do with the square root and principal square root. I hope I can finally grasp the concept with the following two questions...
Question 1:
$\sqrt{12x^2} = \sqrt{4\cdot3x^2} = 2|x|\sqrt{3}$
Why do we only have to keep in mind that $x$ could be negative, yet we just factorize $\sqrt{12}$ to $\sqrt{3\cdot4}$ and put the 2 right in front of the radical sign?
$\sqrt{12}$ could be factorized as $\sqrt{-3 \cdot -4}$ as well.
Question 2:
I've been told to take the square root of both sides in the following equation, but the square root doesn't have its own symbol right? Only the principal root does... So if you'd use the radical sign you're using the principal root and therefore you're missing out on a solution in the following scenario:
$(x + 8)² = 1 \iff \sqrt{(x + 8)^2} = \sqrt1 \iff x + 8 = 1 \iff x = -7$
Does that mean that whenever we want to take the square root of something (not the principal) we just use a $\pm$ symbol instead? For example:
$(x + 8)^2 = 1 \iff x + 8 = \pm 1 \iff x = -8\pm1$

For question 1, you are correct that you could factor $12=-3 \cdot -4$, but it is not useful.  You are choosing to use $12=3 \cdot 4$ because that is useful.  You know $3,4 \gt 0$ because you chose them that way.  $x$ is a variable, and we don't know its sign, so we need the absolute value signs.  We pull out the $2$ because we like $2 \sqrt 3$ better than $\sqrt {12}$.  It is a matter of taste, and in some intermediate calculations we would prefer $\sqrt {12}$ or some other form.

For question 2, yes I would just go to $x+8 = \pm 1$

We have to keep in mind that $x$ might be negative because we don't know what $x$ is and it might by negative.

But we don't have to keep anything in mind about $12$ because we know exactly what $12$ is.
But perhaps we need to understand why we are allowed to split $\sqrt {ab}=\sqrt{a}\sqrt{b}$ in the first place.
Real Number Answer (Assumes Since $m^2 \ge 0$ for all $m \in \mathbb R$ then $x^2 = $a negative number is impossible)
If $k=\sqrt{a}\sqrt{b}>0$ (presuming $a,b> 0$) then $k^2 = (\sqrt{a}\sqrt{b})^2 = \sqrt{a}^2 \sqrt{b}^2 = ab$ so if we ask "What is $\sqrt {ab}$, that is what is the unique non-negative number so $x^2=ab$".  Well, we just demonstrated that $k = \sqrt{a}\sqrt{b}$ is such a number.
So if we have $M > 0$ and $M$ could be written as $M = a*b$ where $a,b\ge 0$ we would have $\sqrt M = \sqrt{a}\sqrt {b}$.
Now you are correct that $M = (-a)*(-b)$.  But how does that help us in anyway.  We explained why $M =a*b$ works but $M=(-a)*(-b)$ is as useful for find square roots as noting that $M = j+k$ for some $j$ and $k$.  It's just not relevant.
We need $\sqrt{a}$ so that $\sqrt{a}^2 =a$ as well as a $\sqrt{b}$ so that $\sqrt{b}^2 = b$ for this to work and if we tried it with $-a*-b$ we'd need $\sqrt{-a}$ so that $\sqrt{-a}^2 = -a$ and there's just no such thing.
Unless we use complex numbers...
Complex number:  There exist $i$ and $-i = (-1)*i$ so that $i^2=-1$. (And $(-i)^2 = i^2 = -1$ as well.
So if $k^2 = 12$ we could do $(\sqrt{3}2)^2 = 3*4 = 12$ so $k=\sqrt 3\sqrt 4=2$ is an acceptable answer.  So is $(\sqrt{3}i*2i)^2 = (\sqrt\3i)^2*(2i)^2 = (-3)*(-4) = 12$ but $\sqrt{3}i*2i$ is just another way of writing $\sqrt{3}i*2i = (\sqrt{3}2)*(i^2)= -2\sqrt 3$.
So the two values of $k$ are $2\sqrt 3$ and $-2\sqrt 3$ just as we wanted.
But we can do $\sqrt{12} = \sqrt{-3}\sqrt{-4} = \sqrt 3i* 2i=-2\sqrt 3$.
(I suppose I should point out we tend not to worry about primary square roots in complex numbers because it just opens a can of worms.)

"I've been told to take the square root of both sides in the following equation"

Well, short answer, since we do not know whether $x+8$ is positive or negative we don't know that $\sqrt{(x+8)^2} = x+8$.  If $x = -17$ that is false.
$\sqrt{(-17+8)^2} = \sqrt{(-9)^2} =\sqrt{81}= 9\ne -17+8$.
Instead all we know is that $\sqrt{(x+8^2} = |x+8|$
$\sqrt{(-17+8)^2}=\sqrt{81} = 9 $ which does $|-9| = |-17+8|$.
So to solve
$(x + 8)^2 = 1 \implies \sqrt{(x+8)^2} = \sqrt 1\implies |x+8| = 1$
$\implies -(x+8) = 1$ OR $x+8 = 1$ $\implies$
$x+8 = -1$ ** OR ** $x+8 = 1$
And we can write that last line as
$x+8 = \pm 1$.
Or we can take a short cut and go directly to:
$(x+8)^2 = 1$
$\sqrt{(8+x)^2} = \sqrt 1$
$8+ x = \pm 1$.
.... and from here we can go
$x = \pm 1 -8$ so $x= -1 -8=-9$  OR $x=1-8=-7$.


Number of walks on triangular latice Let $a_n$ denote the number of self-avoiding walks of length $n \in \mathbb{N}$ on the triangular lattice, it is the two-dimensional lattice where unit triangles are packed together. I want to show that
$$2^n \leq
a_n ≤ 6 \cdot 5^{n−1}, ~~ n \in \mathbb{N}.$$
I am looking for a rigorous way of doing this instead of an intuitive explanation.

I think you forgot to specify that the walks all start at some origin; otherwise there are countably infinitely many of them.

As in your other question, since every vertex has $6$ neighbours, $6\cdot5^{n-1}$ is the number of walks that don't immediately return to the same place, and thus an upper bound for the number of self-avoiding walks.

For the lower bound, choose, say, two directions to the upper right and to the lower right; then in each step you can use one of them, and you'll always be going right and therefore never returning.


maximum number of maximal cliques I would like to know about upper bounds on the number of maximal cliques in graphs with small degrees. More precisely, how does the number of maximal cliques scale with graph size (i.e., number of nodes and links) and maximum degree (maximum of the degrees of all the nodes)?

If we know the size of the maximal clique $k$ then a trivial upper bound is $n/k$. An upper bound on the number of maximal cliques of all graphs is the number of vertices. An upper bound on the size of maximal clique is the maximum node degree $d$. Suppose $d$ is small relative to $n$, then the number of maximal clique is bounded by $n /d$ from below and $n$ from above.


Recursion formula for: $a_n = \left(\frac12\right)^n+2n−1$ How do I write a recursive formula for: $$a_n = \left(\frac12\right)^n+2n−1$$ so we were taught to expand the recursion and the multiply all together. So is the answer:

$$a_1=\frac32,\quad a_n=a_{n-1}+\frac52$$

How do you show the expanding and multiplying?

This answer is probably not correct.

Going from explicit to recursive is easy for any sequence; just compute $a_{n+1} - a_n$:

$a_{n+1} - a_n = \left[\left(\dfrac {1}{2} \right)^{n+1}+2(n+1)-1\right] - \left[\left( \dfrac {1}{2} \right)^n+2n-1 \right]$

There we go, we've already god a recursive sequence. Can you simplify it from here?

EDIT: To simplify $a_n-a_{n-1} = \left(\dfrac12 \right)^n - \left( \dfrac12 \right)^{n-1}+ 2$

Rewrite  $\left(\dfrac12 \right)^n$ as $\left(\dfrac12 \right)^{n-1} \cdot \left(\dfrac12 \right)^1$

$$\left(\dfrac12 \right)^{n-1} \cdot \left( \dfrac 12\right) - \left( \dfrac12 \right)^{n-1}+ 2$$

The first two terms both have $\left(\dfrac12 \right)^{n-1}$, so factor it out:

$$ \left( \dfrac {1}{2} \right)^{n-1} \left(\dfrac 12 - 1 \right) +2 $$

$\dfrac 12 - 1$ is equal to $- \dfrac 12$

$$ = \left( \dfrac {1}{2} \right)^{n-1} \left(-\dfrac 12 \right)^1 +2$$

Take the negative outside and add the powers

$$ -\left(\dfrac {1}{2} \right)^{n}+2$$


Is the following a Cauchy sequence? My working:

Fix epsilon>0. I need to show whether an N can be found such that

$|a_m - a_n| \lt \epsilon$ for all $m,n \ge N$

$|a_m - a_n| = |(-1)^m(2+1/m^2) - (-1)^n(2+1/n^2)|$

by taking m even, n odd (is this a good approach?) we get

$|4 + 1/m^2 + 1/n^2| \le |4+2/N^2|$

from here I'm not sure where to go.

Thanks for your help

Simply proove that both sequences have different limits. This is not possible for a Cauchy sequence.

Why? if $a_{2n}\to l_1$ and $a_{2n+1}\to l_2$m take $$\epsilon = \frac{|l_1 - l_2|}3$$

There is a $N$ such as $\forall n\ge N \  |a_{2n} - l_1| + |a_{2n+1} - l_2| < \epsilon$.

Then, for instance, $|a_{2n} - a_{2N+1}| \ge |l_1 - l_2| - (|a_{2n} - l_1| + |a_{2n+1} - l_2|)\ge \epsilon $.

Here, $a_{2n}\to 2$ and $a_{2n+1}\to -2$.

Short answer, no.  To verify, evaluate the limit of this sequence


Can anyone explain the final step in the following proof. Suppose $f:\mathbb{R} \to \mathbb{R}$ is twice differentiable and that $f''(x)+f(x)=0$ for all $x$
Prove that if $f(0)=0$ and $f′(0)=0$, then $f(x)=0$ for all x.

Proof:
$$f''(x)+ f(x) = 0 $$

$$ f'(x) f''(x) +f'(x)f(x) =0 $$

$$ (1/2)(f^2 + f'^2 )' =0$$

$$f^2 + f'^2=C$$

Since  $$ (f^2 + f'^2)(0)=0$$

We get $C=0$,  that is $f(x)=0$

Can anyone explain how they deduced that $f(x) = 0$, I am really confused.

It comes from
$$
0\le[f(x)]^2=\color{red}{-}[f'(x)]^2\le0,\quad x \in \mathbb{R},
$$ which is only possible with
$$
f(x)=0,\quad x \in \mathbb{R}.
$$

Let $f\left(x\right)=a_0+a_1x+a_2x^2...$
then, $$f\left(x\right)+f'\left(x\right)=(a_0+a_1)+(a_1+2a_2)x...=0$$
but $a_0=f\left(0\right)=0$ and $a_1=f'\left(0\right)=0$
Therefore, it must be zero at all points to satisfy them.


Integration of a square root of a reciprocal function I've been trying to solve this integration for an investigation but have no idea where to start:
$$\int\sqrt{\frac{a+x}{b-x}}dx$$
I've tried integrating by parts, but seem to end up with an even more complicated integral.
I also tried to move the $x$ to the denominator:
$$\int\sqrt{\frac{a+b}{b-x}-1}dx$$
But I don't think I've ever solved an integral like this before.

with $$t=\sqrt{\frac{a+x}{b+x}}$$ we get
$$x=\frac{t^2b-a}{1+t^2}$$ and $$dx=\frac{2t(a+b)}{(t^2+1)^2}dt$$ and our integral is given by now as $$\int\frac{2t^2(a+b)}{(t^2+1)^2}dt$$

Hint:

If $x\ne -a,$ we need $$(x+a)(x-b)<0$$

Assuming $b>-a,$ we need $$-a<x<b$$

$\iff p-a<p+x<p+b$

If we set $p+b=-(p-a),2p=a-b\implies\dfrac{a-b}2-a<\dfrac{a-b}2+x<\dfrac{a-b}2+b$

Set $\dfrac{a-b}2+x=\dfrac{a+b}2\cdot\cos2t$

$\iff x=b\cos^2t-a\sin^2t$

$x+a=?,b-x=?$


Points of Inflection Does $f(x)=|x^2−1|$ have an inflection point at $x=(-1)$ and $x=1$? It seems like the concavity is changing, but I just wanted to make sure.

No. A well-defined derivative does not exist at $x=\pm 1$. It has a local minimum at these points.

Yes, the concavity is changing, so you could say that. But be careful, since the first derivative at those points does not exist (hence, the second neither), so you can't test the signs of $f''(-1)$ and $f''(1)$. Here I asked a question about this same function, and you can see the graph of the function to convince yourself.


Is Euclidean Geometry complete and unique Please help me understand this concept of completeness of geometry and set me on the right path.

This is my context: From wikipedia, a formal system is complete if every tautology is also a theorem. For me this means that a system of axioms is complete if every statement that is true can be proved from the axioms (the question is tautologies in what kind of logic, first order, second order,...?). Anyway, it seems then that the concept of completeness depend on the system of axioms.

So, in the case of Euclidean Geometry, its completeness depends on their axioms (For example Euclid's Axioms, Hilbert Axioms, Tarki's Axioms,etc). Comparing for example the axioms of Hilbert and the axioms given by Tarski, I can see that they are essentially different in that Hilbert uses second order logic and Tarski's only first order logic. Also Tarski proved that his system of axioms is complete.

Now, the incompleteness theorem of Gödel stated that the system of axioms of arithmetic is incomplete. So, when we consider $R^2$ and $R^3$ as a representation of Euclidean Geometry we are talking about a system that is not complete because we are using the real numbers. So, are we speaking of different Euclidean Geometries? I'm totally confused.

It sounds like you might get a lot out of Greenberg's short paper Old and New Results in the Foundations of Elementary Plane Euclidean and Non-Euclidean Geometries.

In it he discusses how the ordinary theory of plane geometry is incomplete (pg 202), but why Tarski's geometry is complete (section 3.3, page 215).

For any fragment of euclidean geometry seems to be topologically and notionally equal.  R2 and R3 are different only in dimension.

One should be careful about what is being implied by Euclid's geometry.

Just as the curvature of space (the sum of angles of a triangle for example), leads to euclidean, hyperbolic and spherical geometries, there is a second statement for which any two, but not all three parts, can be true.  These are set in italics.

In a complete  and orientable space, lines cross once.

Euclid's geometry deals with a fragment of the plane, and never considers completeness.

The spheric geometries are necessarily complete, so one either rejects orientable to give elliptic, or once, giving spherical geometry.

So when one starts talking of the nature of the euclidean geometry and the horizon at infinity, one gets a presentation as a line (projective) or point (inversion).  Some people consider what happens on the other side of infinity, that is, they try to complet the euclidean plane.

A complete euclidean plane is perfectly modelled by the sphere at infinity.  This is the centre of a planes through a point, and thus must be a the centre of a plane centred on a real point.

The actual geometry is called "Möbius" geometry, but corresponds to the implemetation of 0/0 as undefined, that any circle is a straight line, and crosses other circles twice.  If one supposes that 'any point on the surface' is 'a point at infinity', and that straight lines contains the point at infinity, this folds down to the inversion model of the euclidean plane.

In short, Möbius geometry is oE, or orientable-complete-euclidean, whereas Euclid studied the geometry of orientable-fragment-euclidean.

Hyperbolic geometry is necessarily a fragment of space, even at 'infinity'.  This is because there is no meaning for a plane in 3D centred on a real point, unless you get Meta-infinities come to hand.


How many roots does the polynomial $p(z) = z^8 + 3z^7 + 6z^2 + 1$ have inside the annulus $1 < |z| < 2$? How many roots does the polynomial $p(z) = z^8 + 3z^7 + 6z^2 + 1$ have inside the annulus $1 < |z| < 2$?

I know I can use Rouche's Theorem. I'm just not sure how. It states that $|f(z) − g(z)| < |f(z)|$ where $f(z)$ is the given polynomial and we choose $g(z)$. Any solutions or hints are greatly appreciated.

General approach, you use Rouche theorem to count the number ofroots inside the circles of radius $1$ and $2$ and by substracting this quantity, you obtain the number of roots in the annulus.

How to apply Rouches theorem to find the number of roots in the circle of radius $1$. You can chose $f(z)=6z^2$ and $g(z)=z^8+3z^7+1$. On the considerd contour we have : $|g(z)|<|f(z)|$. So $f$ and $f+g$ have the same number of roots inside the contour counting multiplicity. So $p$ has $2$ roots in the circle of radius $1$.

I will let you do the same with the circle of radius $2$.

I think total zeros are 7-2=5 inside the annulus. Because, for |z|<1: 2 roots and |z|<2: 7 roots. Is it ok.


An inequality: $1+\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt\frac53$ $n$ is a positive integer, then

$$1+\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt\frac53.$$

please don't refer to the famous  $1+\frac1{2^2}+\frac1{3^2}+\dotsb=\frac{\pi^2}6$.

I want to find a better proof.

My  stupid method:

$$1+\frac1{2^2}+\frac1{3^2}+\dotsb+\frac1{n^2}\lt \left(1+\frac1{2^2}+\dotsb+\frac1{10^2}\right)+\frac1{10\cdot11}+\dotsb+\frac1{n(n-1)}\\<1.549768...+\frac1{10}\lt\frac53$$

I don't know if this is "better," but since $1/x^2$ is strictly decreasing, we have

$$\sum_{n=N+1}^\infty{1\over n^2}\lt\int_N^\infty{1\over x^2}\,dx={1\over N}$$

after which a little systematic trial and error gives

$$\sum_{n=1}^\infty{1\over n^2}\lt 1+{1\over4}+{1\over9}+{1\over16}+{1\over25}+{1\over5}={5989\over3600}\lt{6000\over3600}={5\over3}$$

Added later:  It occurs to me that my approach is not inherently different from the OP's "stupid" method.  We each basically argue that

$$\sum_{n=1}^\infty{1\over n^2}\lt1+{1\over4}+\cdots+{1\over N^2}+{1\over N}$$

for any $N$.  If there's anything stupid in the OP's solution (which there isn't, really), it's just that it's not necessary to go all the way out to $N=10$.


Why do you need absolute value when taking $\sqrt{\cos^2(x)}$ $$\sqrt{\cos^2(x)} = |\cos(x)|$$

Is this on the right track?  If you have an underlying $\cos(x)$ that is negative, and then you square it, you will now have $\cos^2{x}$, which is positive.  But, if you now square root that positive squared term, you will just have a positive value of the $\cos(x)$, which is not what you started with?

As $(-t)^2 = t^2$, the square root doesn't know which is the sign of the thing-inside-the-square.


Nonlinear equation in two unknowns with unique solution over the field of complex numbers We know that a linear equation with two unknowns can not have a unique solution. On the other hand a nonlinear equation in two unknowns such as $$a^2+b^2=0$$ has a unique solution $a=0$ and $b=0$ over the real numbers.

But does there exist a nonlinear equation in two unknowns which has a unique solution in the field of complex numbers?

Restrict the space of functions to the elementary functions: http://en.wikipedia.org/wiki/Elementary_function.

It can be any complex number. For example a special case of imaginary number as sum of two squares:

$ a = x + i y,  b = u + i v ,$

$  a^2 + b^2 =  ( x^2 - y^2 + u^2 - v^2) + 2 i (x y + u v) $

if $ ( a,b) = ( (1,1), (-1,-1)),$  then $ a^2 + b^2 = 4 i. $


Why a inconsistent system can be considered as a span? (updated, example added) I have some problems in learning about span.
The definition of $\operatorname{span}\{\vec{v}_1,\vec{v}_2,\vec{v}_3\}=$ all linear combinations of $\vec{v}_1,\vec{v}_2,\vec{v}_3$.
However, when I deal with question like asking a given vector where it is inside the span of given vector $\{\vec{v}_1,\vec{v}_2,\vec{v}_3\}$. I try to solve it as a system $\begin{bmatrix}\vec{v}_1&\vec{v}_2&\vec{v}_3\end{bmatrix}\cdot\begin{bmatrix}t_1&t_2&t_3\end{bmatrix}^T=v$.
I found the system is inconsistent, but $v$ is still an element of $\operatorname{span}\{\vec{v}_1,\vec{v}_2,\vec{v_3}\}$, why?
an example:

If $v$ is in the span of the set you mentioned, then it is expressible as a linear combination of vectors in the original set. This linear combination solves the equation you say is "inconsistent".

If what you mean by "inconsistent" is that a matrix made of the $v_i$ has zero determinant, this simply expresses the fact that they are linearly dependent, and that they will span a space of fewer than $3$ dimensions. But the span still exists. If $v_1\neq 0$ then $v_1$ is definitely in the span, for example.

the only v is in the span {v1, v2, v3} if the system you mentioned is consistent. Not sure if your last sentence is true.


Relation Between Girth and Diameter of $G$ I have difficulties in understanding the proof for the following theorem.

Theorem. Every graph $G$ containing a cycle satisfies $\def\diam{\operatorname{diam}}g(G) \leq 2\diam(G)+1$.

  Q:The first question why is actually $2\diam(G)+1$, intuitively it should be at most $2\diam(G)$, just the longest distance between two vertices back and forward.

Proof. Let $C$ be a shortest cycle in $G$. If $g(G) \geq 2\diam G+2$, then $C$ has two vertices whose distance in $C$ is at least $\diam G+1$. In $G$, these vertices have a lesser distance;

  Q: Why the distance in $C$ should be different from distance $G$, by definition $C$ is a shortest circle in $G$, so $C$ is based on vertices of $G$.

any shortest path $P$ between them is therefore not a subgraph of $C$. Thus, $P$ contains a $C$-path $xPy$. Together with a shorter of the two $x-y$ paths in $C$, this path $xPy$ forms a shorter cycle than $C$, a contradiction.

I would appreciate if anyone would give more details regarding this proof.

I apologize in advance for my proof seeming pedantic, but virtually every "proof" I have seen in textbooks, or posted online, is incorrect.  The idea behind the proof is straightforward enough, but one must exercise great care in its execution.  So here goes ...

Set $g=girth(G)$ and $d=diam(G)$.  We need to prove that $g$ is at most 2d+1.  By way of contradiction, assume that $g$ is at least $2d+2$.

Let $C$ be a $g$-cycle in $G$, with consecutive vertices $x_0, x_1, x_2,..., x_{g-1}$. Denote by $P_C$ a shortest path along $C$ that has end-vertices $x_0$ and $x_{d+1}$. Note that when $g>2d+2$ this path is uniquely defined, whereas for $g=2d+2$ there are two such shortest paths.  In either case, we may always choose $P_C$ to be the path having consecutive vertices $x_0, x_1, x_2,...,x_{d+1}$.

Set $t=dist_G(x_0,x_{d+1})$, and let $P$ be a path in $G$ from $x_0$ to $x_{d+1}$ with consecutive vertices $y_0(=x_0), y_1, y_2,...,y_{t-1}, y_t(=x_{d+1})$.  By definition of diameter, $t \leq d$.

Let $j$ be the minimum value for which $y_j \neq x_j$.  (Note that this implies that $x_i = y_i \forall i < j$.)  Such a $j$ must exist, since otherwise we would have $P=P_C$, a contradiction since $P$ and $P_C$ have different lengths.

Now, among all values strictly greater than $j$, let $k$ be the minimal value for which $y_k$ lies on $P_C$.  Such a $k$ must exist because $y_t (=x_{d+1})$ lies on $P_C$.

Thus $y_k= x_m$ for some $m \leq d+1$.

Now suppose $m < j$. Then by minimality of $j$, we have $y_k=x_m=y_m$.  But this is impossible since $k>j>m$, and a path cannot have repeated vertices. We conclude that $m \geq j$.

At this point, we may construct the cycle with consecutive vertices

$$y_j, y_{j+1},..., y_k(=x_m), x_{m-1}, x_{m-2},...,x_{j-1}(=y_{j-1})$$

The length of this cycle is $m-j+1+k-j+1$ which is smaller than $g$.  The easiest way to see this, is to note that $x_{j-1}, x_j,..., x_m$ is a portion of the path $P_C$, hence its length is at most $d+1$.  Also note that $y_{j-1}, y_{j},..., y_k$ is a portion of the path $P$, hence its length at most $d$.  Thus this cycle has length at most $2d+1$, a contradiction to our assumption that $g$ is at least $2d+2$.

$\def\diam{\operatorname{diam}}$For the first question:  If $n$ is odd then $g(C_n)=2\diam(C_n)+1$ and if $n$ is even then $g(C_n)=2\diam(C_n)$.  In any case, $g(C_n)\leq 2\diam(C_n) +1$.  For any subgraph $H$ of a graph $G$, if $d_H(u,v) = d_G(u,v)$ for all $u,v \in H$ then $\diam(H) \leq \diam(G)$.  If $C_n$ is a girth cycle of $G$ then $d_{C_n}(u,v) = d_G(u,v)$ for all $u,v \in C_n$ (this needs proof but is straightforward).    Suppose that $C_n$ is girth cycle of $G$.  Then

$g(G) = g(C_n)\leq 2\diam(C_n) + 1 \leq 2\diam(G)+1.$


What is the total number of possible chess moves at White's second turn? At White's first turn there are $20$ possible moves: each pawn can move forward one or two spaces, or a knight can jump over the pawns to one of two positions each. Some moves are of course likelier than others (for example, in all the novice games I've seen, I've never seen anyone open with the king's pawn). Black also has $20$ possible moves at this stage.

At White's second turn, there is a small number of available moves based on what the first move was, plus a good player should take Black's first move into account. But what I want to know is, if we follow each of the $20$ scenarios created by the first move, count the number of available moves in each, no matter how strategically inadvisable the move may be (e.g., move the king forward) as long as it's valid by the rules of the game, delete duplicates (that is, count each move just once, not again and again in each branch that it's still possible to make it), how many moves are there?

Also, what is the most efficient way to tally up these moves? How far can we take this method, say, Black's second move, White's third, or further still?

Based on your comment, you are really counting the number of single moves possible in white's first two.  Each pawn has three possiblities, one space, two spaces, and (if it was advanced two on the first move) from rank 4 to rank 5, for a total 24.  Each knight has two for a first move and eight for a second move (including back where it started-is 1 Na3 2Nb1 different from 1Nc3 2Nb1?  Does that count as four moves?), total 20.  Each rook has two for a second move.  Each bishop has seven.  The queen has nine and the king three.  Overall total $24+20+2\cdot 2 + 2 \cdot 7 + 9+3=74$ possible moves in white's first two.  It depends also whether you think 2Bh6 is different from 2BxNh6 and 2Bxh6.  I have counted it only once.

I just worked out how many possible ways a chess game can progress through White's second move.  I got $8,902$, but I was outdoors and didn't use any paper so I may have made an error.

If we're just looking at White's second move and ignoring Black, I guess it would be... figuring backwards... it was $8,902$ and not $8,900$ because of the small number of moves Black can make that interfere with White, giving or subtracting a move.  So, $8,900$ divided by $20$ is $445$.  That's how many positions White can have after move $2$, but it's not clear whether that's exactly what you were asking.  The poster above may have answered what you meant to ask.


What type of function can I apply to both sides of an inequality in order to obtain an equivalent inequality? I'd like to know what are the types of functions that I can apply to both sides of an inequality in order to get an equivalent inequality. Are those functions the ones that are strictly increasing or strictly decreasing? I'm not sure about it.
Also, I'd like to get a recommendation about good sources where I can read about it.
Thanks.

If by "equivalent inequality" you mean $a<b \iff f(a)<f(b)$ then $f$ must be strictly increasing - for example:

$a<b \iff a+k < b+k\\
a<b \iff ka < kb, \space k>0\\
a<b \iff a^3 < b^3\\
a<b \iff \root{3}\of{a} < \root{3}\of b\\
a<b \iff \tan(a) < \tan(b), \space -\frac \pi 2 <a,b<\frac \pi 2$

If $f$ is strictly decreasing then the inequality is reversed i.e. the equivalence becomes $a<b \iff f(a) > f(b)$. For example:

$a<b \iff ka > kb, \space k<0\\
a<b \iff \frac 1 a > \frac 1 b, \space a,b>0$

It is always true that $a=b$ implies $f(a)=f(b)$, no matter what function $f$ is.
But in order to justify that $f(a)=f(b)$ implies $a=b$ (or, taken together, that $a=b$ is equivalent to $f(a)=f(b)$), you better consider only injective functions $f$.

Indeed, $f(a)=f(b)\implies a=b$ for all applicable $a,b$ is precisely the definition of $f$ being injective. Of course, strictly increasing (or decreasing) functions are in particular injective.


Find the $\lim\limits_{x\rightarrow - \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}$ The task is to find $$\lim_{x\rightarrow - \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}$$

What I've tried is dividing both the numerator and the denominator by $x$, but I just can't calculate it completely.

I know it should be something easy I just can't see.

Thanks in advance.

From
$$\lim_{x\rightarrow - \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}$$
one can factor an $x$ from each term as follows:
\begin{align}
\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x} &= \frac{x \left(1 + \sqrt{1 + \frac{a^{2}}{x^2}} \right)}{x \left( 1 + \sqrt{1 + \frac{b^2}{x^2}} \right)} =
 \frac{1 + \sqrt{1 + \frac{a^{2}}{x^2}} }{ 1 + \sqrt{1 + \frac{b^2}{x^2}} }.
\end{align}
From here the limit can be taken or one can expand the one more time. Using
\begin{align}
\sqrt{1 + t} = 1 + \frac{t}{2} - \frac{t^2}{8} + \mathcal{O}(t^3)
\end{align}
then
\begin{align}
\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x} &= \frac{2 + \frac{a^2}{2 x^2} - \frac{a^4}{8 x^4} + \mathcal{O}\left(\frac{1}{x^6}\right)}{2 + \frac{b^2}{2 x^2} - \frac{b^4}{8 x^4} + \mathcal{O}\left(\frac{1}{x^6}\right)} = 1 + \frac{a^2 - b^2}{2 x^2} + \frac{2 b^4 - a^2 b^2 - a^2}{8 x^4} + \mathcal{O}\left(\frac{1}{x^6}\right).
\end{align}
Upon taking the limit the result becomes
\begin{align}
\lim_{x \to \pm \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x} = \lim_{x \to \pm \infty} 1 + \frac{a^2 - b^2}{2 x^2} + \frac{2 b^4 - a^2 b^2 - a^2}{8 x^4} + \mathcal{O}\left(\frac{1}{x^6}\right) = 1.
\end{align}


ODE $y(1+2xy)dx+x(1-xy)dy=0$ $$y(1+2xy)dx+x(1-xy)dy=0$$

I have tried to isolate $\frac{dy}{dx}$ and got the following:

$$\frac{dy}{dx}=-\frac{y(1+2xy)}{x(1-xy)}$$

but I understand that the terms have to be in the same order and that is not the case.

What should I do?

Set $u=xy$. Hence $\frac{du}{dx}=x\frac{dy}{dx}+y.$ Hence your equation became $$\frac{x}{u}\frac{du}{dx}-1 = \frac{1+2u}{u-1}.$$ This can be rewritten $$x\frac{du}{dx} = u\frac{3u}{u-1}$$ or else $$\frac{u-1}{3u^2}du=\frac{1}{x}dx.$$ Now you can integrate both sides.

Given equation can be written as,
ydx +xdy+2xy²dx- x²ydy=0

Dividing both side by (xy)², we get

d(xy)/(xy)² + 2dx/x - dy/y=0

Now it can be integrated easily...


Triangle - What is the length of the hypotenuse? Triangle ABC has a right angle at corner C. It has a height from C to a point D on side |AB|. If |CD|=5 and |AD|=7 then what is the length of the hypotenuse? (|AB|=?)

Correct Answer: 74/7

I have tried solving the question above by the help of the law of cosines and sines, pythagoras theorm and uniform triangles but without much success. I always tend to get to many unknown variables. Thanks in advance.

Draw a picture.  You just need similar triangles.  $ABC,ACD,$ and $CBD$ are similar.  $\frac {BD}{DC}=\frac {CD}{AD}=\frac 57$ so $BD=\frac {25}7$ and $AB=AD+BD=7+\frac {25}7=\frac {74}7$


If $E[X\mid Y]=0$ a.s., does it follow that $E[X\mid f(Y)]=0$ a.s.? If $E[X\mid Y]=0$ a.s., does it follow that $E[X\mid f(Y)]=0$ a.s., for any measurable function $f$?

Let $\mathcal{F}$ be the $\sigma$-algebra generated by $Y$ and $\mathcal{G}$ the $\sigma$-algebra generated by $f(Y)$.  By definition, if $A\in\mathcal{G}$, then there is a Borel set such that $A=(f\circ Y)^{-1}(B)$. But $(f\circ Y)^{-1}=Y^{-1}\circ f^{-1}$ and so
$$
A=(f\circ Y)^{-1}(B)=Y^{-1}(f^{-1}(B)).
$$
Because of $f$'s measurability, $f^{-1}(B)$ is also Borel-measurable. We infer that $A\in\mathcal{F}$ and, more importantly, $\mathcal{G}\subset\mathcal{F}$. Now the claim follows:
$$
E[X|f(Y)]=E[X|\mathcal{G}]=E[E(X|\mathcal{F})|\mathcal{G}]=E[E(X|Y)|\mathcal{G}]=E[0|\mathcal{G}]=0.
$$
The second equality above makes use of iterated conditioning and $\mathcal{G}\subset\mathcal{F}$.

If you take $f(x)=x$ then $E[X|f(Y)]=E[X|Y]=0$.

Is this really the problem you wanted to ask?


A zero-dimension set and self-referencial equation Let $K$ be a compact set in $\mathbb{R}^2$. Let $f_1,..., f_n$ be contracting similarities of $\mathbb{R}^2$ to itself. Suposse $K$ satisfies the self-referencial equation
  $$K=\bigcup_{i=1}^{n}f_i[K]$$
  and $f_i[K]\cap f_j[K]=\emptyset$. Prove that $K$ is zero-dimensional.

Since $f_i[K]\cap f_j[K]=\emptyset$ I know that the distance between both set is positive, because those sets are compact. Then each $f_i[K]$ is a clopen set of $K$. I don't know how to use that $f_1,..., f_n$ are contraction. Any hint?

Thanks!

I will used all the time the notation given in the problem. Then the function $f_i$ will always mean that it supposed to mean in the hypothesis.

To prove the theorem we will find a clopen basis. For simplicity define
$$\bigcirc_{k=1}^sf_k =f_s\circ f_{s-1}\circ\cdots\circ f_{1}$$
That is a successive composed functions. Now take a sequence $\{i_k\}_{n=1}^\infty$  such that all its member are integers between $1$ and $n$ ($1\leq i_k\leq n$ for every $k$).

The letters $r_1,..., r_n$ will denote the constant of contraction (that is the real number such that $|f_i(x)-f_i(y)|=r_i|x-y|$ for every $x,y$ in the domain).

Since $K$ is compact then $A=\text{diam } K$ is finite.

First claim: Each $f_{i}[K]$ is a clopen set.

proof: Each $f_{i}[K]$ is compact, since $K$ is compact. By the condition $f_{i}[K]\cap f_j[K]=\emptyset$ the distance between each $f_i[K]$ is positive furthermore this implies too that $f_i[K]$ is closed being the image of a compact set in a eucliden space. Its complement is
$$\bigcup_{j\not= i}f_j[K] $$
and this set is closed, being the finite union of closed sets, then $f_i[K]$ is open therefore clopen.

Second claim: For every sequence $f_{i_k}$ where $\{i_k\}_{n=1}^\infty$ is a sequence defined as above the following is true:
$$\lim_{s\rightarrow \infty}\text{diam }\left[\bigcirc_{k=1}^sf_k(K)\right]=0 $$
proof: Defined $ g_s=\bigcirc_{k=1}^sf_{i_k}(K)$. For each $x,y\in f_{i_k}(K)$, the equality $|f_{i_k}(x)-f_{i_k}(y)|=r_{i_k}|x-y|\leq r_{i_k}A $ is the basic step on the proof by induction of
$$|g_{s+1}(x)-g_{s+1}(x)|=r_{i_s}r_{i_{s-1}}\cdots r_{i_1}|x-y|\leq r_{i_s}r_{i_{s-1}}\cdots r_{i_1} A $$. Which implies
$$\text{diam }g_s[K]\leq r_{i_s}r_{i_{s-1}}\cdots r_{i_1} A $$
and the sequence $r_{i_s}r_{i_{s-1}}\cdots r_{i_1} A$ is bounded by $AR^s$ where $R=\max\{r_1,...,r_n\}<1$ then $AR^n\rightarrow 0$ when $n\rightarrow\infty$. This proves the second claim.

Third claim: Taking indices $i_1,..., i_n$ varying between $1$ to $n$. We will have that
$$K=\bigcup_{i_n,...,i_1=1}^n \bigcirc_{k=1}^nf_{i_k}[K] $$
and that $f_{i_n}\circ\cdots\circ f_{i_1}[K]\cap f_{i_n'}\circ\cdots\circ f_{i_1}[K]=\emptyset$ if $i_n\not= i_n'$. Which implies that this set are clopen.
As ilustration i will show the case $n=2$. We have that
$$K=\bigcup_{i=1}^nf_i[K]$$
Then $f_jK=\bigcup_{i=1}^nf_jf_i[K]$, consequently
$$K=\bigcup_{j=1}^n\bigcup_{i=1}^nf_jf_i[K]$$
And, since $f_i[K]\subset K$ then $f_jf_i[K]\subset f_j[K]$  then $f_jf_i[K]\cap f_{j'}f_i[K]=\emptyset$ if $j\not=j'$. And it is clear that the set $f_jf_i[K]$ are clopen (the reason is analagous to the proof in the first claim).

This three claims proof the theorem, since for each neighborhood of a pint in $K$ I can take a sufficiently small  (by the firts claim) set of the form $f_{i_k}\circ\cdots\circ f_{i_1}[K]$ wich is clopen by the second claim.

Then the family of sets $f_{i_k}\circ\cdots\circ f_{i_1}[K]$ form a clopen basis, then $K$ is zero dimensional.

Hint: Prove that $K$ is a Cantor set (of Hausdorff dimension $< 2$).


Find the local linearization of f(x) near 0 and use this to approximate the value of a. $$ f(x) = (1+x)^\frac35, a= (1.2)^\frac35 $$

I got the linear approximation equation of $$1+ \frac35 x$$
What do I do with the value of a?

Hint: $f(x)=a$ when $x=\cdots\,$.

Find $f(x)$ approximately by using this $x$ value in the linear approximation.


Proof that non-monotonic sequence is unbounded I am looking at a non-monotonic recursive sequence given by
$$a_0 =2,$$
$$a_{n+1}  = \frac{1}{2}\left( a_n - \frac{1}{a_n}\right).$$
Is the sequence bounded or unbounded?
I suspect that the sequence is unbounded.
Easy concepts like finding another unbounded sequence as a lower bound fails. This is because the sequence always approaches 0 until $a_n$ gets very small at which point the sequence blows up again. Any ideas how to go from here?
Note: I saw this problem in some public talk for high school students around 12 years ago, and it got stuck in my head ever since (the speaker even set a reward for solving this). Maybe this is some kind of trick question that relates to some famous unanswered problem so there will be no hope for mere mortals like me. Any hints in that direction would be nice as well.

Hint:
Establish that liminf an = -∞ as n->∞ and
            limsup an = +∞ as n->∞.

knowing these two facts, what can we say about the boundedness of sequence $a_n$

The sequence $a_n$ is unbounded,  -∞ < $a_n$ < ∞.
and liminf $a_n$ = -∞ as n->∞ and limsup $a_n$ = +∞ as n->∞.
Proof
The solution for $a_n$ is   $a_n$ =  cot(arccot(2)$2^n$)
or                          $a_n$          ~  cot(.4636476*$2^n$)
consider the the function a(x) in interval (nπ, (n + 1)π) as x->∞.
Define a(x) as a(x) := cot ((arccot(2)2x)), x ε R,
then a(x)-> -∞ as x->(nπ)+, and a(x)-> ∞, as x-> ((n + 1)π)-.
The process is repeated in every interval (n π, (n+ 1)π) of duration π, n = 0, 1, 2, 3,...
It follows that a_n -> -∞ infinitely often and a_n -> +∞ infinitely often.
Therefore
liminf $a_n$ = -∞ as n->∞, and limsup $a_n$ = +∞ as n->∞ as asserted.
Graph of $a_n$ for n= -1023 to 1063


Conformal transformation of the divergence Let $f$ be a smooth function on a $n$-dimensional Riemannian mainfold $(M, g)$, so that $\tilde{g} = e^{2f} g$ is a conformal transformation of $g$. I'm trying to show that the divergence operator transforms as
$$
\text{div}_{\tilde{g}} X = \text{div}_{g} X + n e^{-2f} X(f)
$$
but I'm a little stuck. Dropping down to coordinates, the divergence looks like
$$
\text{div}_\tilde{g} X = \sum_i \frac{1}{\sqrt{\det \tilde{g}}} \frac{\partial}{\partial x^i} \left( X^i \sqrt{\det \tilde{g}} \right) = \sum_i \frac{e^{-nf}}{\sqrt{\det g}} \frac{\partial}{\partial x^i} \left( X^i e^{nf} \sqrt{\det g} \right)
$$
An application of the product rule to the differential term gets you one term looking like $\text{div}_g X$ (hooray!) and one term looking like $nX(f)$. This isn't what I want - it's off by the factor of $e^{-2f}$. What's going wrong with this very simple calculation? I'm pretty sure the formula at the top is correct.

Also, is there a coordinate-free way to do this? Maybe that would help.

Here's an relatively straightfoward, coordinate-free way to approach the problem:

The divergence of a metric $g$ actually depends only on its volume form $\mu$ (for either orientation, it doesn't matter which). It can be characterized by
$$\mathcal{L}_X \mu = (\text{div } X) \mu.$$

Now, the volume form $\widehat{\mu}$ of a conformally related metric $\widehat{g} = e^{2f} g$ is
$$\widehat{\mu} = e^{n f} \mu;$$
substituting in the above characterization of the divergence---and using the Leibniz identity $$\mathcal L_X (h \alpha) = (X \cdot h) \alpha + h \mathcal L_X \alpha$$ for a function $h$ and form $\alpha$---gives an identity relating the divergences of the two metrics.

Your computation is correct but I think what you want to show is incorrect: divergence does not change if you do a simple rescale $g \to c^2g$ when $c$ is constant, but the term $ne^{-2f}X(f)$ does change if you replace $f$ by $f+C$ (that is one more simple rescale after the conformal change).


Representing the multiplication of two numbers on the real line There is a simple way to graphically represent positive numbers $x$ and $y$ multiplied using only a ruler and a compass: Just draw the rectangle with height $y$ in top of it side $x$ (or vice versa), like this

But is there a way to draw the number $xy$ directly on the real line (i.e. not as an area on top of the real line) by using only some standard drawing means like using a compass, a ruler, a straightedge etc. (i.e. not multiplying $x$ and $y$ out and then putting the number $xy$ at right spot), like indicated above ?

(I think this question actually asks if the multiplication is representable as a composition of (the mathematical translation of) operations of drawing circles using a straightedge etc.)

You can do so, but if you want to represent the result of the multiplication again as a length you have to choose a unit. The ancient Greeks didn't come up with this idea; whence their products were always areas.

Responding to temo's comment: When the point $O\in g$ representing the number $0$ has been chosen the geometric construction for the sum of two numbers is scale invariant, as a consequence of $(\lambda x)+(\lambda y)=\lambda(x+y)$; but a similar identity for multiplication does not hold: $(\lambda x)\cdot(\lambda y)\ne\lambda(x\cdot y)$. You can test the effect in the figure by keeping $x$ and $y$ but choosing a new unit $1'$. The point $xy$ will now be at a different location.

This is an old question that I myself asked; I saw this thread and then thought about the answer, and here is what I came up with.

Suppose I have three segments that have common endpoint A: a segment with length 1 and endpoint B, a segment of length x with endpoint X, and a segment of length y with endpoint Y. All of these points are collinear.

Construct three circles centered at A that have a radius 1, x, and y. Construct C on the unit circle so that AC is perpendicular to AB. The points where line containing A and C intersect the other circles are called Cx and Cy in the obvious way.

Now create a point M on the line containing A and Y so that angle ACxX is congruent to ACyM. Since ACxX and ACyM are similar triangles, you can show that the length of AM is x*y. (This was inspired by the proportion guy posting elsewhere in this thread)


Age Problem (Equation Writing Help) I am new to writing equations and have to write and solve the following:

Kevin is $7$ less than three times Maria's age.  If Kevin is $17$, how old is Maria?

I think Maria is $30$ if Kevin is $17$ because Kevin is $7$ less than $3$ times Maria's age but I don't know how to put it in the correct equation format.  Do I need more than one variable?  This is where my confusion is.

Thanks for any help or tips!

Edited to add - I think Maria is $30$ because if Kevin is $17$ and he is $7$ less than three times Maria's age $(17-7=10)$ and $10\times 3 =30$.

You only need one variable.

Let Maria's age be $x$. So $3x-7=17$ since Kevin is $17$. Solving, we get $x=8$. So Maria is $8$ years old.

In other words, $3x-7=17$ is the equation you want.

(If you try to have Maria being $30$, then Kevin's age would be $3\times30-7=83$!! Which I don't think is $17$. Wait, am I wrong?)

Kevin is $17$, so add $7$ to $17$, then divide by $3$.


Question about finding the value of an infinite sum What is the value of:
$$\sum_{k=0}^\infty \frac{1}{(4k+1)^2}?$$
I realised that
$$\sum_{n=2,4,6,8,...} \frac{1}{n^2} + \sum_{n=1,3,5,7,...} \frac{1}{n^2} = \sum_{n \geq 1 } \frac{1}{n^2}$$
$$\sum_{n \geq 1 } \frac{1}{4n^2}+\sum_{n \geq 0 } \frac{1}{(2n+1)^2} = \sum_{n \geq 1 } \frac{1}{n^2}$$
$$\sum_{n \geq 0 } \frac{1}{(2n+1)^2} = \frac{3}{4}\frac{\pi^2}{6} \qquad \Rightarrow \qquad \sum_{n=1,3,5,7,...} \frac{1}{n^2} = \frac{1}{8} \cdot \pi^2$$
So
$$\sum_{k=0}^\infty \frac{1}{(4k+1)^2}  + \sum_{k=0}^\infty \frac{1}{(4k+3)^2} = \frac{\pi^2}{8}$$
But I cannot find the value of the second summation.
Any suggestions?

Recall the definition of Catalan's Constant:
$$K=\sum_{k=0}^{\infty} \frac 1 {(4k+1)^2}- \sum_{k=0}^{\infty} \frac 1 {(4k+3)^2}$$
Combining that with the equation you have already derived we immediately get $$\sum_{k=0}^{\infty} \frac 1 {(4k+1)^2}=\frac K2+\frac {\pi^2}{16}$$

The sum can be expressed in terms of Catalan's constant. In the following video, the sum is computed in a step by step manner: https://www.youtube.com/watch?v=r2OJtsHNDZA.


Why is $n^2+4$ never divisible by $3$? Can somebody please explain why $n^2+4$ is never divisible by $3$?  I know there is an example with $n^2+1$, however a $4$ can be broken down to $3+1$, and factor out a three, which would be divisible by $4$.

Write $n=3r+q$ with $q\in\{0,1,2\}$. Then
$$
n^2+4=9r^2+6rq+q^2+4=(9r^2+6rq+3)+q^2+1.
$$
The expression in the parentheses is divisible by $3$ whereas inspecting the three possibilities for $q$ reveals that $q^2+1$ is never divisible by $3$.

Hint: What does $-4$ equal to in $\mathbb{Z_3}$?


Diffeomorphism: Unit Ball vs. Euclidean Space In my differential geometry class we are being asked to prove that the open unit ball
$B^n$ = { $x$ $\in$ $\mathbb{R}$$^n$ such that |$x$| < $1$}
is diffeomorphic to $\mathbb{R}$$^n$

I am having a hard time with this as I am brand new not only to differential geometry, but also topology.

I know that I need to construct a smooth, differentiable bijection between the two with a differentiable inverse, but beyond that, I am unsure of where to start. Some guidance in the right direction would be greatly appreciated.

Let $\phi \colon [0,1) \to [0, \infty)$ a diffeomorphism with inverse $\psi$. Some possible choices: $t \mapsto \frac{t}{1-t}$, $t \mapsto \tan (\frac{\pi}{2}\cdot t)$.

The map
$$x \mapsto \phi(||x||) \cdot \frac{x}{||x||}$$

is a diffeomorphism from $B^n$ to $ \mathbb{R}^n$ with inverse
$$y \mapsto \psi(||y||) \cdot \frac{y}{||y||}$$

$\bf{Added:}$  It turns out that the choice of the diffeomorphism from $[0,1)$ to $[0,\infty)$ matters a lot, since $x \mapsto ||x||$ is not smooth at $0$. This was brought to my attention by @Freeze_S and I thank him a lot! One can check that the map obtained for $\phi(t) = \frac{t}{1-t}$ is only $C^1$ at $0$... However, we can use the map so kindly suggested by @Jesus RS: ( big thanks! ) $\phi(t) = \frac{t}{\sqrt{1-t^2}}$ with inverse
$\psi(s) = \frac{s}{\sqrt{1+s^2}}$ and it will work just fine. The diffeomorphisms are, as written by @Jesus RS:
$$x \mapsto \frac{x}{\sqrt{1-||x||^2}} \\
y \mapsto \frac{y}{\sqrt{1+||y||^2}}$$

In fact, as long as $\phi(t)$ is an odd function of $t$ things will work OK. So, another example is

$$x \mapsto \frac{\tan (\frac{\pi}{2} \cdot ||x|| )}{||x||} \cdot x $$

As to the book recommendations I suggest the following:

(i) A course in Differential Geometry (W. Klingenberg)

(ii) Elementary Differential Geometry (A. Pressley)

(iii) Differential-Geometrie und Minimal-Flächen (J. Jost) (in german but it's a wonderful book).

Some more advanced books dealing with differential geometry in the context of differentiable manifolds I'd suggest:

(v) Differential Geometry: Manifolds, Curves and Surfaces (M. Berger, B. Gostiaux)

(vii) A Comprehensive Introduction to Differential Geometry (M. Spivak) (all the volumes)

I hope it helps.

As to your question about the diffeomorphism between the open unit ball and $\mathbb R^n$ the idea is defining a function which will stretch the open ball in all directions.


On uniformly continuous, bounded, hermitian functions whose product is continuously differentiable Let $f,g: \mathbb R \to \mathbb C$ be uniformly continuous, such that $f(0)=g(0)=1$, $|f(x)|\le 1, |g(x)|\le 1,\forall x\in \mathbb R$, $f(-t)=\overline {f(t)}, g(-t)=\overline {g(t)}, \forall t \in \mathbb R$ and $f.g$ is continuously differentiable on $\mathbb R$.

Then is it true that $f,g$ are differentiable at $0$ ? If this is not true in general, what if we also assume that $f,g$ are real-valued ?

Define $r(x)=\frac1{|x|+1}$ and take $f(x)=r(x)+i\sigma(x)\sqrt{1-(r(x))^2}$ and $g(x)=\overline{f(x)}$. Here $\sigma(x)$ is the sign of $x$. Notice that $r$ is not differentiable at $0$, so neither are $f$ and $g$. However $|f(x)|=1$, so $fg=1$, so $fg$ is continuously differentiable. All the other properties are easier to check.

Now, to prove that the conjecture does hold for $f,g:\mathbb{R}\to\mathbb{R}$. If $f$ is not differentiable at $0$, then there exists $\varepsilon>0$ such that for all $\delta>0$ there exists $x\in(-\delta,\delta)$ with $1-f(x)>\varepsilon|x|$. Since $|g(x)|\leq1$, we then also find $1-(fg)(x)>\varepsilon|x|$ (for $x$ small enough such that $f(x)>0$). Hence $fg$ is not differentiable at $0$.

Take $f(x)=(1+|x|)^{-0.5}$ then $f$ is even and hence $f(-x)=\overline{f(x)}$ since it is a real valued. It is easy to see that $|f(x)|\leq 1$ but $f$ is not differentiable at $0.$ Since its derivative is bounded for $|x|\geq 1$ the function is uniformly continuous.